advertisement

Module in Inferential Statistics by Fernando P. Tataro Introduction Module has been around for decades as an aid to the delivery of teaching. With the proliferation of the Covid 19, module has become a necessity. In fact, many authors have already written this type of manuscript. As the Covid 19 spreads out worldwide, it causes many institutions/sectors (private or state) to be locked down. Its spread prompted many countries in almost all parts of the world to issue ECQ, GCQ, or Modified GCQ or whatever they may call it. The top 5 countries hardly hit by this Pandemic include the United States (USA), Brazil, Russia, India and the United Kingdom (UK). The number of infected (total cases) in the USA has already reached more than 2.6 million and is still growing. In the Philippines, though not that serious compared to the above-mentioned countries, for fear that the Medical Sector may not be able to handle the great number of infected persons and to prevent the rapid increase of infection, the government, as advised by the Department of Health (DOH), has issued a lockdown as well. One of the sectors greatly impacted by this Pandemic Covid 19 is the Educational System. The issuance of lockdown due to the virus, hindered educators and students to meet face-to-face because they have to stay at home. But all concerned individuals must not succumb to the adversities brought about by it. Thanks to the new technology at hand. The advent of this technology has somehow enabled us to cope up with the current challenges, especially that caused by Covid 19 pandemic. The influx of computers, TVs, cellphones, etc., has grown in gigantic numbers, and so with the application programs / softwares (You-Tubes, Zoom, Google Classroom, Google Forms, etc.), running in various platforms. The role of Modules however will remain essential to the Educational System. General Objectives: After finishing Modules 1 to 3, you are expected to 1. 2. 3. 4. 5. 6. 7. Distinguish between parametric and nonparametric tests Categorize the type of statistics you will be using Calculate and compare results to tabular values to make a conclusion Arrive at the conclusion using p-value instead of the computed value Organize results in tables or graphs Interpret tables or graphs and describe its characteristics Achieve mastery in following the Stepwise Method in Solving Statistical Problems 8. Apply to gained knowledge to real life situations Module 1 SKEWNESS AND KURTOSIS Skewness and kurtosis are measures that give information whether the distribution is normal or abnormal. When skewness is positive, the distribution is said to be positively skewed or skewed right, meaning, the right tail of the distribution is longer than the left. When the skewness is negative, the distribution is negatively skewed or skewed left, which means that the left tail of the distribution is longer than the right. And If skewness is zero, the distribution is perfectly symmetrical. The curve representing the distribution is bell shape. When the skewness is zero (perfectly symmetrical) and kurtosis is 0.265 (mesokurtic), the distribution is said to be normal. These two measures are very important in the succeeding lessons on Inferential Statistics. These enable us to determine if the distribution is normal or non-normal. Objectives: When you have completed the lessons, you will be able to: • • • • Calculate quantities such as the mean, median and standard deviation for use in the skewness. Use skewness to determine whether the distribution of data is positively skewed or negatively skewed (abnormal) or neither (normal). Calculate the quartiles, quartile deviation, percentiles and kurtosis to distinguish whether the distribution of data is mesokurtic, leptokurtic and platykurtic then tell the abnormality or normality of the distribution based on these criteria. Identify the different types of graphs of skewness and kurtosis. LESSON 1 - The SKEWNESS The formula for skewness is ππ = 3(πΜ − Md) SD Where: πΜ = Mean Md=Median SD= Standard Deviation Example 1. Consider whether this frequency distribution is normal or abnormal. Scores f 45-50 3 40-44 6 35-39 9 30-34 11 25-29 12 20-24 5 15-19 4 n=50 Expanded Class Frequency Table Scores π 45-50 40-44 35-39 30-34 25-29 20-24 15-19 3 6 9 11 12 5 4 50 Class Mark x 47.5 42.5 37.5 32.5 27.5 22.5 17.5 a) Calculation of the Mean: 2 ππ₯ ππ₯ 142.5 255 337.5 357.5 330 112.5 70 1605 6768.75 10837.5 12656.25 11618.75 9075 2531.25 1225 54712.5 Class Boundary Lb Ub 44.5 50.5 39.5 44.5 34.5 39.5 29.5 34.5 24.5 29.5 19.5 24.5 14.5 19.5 Cum. Frequency CF 50 47 41 32 21 9 4 ∑ππ₯ πΜ = π = 1605 50 πΜ =32.1 b) Calculation of the Median, Md: Using the formula: π −πΆπΉ Md=Lme+2πΉππ π π To determine values of Lme, CF, and Fme, first calculate 2. π 2 50 = 2 = 25 Then, identify the value of Lme under the Cumulative Frequency (CF) Column. Starting from the bottom, search for the first value that is equal to or greater the π calculated 2 = 25. This value is 32. The value 32 enables us to locate the Median Class. Thus, the Median Class is 29.5 – 34.5. So, Lme=29.5, CF=21, and Fme=11. The class interval i is 5, the difference between two subsequent values, say 45 and 40. So 45-40 =5. i =5 Substituting values in the formula for Median above: Thus π −πΆπΉ ππ = πΏππ + 2πΉππ π ππ = 29.5 + 25−21 11 (5) ππ = 31.318 c) Calculation of the Standard Deviation: From the formula: 2 2 (∑ ππ₯) ∑ ππ₯ − π ππ· = √ π−1 16052 50 54712.5− ππ· = √ 50−1 ππ· = 8.071 d) Calculation of the SK ππΎ = ππΎ = 3(πΜ −ππ) ππ· 3(32.1−31.318) 8.071 ππΎ = 0.291 The SK value of 0.291 indicates that the distribution is abnormal. Being positive, the distribution is said to be positively skewed or skewed right. DIFFERENT GRAPHS OF SKEWNESS Skewness is the shape of the distribution. The distribution is negatively skewed when the thinner tail is deviated to the left side. It is positively skewed when the thinner tail is deviated to the right, but it is normal when it is a bell shape. (Broto, 2006) SK=+ SK=0 SK=- Positively Skewed Normal Negatively Skewed Normal LESSON 2 - THE KURTOSIS Kurtosis is a statistical measure used to describe the degree to which scores cluster in the tails or the peak of a frequency distribution. The peak is the tallest part of the distribution, and the tails are the ends of the distribution. There are three types of kurtosis: mesokurtic, leptokurtic, and platykurtic. The distribution is said to be normal when the value of kurtosis is equal to 0.265 mesokurtic. When its value is less than 0.265 leptokurtic, and when its value is greater than 0.265 platykurtic. Both leptokurtic and platykurtic types of kurtosis indicate that the distribution is abnormal. The formula for kurtosis: ππ πΎπ’ = (π 90 −π10 ) Where: Ku=Kurtosis Q=Quartile Deviation π90 = πππππππ‘πππ 90 π10 = πππππππ‘πππ 10 Example: Consider the frequency distribution of example 1 for the problem in Skewness. Determine whether it is normal or abnormal by solving the value of Kurtosis. a) Solve for Qd. π3 − π1 2 But Qd has the formula above. And Q1 and Q3 must be calculated first. ππ = For Q1: π1 = πΏπ1 + 1) Determine π 4 = 50 4 π −πΆπΉ 4 πΉπ1 π = 12.5. 2) Identify the first quartile class: Look for the first value under Cumulative Frequency Column from the above table greater than or equal to 12.5, this value is 21. The value 21 is (21>=12.5). This value is located in the third row from the bottom of the table. The 1st Quartile Class (24.5 – 29.5) belongs to this row. LQ1 therefore is 24.5; CF=9, the cumulative frequency before; the FQ1 =12; and i=5. π1 = 24.5 + 12.5 − 9 (5) 12 Q1 = 25.958 For Q3: π3 = πΏπ3 + 1) Determine 3π 4 = 3(50) 4 3π −πΆπΉ 4 πΉπ3 π = 37.5. 2) Identify the third quartile class: Look for the first value under Cumulative Frequency Column from the above table greater than or equal to 37.5, this value is 41. The value 41 is (41 >= 37.5). This value is located in the fifth row from the bottom of the table. The Quartile Class (34.5 – 39.5) belongs to this row. LQ3 therefore is 34.5; CF=32, the cumulative frequency before; the FQ3 =9; and i=5. π3 = 34.5 + 37.5 − 32 (5) 9 Q3 =37.556 ππ = 37.556−25.958 2 ππ = 5.799 For P10: π10 = πΏπ10 + 1) Determine 10π 100 = 10(50) 100 10π −πΆπΉ 100 πΉπ10 π = 5. 2) Identify the P10 class: Look for the first value under Cumulative Frequency Column from the above table greater the or equal to 5, this value is 9. The value 9 is (9 >=5). This value is in the 2nd row from the bottom of the table. The P10 Class (19.524.5) belongs to this row. LP10 is 19.5, CF=4, and FP10 = 5. Substituting values, π10 = 19.5 + π10 = 20.5 5−4 5 (5) 90π π90 = πΏπ90 + 100 πΉ P90: −πΆπΉ π90 90π 1) Determine 100 = 90(50) 100 π = 45. 2) Identify the P90 class: Look for the first value under Cumulative Frequency Column from the above table greater the or equal to 45, this value is 47. The value 47 is (47 >=45). This value is in the 2nd row from the top of the table. The P90 Class (39.5-44.5) belongs to this row. LP10 is 39.5, CF=41, and FP90 = 6. Substituting values, π90 = 39.5 + 45−41 6 (5) π90 = 42.833 Since the item values in the formula for Ku were all determined, thus πΎπ’ = (π ππ 90 −π10 ) 5.799 πΎπ’ = (42.833−20.5) πΎπ’ = 0.260 The value of Ku equal to 0.260 is less than 0.265. This means that, the distribution is Leptokurtic or the data cluster to the peak. The distribution is abnormal. Note: The arrangement of the data in the Frequency Distribution Table is in descending order. The cumulative frequencies obtained started from the frequency below the table towards the top by adding them together one after the other. Had the data were arranged in ascending order, cumulative frequencies shall be obtained in reverse order (i.e., by starting from the top). Exercise: Consider the following Frequency Distribution. Determine if the distribution is normal or abnormal by solving for skewness and kurtosis. Scores f 16-20 2 21-25 5 26-30 11 31-35 13 36-40 10 41-45 6 46-50 3 n=50 LESSON 3 - INTRODUCTION TO INFERENTIAL STATISTICS Inferential statistics deals with the analysis and interpretation of data based on a sample or representative of the population. This statistic consists of differential statistical tools/tests used in the analysis of interval, ratio, nominal and ordinal data. These tests are used in making inferences about populations using data drawn from the populations. The extent to which the use of this statistics can be done with accuracy depends on the goodness of samples. The sampling techniques / procedures are also of great importance with regard to the use of these different statistical tests. Objectives: When you have completed the Lessons, you will be able to: • • • • • • • • • Become familiar with the kinds of statistical tests. Illustrate the difference between a Null Hypothesis and an Alternative or Research Hypothesis. Differentiate a Type I error from a Type II Error. Explain the meaning and the use of the Level of Significance. Identify the conditions that will require the use of either a One-tailed Test or a Two-tailed Test. Explain the concept of the critical or rejection region in a Normal Curve. Identify which formula to apply when the sample size is Large and when the size is Small; alternatively, when to use a Z test or a t test. State the conclusion to a Test of Hypothesis of rejecting or accepting the Null Hypothesis. Become familiar with solving the problems by Stepwise Method. KINDS OF STATISTICAL TESTS Statistical tests can be grouped into two. The parametric and the nonparametric tests. The parametric tests. To use the parametric tests, there are some conditions that should be met. The data must be normally distributed and the level of measurement must be either interval or ratio. The data are said to be normal when the value of skewness equals zero and the value of kurtosis is 0.265. The interval data provide numbers that reflect difference among items. With interval scales, the measurement units are equal. Examples are scores of intelligence tests, and time reckoned from the calendar. They have no true zero value. The ratio scale is the highest type of scale. The basic difference between the interval and the ratio scales is that, the interval scale has no true zero value while the ratio scale has an absolute zero value. Common ratio scales are measures of length, width, weight, capacity and loudness and others. The nonparametric tests. The nonparametric tests do not require normality of the distribution. Skewness and Kurtosis are the measures that tell whether the data is normal or abnormal. If the value of skewness is either positive or negative, distribution is said to be abnormal. When the Kurtosis is greater than or less than 0.265, which means, it is not equal to that value, the distribution is abnormal as well. Under these tests, the levels of measurement are the nominal and ordinal data. Nominal data are data such as: male or female, yes or no responses, political affiliations like, LP, LDP, Lakas, and religious groupings, Christians and Non-Christian and other organizations. Ordinal data are data such as: Strongly Agree, Agree, No Opinion, Disagree and Strongly Disagree and also other data which employ rankings. LESSON 4 - TESTS OF HYPOTHESES (Garcia, 2003) The Null Hypothesis A hypothesis is a statement about our expectation with regard to some characteristic of a population. It is something that we assume about a population. With the passing of time, however, population characteristics as measured by the mean, can change. This is where we test a hypothesis about the population. The hypothesis is a statement that, at the end of the solution to the problem, is either rejected or accepted. This statement is called the Null Hypothesis and is denoted by Ho. On the other hand, an alternative claim is the Alternative Hypothesis and uses the symbol Ha. The options with regard to these two hypotheses are: • • If Ho is rejected, then Ha is accepted If Ho is accepted, then Ha is rejected Put another way, when we reject Ho, we affirm the statement that is the alternative to Ho (that is, we accept Ha). On the other hand, if we accept Ho, we affirm the statement expressed by the Null Hypothesis (therefore, we reject Ha). The Null Hypothesis is a statement of our expectation with regard to the population and is expressed in the negative form. The following examples will illustrate this point. • Expectation: The IQ level of first year college students in a particular university is still the same. Ho: The average IQ level of first year college students has not changed. Ha: The average IQ level of first year college students has changed. • Expectation: Despite a number of economic problems, the average annual income of Filipino families is still the same. Ho: The average annual income of Filipino families has not changed. Ha. The average annual income of Filipino families is lower than that of the previous years. • Expectation: The weight (in grams) of a can of biscuits is consistent with the manufacturer’s claim in the label. Ho: The average weight of a can of biscuits has not changed. Ha: The average weight of a can of biscuits is less than what is claimed by the manufacturer. • Expectation: The average monthly allowance of the morning and afternoon students in a high school are the same. Ho: There is no significant difference in the average monthly allowance of the morning and afternoon students. Ha: There is a significant difference in the average monthly allowance of the morning and afternoon students. Types of Test There are a number of tests involved when a hypothesis is being tested. In this chapter, we shall discuss two types. These are: 1. Test of Means 2. Test of Difference of Means The first type aims to find out if a population characteristic, as indicated by the Mean, has changed. The second seeks to determine if the same characteristic between two populations is significantly different. Types of Error The Null Hypothesis as stated, can either be correct (that is, the statement is true) or wrong (that is, the statement is false). Two types of errors can be committed when a conclusion is made after a hypothesis is tested. These are: • • Type I error (or alpha error): The Null Hypothesis is rejected when, in fact, it is true. Type II error (or beta error): The Null Hypothesis is accepted when, in fact, it is false. Since we can never be completely certain about the claim made in the Null Hypothesis and the corresponding conclusion, therefore, there is always the risk of making either a Type I or a Type II error. Between these two types, the more serious error is Type II. We certainly would avoid making a Type II error, if any error should be made at all. The chance of making an error of this type can be minimized by using what is referred to as the Level of Significance. The Level of Significance One reason for taking a sample from a population is to compute for the mean of the sample and to use this mean as an estimate of the population mean. There are many samples that one can get from a given population. For some of these samples: the computed mean will be close to the population mean; for some, it may not be very close; and for quite a few exceptional samples, the computed Mean may be quite far from the population mean, making the sample mean a poor (or even wrong) estimate of the population mean. Despite the effort to choose a representative sample, such as situation, although unlikely, is still possible. In Statistics, the dividing line between what is acceptable and not acceptable is the Level of Significance. It is a level of probability that separates the sample results that are acceptable and those that are not. More specifically, it is the level of probability of making a Type I error. The customary Levels of Significance are 5 % and 1 %. A level of Significance of 5% means, that there are 5 chances in 100 of rejecting the Null Hypothesis when in fact it is true. Correspondingly, we are 95% confident that we are making the right decision (in terms of our conclusion). Reducing the Level of Significance to 1 % makes us 99% confident that we are doing the right or correct decision. This is especially true in the medical profession, where “life-anddeath” situation is at stake. The probability of making a Type I error could be minimized if we lower the Level of Significance to 1% instead of 5%. A basic principle in Statistics states that the mean of each of the so many samples that can be taken from a population, when organized into a Frequency Distribution, tends to approximate a Normal Curve. This is called the Central Limit Theorem which states that: when all possible samples of a given size are taken from any population, the distribution of the means of Samples will approximate a Normal Distribution. This approximation improves as the Sample Size becomes larger. The Means of most samples will be clustered around the population mean in the Normal Curve. For a few exceptional samples, that, by chance, yield sample means that are far from the population mean, their means are close to the tail ends of the Normal Curve. The region where such values are situated is called the Critical Region. Critical or rejection region Critical or rejection region Level of Significance As shown above, the two critical regions on both ends of the Normal Curve are also called the rejection regions. The area in the Middle is the acceptance region. The Level of Significance is the dividing line between the acceptance region and the rejection region. Important Note: Another way of knowing when to reject or accept the Null Hypothesis is to solve for the p-value. When the p-value is less than the Level of Significance, then the Null Hypothesis is rejected. Use of Symbols We need to make a distinction between the symbols used for a population and those for a sample. These symbols are used in the formulas applied in testing a Hypothesis. Let us summarize them below: Sample Population Size n N Mean πΜ π (mu) Standard Deviation s π (π ππππ) (small letter s) One-Tailed and Two-Tailed Tests In testing a hypothesis, we have to decide whether to use a One-tailed Test or a Two-tailed Test. Whether it is one-tailed or two-tailed will depend on the nature of the problem which, in turn, will determine how the Null Hypothesis (Ho) and Alternative Hypothesis (Ha) are stated. Before we cite examples to determine which test to use, let us point out that in a two-tailed test, the critical regions are on both sides of the Normal Curve, that is: Critical or rejection region Critical or rejection region On the other hand, in one-tailed test, the critical region is only on one side of the Normal Curve, either on the left or right side of it, that is: Critical or rejection region And Critical or rejection region Let us now cite examples to illustrate when to use each type of test. Without solving the problem, the following example will illustrate the use of a Two-tailed Test: • A machine is used to produce a precision part of an automobile. The diameter of one particular item should be 5 centimeters with a slight tolerable allowance on both sides, that is, the diameter can be slightly wider or narrower but for as long as the average diameter of the parts being produced stays at 5 cm, the machine is functioning well. Since the matching has already been used for several years, the manufacturer feels that it is time to check the machine. If it is no longer functioning well, the average diameter could be wider or narrower than 5 cm. One way of checking is to test is a hypothesis by taking a random sample of the parts being manufactured. This problem will require a Two-tailed Test with critical regions on both sides of the Normal Curve because if the machine is not functioning properly it will produce parts whose diameters will be wider and narrower. It is not a case of the malfunctioning machine producing parts with consistently narrower diameters only (or with consistently wider diameters only). This situation determines now how the Null and Alternative Hypotheses are stated, which are: Null Hypothesis: The average diameter of the part being manufactured has not changed (that is, Ho: π = 5 ππ). Alternative Hypothesis: The average diameter has changed (that is, Ha: π ≠ 5 ππ). Note: The manner in which the Alternative Hypothesis is stated as an inequality, confirms the Two-tailed Test that is to be used. The following example will illustrate the use of a One-tailed test (where the critical region is on the right side only of the Normal Curve), that is, specifically a Righttailed test: • A barangay chairman feels that, on the whole households in his barangay, are better-off financially now compared to several years ago. He feels this because he has observed that mothers seem to be visiting supermarkets more often and that he has been receiving less complaints. Furthermore, he knows that a quite a number of husbands in his barangay have gone abroad to work as OFWs (overseas foreign workers) where incomes are much higher than when you are just in the Philippines. He knows from a survey conducted several years back that the average monthly income in his barangay was Php 20,000. He decides to find out if the average monthly income may have possibly increased by taking a sample of households. He tests his hypothesis. The Null and Alternative hypotheses for his problem are stated as follows: Null Hypothesis: The average monthly family income in the barangay has not changed. (that is, Ho: π = πβπ 20,000). Alternative Hypothesis: The average monthly family income has increased (that is, Ha: π > πβπ 20,000). Note: The inequality statement of the Alternative Hypothesis confirms a Right-tailed test. Clearly, the indicators to the barangay chairman suggest that incomes may have possibly increased. There is no suggestion at all that incomes may have decreased. The following example will illustrate a one-tailed (specifically a Left-tailed) test. • A university requires an IQ test for all incoming freshmen students and accepts only those who qualify, with an IQ above the minimum IQ requirement. It is known that the average IQ of the university’s freshmen students is 95. However, for economic reasons (on the part of the university and on the part of the families of students), the IQ test for incoming first-year students has been suspended for the past three years. Lately, some teachers have been complaining that students in their freshmen classes have been getting somewhat lower grades and this appears to be a persistent claim to find out if the IQ level of their first-year students has decreased. They take a sample of freshmen students and test a hypothesis. Let us state the Null and Alternative Hypotheses: Null Hypothesis: The average IQ of freshmen students has not changed (that is, Ho: π = 95). Alternative Hypothesis: The average IQ of freshmen of students has decreased (that is, Ha: π < 95). Note: The evidence based on claims suggests that the IQ has decreased. It cannot be stated as the inequality (π ≠ 95). Definitely, neither (π > 95). Thus, this Alternative Hypothesis suggests a Left-tailed test. Solving Problems by Stepwise Method: (Broto, 2006) Here are the steps: I. II. III. Problem: Hypothesis: hypothesis. Level of Significance: Formulate the Statement of the Problem. State the null hypothesis and/or the research Specify the level of significance, denoted alpha πΌ. Usually taken as 0.05 or 0.01 depending on the problem or study. IV. V. VI. Statistics: Identify the Statistics Tools to use. Decision Rule: In general, the Decision Rule goes this way: If the computed value of the statistics is greater than or beyond the tabular value/critical value, reject the null hypothesis. Conclusion: State the conclusion based on the analysis and interpretation arrived. MODULE 2 THE PARAMETRIC TEST Parametric tests require normal distribution and the levels of measurement are expressed in interval or ratio data. Types of parametric tests are the t-test, z-test, F-test, analysis of variance for the test of difference and the Pearson Product Moment Coefficient of Correlation for the tests of relationship / association, and the tests for prediction and forecasting are the Simple Linear Regression Analysis and the Multiple Regression Analysis. Objectives: When you have completed the chapter, you will be able to: • • • Use lessons in module 1 to determine the types of distribution of data. Distinguish parametric from nonparametric tests and their relationship to normality of distribution. Apply Stepwise Method in solving problems using parametric tests. LESSON 5 - THE t-TEST The t-test is used to compare two means, the means of two independent samples or groups, and the means of correlated samples before and after the treatment. Ideally the t-test is used when there are less than 30 samples, but some researchers use the t-test even if there are more data 30 samples. The formula for the t-test is π‘= Μ Μ Μ Μ Μ Μ Μ Μ π1 −π 2 ππ1 +ππ2 1 1 )( + ) √( π +π 1 2 −2 π1 π2 Where: t = the t-test Μ Μ Μ 1 = the mean of group 1 π Μ Μ Μ 2 = the mean of group 2 π ππ1= sum of squares of group 1 ππ2 = sum of squares of group 2 π1 = number of observations in group 1 π2 = number of observations in group 2 Example 1. The following are the scores of 10 male and 10 female BSBM students in mathematics. Test the null hypothesis that there is no significant difference between the performance of male and female BSMB students in the said test. Use the t-test at 0.05 level of significance. Male 15 19 17 15 6 12 14 10 11 16 Male π₯12 225 361 289 225 36 144 196 100 121 256 1953 π₯1 15 19 17 15 6 12 14 10 11 16 135 Female 13 9 12 5 9 3 8 4 6 13 Female π₯2 π₯22 13 169 9 81 12 144 5 25 9 81 3 9 8 64 4 16 6 36 13 169 82 794 From the table above, Μ Μ Μ 1 = ∑ π₯1 = 13.5 ∑ π₯1 = 135, ∑ π₯1 2 = 1953, n1 = 10, π π1 Μ Μ Μ 2 = ∑ π₯2 = 82, ∑ π₯2 2 = 794, n2 = 10, π 2 ππ1 = ∑ π₯1 − 2 (∑ π₯1 )2 ππ2 = ∑ π₯2 − π1 (∑ π₯2 )2 π2 = 1953 − (135)2 = 794 − 10 822 10 ∑ π₯2 π2 = 130.5 = 121.6 = 8.2 Μ Μ Μ 1 − π Μ Μ Μ 2 π π‘= √( π‘= ππ1 + ππ2 1 1 π1 + π2 − 2)(π1 + π2 ) 13.5 − 8.2 √( 130.5 + 121.6)( 1 + 1 ) 10 + 10 − 2 10 10 π‘= 5.3 √252.1 ( 1 + 1 ) 18 10 10 π‘= 5.3 √14.0056(0.2) π‘= 5.3 √2.80112 π‘= 5.3 1.6737 π‘ =3.167 Adopting the Stepwise Method briefly delineated above: I. II. III. Problem: Is there a significant difference between the performance of the male and female students in mathematics? Hypothesis: Ho: There is no significant difference between the performance of Μ Μ Μ 2 ). male and female BSBM students in mathematics( Μ Μ Μ Μ π1 = π H1: There is a significant difference between the performance of male and Μ Μ Μ Μ Μ Μ Μ female BSBM students in mathematics ( π 1 ≠ π2 ). Level of Significance: πΌ = 0.05 ππ = π1 + π2 − 2 =10+10-2=18 From the table, π‘0.05 = 2.101. This value was obtained by pairing the level of significance, πΌ and the degree of freedom, df, (0.05, 18). IV. V. VI. Statistics: t-test for Two Independent Samples Decision Rule: If the t-computed value is greater than the t-tabular /critical value, reject the null hypothesis Ho. Conclusion: Since the t-computed value of 3.167 is greater than the t-tabular value of 2.101, at 0.05 level of significance with 18 degrees of freedom, the null hypothesis is rejected in favor of the research hypothesis. This means that, there is a significant difference in the performance of the male and female BSMB students in mathematics. It implies that the male performed better than the female students, considering that the mean or average score of 13.5 obtained by the male students is greater than the mean score of the female students of 8.2. Exercise: Two groups of experimental rabbits were injected with tranquilizer at 1.0 mg and 1.5 mg dose respectively. The time given in seconds that took them to fall asleep is hereby given. Use the t-test for two independent samples at 0.01 to test the null hypothesis that the difference in dosage has no effect on the length of time it took them to fall asleep. 1.0 mg 5.3 3.4 7.2 6.7 5.6 3.1 4.8 7.8 13.1 8.2 6.4 12.3 1.5 mg 12.1 7.8 15.4 14.2 13.6 9.7 10.4 17.2 20.3 19.7 LESSON 6 - THE t-TEST FOR CORRELATED SAMPLES The t-test for correlated samples is used when comparing the means before and after the treatment such as pretest and posttest. The formula is, π‘= Μ π· 2 2 − (∑ π·) ∑ π· √ π π(π − 1) Where: Μ π· = the mean difference between the pretest and the posttest. D = the difference between the pretest and the posttest n = the sample size Example 1. An experimental study was conducted on the effect of programmed materials in Mathematics on the performance of 20 selected college students. Before the program was implemented, pretest was administered and after 3 months the same instrument was used to get the posttest result. The following is the result of the experiment: PRETEST X1 20 35 15 16 18 17 23 22 19 25 28 22 12 15 21 28 25 16 34 32 POSTTEST X2 24 38 20 25 27 24 35 27 23 28 32 28 25 26 32 39 36 28 41 38 D -4 -3 -5 -9 -9 -7 -12 -5 -4 -3 -4 -6 -13 -11 -11 -11 -11 -12 -7 -6 ∑ π· = -153 Μ = −7.65 π· Substituting values in the formula below, D2 16 9 25 81 81 49 144 25 16 9 16 36 169 121 121 121 121 144 49 36 ∑ π·2 =1389 π = 20 π‘= π‘= Μ π· (∑ π·)2 2 √∑ π· − π π(π − 1) 7.65 (−153)2 √1389 − 20 20(19) t =-10.087 10.087 Solving by the Stepwise Method: I. II. III. IV. V. VI. Problem: Is there a significant difference in the performance of 20 selected college students in Mathematics before and after the program (Pretest and Posttest) was implemented? Hypothesis: Ho: There is no significant difference between the performance of the 20 selected students in the Pretest and Posttest. H1: There is a significant difference between the performance of the 20 selected students in the Pretest and Posttest. Level of Significance: πΌ = 0.05 df = n-1 df = 19 t0.05=1.729 Statistics: t-test for correlated samples. Decision Rule: If the t-computed value is greater than or beyond the critical value, reject Ho. Conclusion: Inasmuch as the t-computed value of 10.087 is greater the tcritical value of 1.729 at 0.05 level of significance with 19 degrees of freedom, the null hypothesis is therefore rejected in favor of the research hypothesis. This means that the posttest performance of the students surpassed their pretest performance. It implies that the use of the programmed materials in Mathematics is effective. Pretest 16 18 16 24 Posttest 20 20 24 28 20 25 22 18 15 15 20 30 23 24 19 15 Exercise: Ten subjects were given an attitude test on a controversial issue. Then, they were shown a film favorable to the ten subjects and the same attitude test was administered. Make a directional test at πΌ = 0.05. THE Z-TEST The z-test is another test under parametric statistics requiring normality of the distribution. It utilizes the two population parameters π πππ π. It is used to compare two means: the sample mean, and perceived population mean. It is also used to compare the two-sample means reckoned from the same population. It is used when the samples are equal to or greater 30. The z-test can be applied in two ways: the One-Sample Mean Test and the Two-Sample Mean Test. The tabular value of z-test at 0.01 and 0.05 level of significance are shown below. Test One-tailed Two-tailed Level of Significance 0.01 0.05 ± 2.33 ± 1.645 ± 2.575 ± 1.96 LESSON 7 - THE ONE-SAMPLE MEAN TEST The One-Sample Mean Test is used when the sample mean is being compared to the perceived population mean. Nevertheless, in the absence of the population standard deviation, the sample standard deviation can be used for this value. The formula is π§= (π₯Μ − π)√π π Where: π₯Μ = π πππππ ππππ π = βπ¦πππ‘βππ ππ§ππ π£πππ’π of the population mean π = population standard deviation n = sample size Example 1. XYZ company claims that the average lifetime of a certain tire is at least 26,500 km. To check the claim, a taxi company puts 40 of these tires on its taxis and gets a mean lifetime of 24,430 km. With a standard deviation of 1240 km, is the claim true? Use z-test at 0.05. Solving by Stepwise Method: I. II. III. Problem: Is the claim true that the average lifetime of a certain tire is at least 26,500 km? Hypothesis: Ho: The average lifetime of a certain tire is 26,500 km. Ha: The average lifetime of a certain tire is not 26,500 km. Level of Significance: πΌ = 0.05 π§ = ±1.645 IV. Statistics: z-test for one-tailed test Computation: π§= = V. VI. (π₯Μ − π)√π π (24430−26500)√40 1240 = −10.56 Decision Rule: If the z-computed value is greater than the z-tabular value, then reject the null hypothesis. Conclusion: Since the z-computed value of 10.56 is greater than the ztabular value of 1.96, at 0.05 level of significance, the research hypothesis is accepted. This means that the average lifetime of a certain tire is no longer 26,500 km. It implies that the average lifetime of these tires is already less than this value. LESSON 8 - THE TWO-SAMPLE MEAN TEST The two-sample mean test is used when comparing two separate samples drawn at random, from a normal population. To test whether the difference between Μ Μ Μ 1 and π Μ Μ Μ 2 is significant or can be attributed to chance, the formula the two values π used is: π§= Μ Μ Μ Μ Μ Μ Μ Μ π1 −π 2 2 2 π1 π2 π π √ 1+ 2 Where: Μ Μ Μ 1 = ππππ ππ π πππππ 1 π Μ Μ Μ 2 = ππππ ππ π πππππ 2 π π 12 = π£πππππππ ππ π πππππ 1 π 22 = π£πππππππ ππ π πππππ 2 π1 = π ππ§π ππ π πππππ 1 π2 = π ππ§π ππ π πππππ 2 Example 1. An admission test was administered to incoming freshmen in the Colleges of Marketing Management and Information and Computer Technology with 100 students. Each was randomly selected. The mean scores of the given samples Μ Μ Μ 2 = 75 and the variances of the test scores were 38 and 32, were Μ Μ Μ π1 = 85 πππ π respectively. Is there a significant difference between the two groups? Use 0.01 level of significance. Solve by Stepwise Method: I. II. Problem: Is there a significant difference between the two groups? Hypotheses: Ho: There is no significant difference between the two groups. Ha: There is a significant difference between the two groups. III. Level of Significance: πΌ = 0.01 π§ = ±2.575 IV. Statistics: z-test for two-tailed test π§= Μ Μ Μ Μ Μ Μ Μ Μ π1 −π 2 2 2 π1 π2 π π √ 1+ 2 π§= 85−75 38 32 + 100 100 √ π§ =11.95 V. VI. Decision Rule: If the z-computed value is greater than the z-tabular value, then reject the null hypothesis. Conclusion: Inasmuch as the z-computed value of 11.95 is greater than the z-tabular value of 2.575 at 0.01 level of significance, so that, the null hypothesis is rejected. This means that the two groups have different performances. It implies that the incoming freshmen in the College of Marketing Management has performed better than the incoming freshmen in the College of Information and Computer Technology in the entrance examination administered to them. THE F-TEST The F-test otherwise known as the Analysis of Variance (ANOVA) is used in comparing the means of two or more independent groups. One-way ANOVA is used when there is only one variable involved. The Two-way ANOVA is used when two variables are involved: the column and row variables. The researcher is interested to know if there are significant differences between and among column and row variables. This is also used to determine if there is an interaction effect between the variables being analyzed. Like the t-test, the F-test is also a parametric test, which requires that the samples are normally distributed and that the data are expressed as interval or ratio. This test is more efficient than the other tests of difference. LESSON 9 - THE F-TEST (ONE-WAY-ANOVA) Example 1. A certain grocery store is selling 4 brands of soap. The owner is interested if there is a significant difference in the average sales for one week. The following data were recorded. Brand A B C D 8 10 2 4 5 12 3 6 4 9 5 7 7 8 6 9 8 7 4 3 5 10 3 4 4 11 3 6 Perform the analysis of variance and test the hypothesis at 0.05 level of significance that the average sales of the four brands of soap are equal. Solving by Stepwise Method: I. II. Problem: Is there a significant difference in the average sales of the four brands of soap? Hypotheses: Ho: There is no significant difference in the average sales of the four brands of soap. Ha: There is a significant difference in the average sales of the four brands of soap. III. Level of Significance: πΌ = .05 ππ = 3 πππ 24 IV. Statistics One-Way-Analysis of Variance (F-test) computation: A B C 2 2 π1 π π π1 π2 π32 2 3 8 64 10 100 2 4 5 25 12 144 3 9 4 16 9 81 5 25 7 49 8 64 6 36 8 64 7 49 4 16 5 25 10 100 3 9 D π4 4 6 7 9 3 4 π42 16 36 49 81 9 16 4 41 16 259 11 67 Μ Μ Μ 1=41 = 5.857 π 7 121 659 3 26 Μ Μ Μ 2=67 = 9.571 π 7 9 108 Μ Μ Μ 3=26 = 3.714 π 7 6 39 36 243 Μ Μ Μ 4=43 = 5.571 π 7 Steps in Solving for F-test (One-Way ANOVA): Calculate the following: 1. Correction Factor, πΆπΉ = CF = (41+67+26+39)2 = 28 (∑ π)2 (173)2 28 π = 1068.893 2. Total Sum of Squares, πππ = ∑ π₯12 +∑ π₯22 +∑ π₯32 +∑ π₯42 − πΆπΉ TSS =259 + 659 + 108 + 243 − 1068.893 = 200.107 (∑ π1 )2 3. Between Sum of Squares, BSS= BSS = 412 7 672 + 7 + 262 7 392 + 7 π1 + (∑ π₯2 )2 π2 + (∑ π3 )2 π3 + (∑ π4 )2 π4 − πΆπΉ − 1068.893=126.393 4. Within Sum of Squares, πππ = πππ − π΅ππ WSS = 200.107 − 126.393 = 73.714 Analysis of Variance Table Sources of Variation Between Groups Within Group Total V. VI. Degrees of Freedom Sum of Squares k-1=4-1=3 126.393 27-3=24 73.714 N-1=28-1=27 200.107 Mean Squares 126.393 = 42.131 3 73.714 = 3.071 24 F-Value Computed Tabular 42.131 3.01 = 13.719 3.071 Decision Rule: If the F-computed value is greater than the F-tabular value, reject the null hypothesis. Conclusion: Since the F-computed value of 13.719 is greater than the Ftabular value of 3.01 at 0.05 with 3 and 24 degrees of freedom, so that the null hypothesis is rejected. This means that there is a significant difference in the average sales of the four brands of soap. It implies that soap B has the greatest average sales. Note: When it was found out that Ho was rejected or that there was a significant difference in the computed means among groups, use ScheffπΜ s Test to identify which groups have significant differences. SCHEFFπ¬ΜS TEST πΉ′ = Μ Μ Μ 1 − π Μ Μ Μ 2 )2 (π ππ 2 (π1 + π2 ) π1 π2 Where: F’ = Scheffe’s Test π1 = ππππ ππ ππππ’π 1 π2 = ππππ ππ ππππ’π 2 π1 = ππ’ππππ ππ π ππππππ ππ ππππ’π 1 π2 = ππ’ππππ ππ π ππππππ ππ ππππ’π 2 ππ 2 = π€ππ‘βππ ππππ π ππ’ππππ Brand A vs B: F’ = (5.857−9.571)2 3.071(7+7) 7(7) Brand A vs C: F’ = (5.857−3.714)2 3.071(7+7) 7(7) F’ = -4.233 F’ = 5.234 Brand A vs D: Brand B vs C: F’ = (5.857−5.571)2 3.071(7+7) 7(7) F’ = (9.571−3.714)2 3.071(7+7) 7(7) F’ = 0.000 F’ = 39.097 Brand B vs D: Brand C vs D: F’ = (9.571−5.571)2 3.071(7+7) 7(7) F’ = 18.235 F’ = (3.714−5.5712 3.071(7+7) 7(7) F’ = 3.930 Comparison of the Average Sales of the Four Brands of Soap Between Brand F’ A vs B A vs C A vs D B vs C B vs D C vs D 4.233 5.234 0.000 39.097 18.235 3.930 (F0.05) (k-1) (3.071) (3) 9.213 9.213 9.213 9.213 9.213 9.213 Interpretation Not Significant Not Significant Not Significant Significant Significant Not Significant The above table shows the F’-computed values for all the four brands of soap under comparison. Since the F’-computed values of brands A vs B, A vs C, A vs D, and C vs D are all less than the F-tabular value of 9.213 at 0.05 level of significance and 3 degrees of freedom, thus it can be said that the means for these brands of soap have no significant differences. On the other hand, inasmuch as the F’-values of 39.097 and 18.235 for B vs C and B vs D respectively are both greater than the F-tabular values, hence the means for brands B and C, and B vs D have significant differences. It implies that the brand B of soap is more saleable than the brands C and D. Exercise. The following data represent the operating time in hours of the 3 types of scientific pocket calculators before a recharge is required. Perform the Analysis of Variance (F-test) at 0.05 level of significance. A 4.8 5.6 4.7 6.2 4.4 6.9 Brand B 6.7 6.9 7.2 5.8 5.4 6.3 4.8 C 7.6 7.2 7.6 6.8 6.3 LESSON 10 - THE F-TEST (TWO-WAY-ANOVA WITH INTERACTION EFFECT) In statistics, the two-way analysis of variance (ANOVA) is an extension of the oneway ANOVA that examines the influence of two different categorical independent variables on one continuous dependent variable. The Two-Way ANOVA uses the following formulas to ultimately calculate the values of F for: Between Columns, Rows and Interaction. (πΊπ)2 • Correction Factor, πΆπΉ = • Total Sum of Squares, πππ = ∑ π₯ 2 − πΆπΉ • Within Sum of Squares, πππ = ∑ π₯ 2 − ∑ • Column Sum of Squares, πππ = ∑ π 2 (∑ ππ€ ) ππ€ ,w = within number 2 (∑ ππ ) ππ − πΆπΉ, c=column number 2 (∑ ππ ) • Row Sum of Squares, πππ = ∑ • Interaction Sum of Squares, SScr = SST-SSW-SSc-SSr ππ − πΆπΉ, r=row number Degrees of Freedom: • • • • • • Total Degree of Freedom, dft = N-1 Within Degree of Freedom, dfw = k(ni-1) Column Degree of Freedom, dfc= c-1 Row Degree of Freedom, dfr= r-1 Interaction Degree of Freedom, dfcr = (c-1)(r-1) ππ Mean of Square, MS = ππ • F-test, F = πππππ¦ πππ€ GT= x grand total, ∑ π₯ 2 = π π’π ππ π ππ’ππππ ππ πππ π₯, ∑ ππ = π π’π ππ π ππ’ππππ ππ π₯ π€ππ‘βππ ∑ ππ = π π’π ππ π ππ’ππππ ππ π₯ ππ π ππππ’ππ π, ∑ ππ = π π’π ππ π ππ’ππππ ππ π₯ ππ π πππ€ Example 1. Thirty language students were randomly assigned to one of three instructors and to one of the two methods of teaching. Achievement was measured on a test administered at the end of the term. Use a Two-Way ANOVA with interaction effect at 0.05 level of significance to test the following null hypotheses: 1. There is no significant difference in the performance of the three groups of students under three different instructors. 2. There is no significant difference in the performance of the two groups of students under two different methods of teaching. 3. Interaction effect is not present. TWO-WAY ANOVA with Significant Interaction Teacher Factor (Column) Method of Teaching 1 (row) Method of Teaching 2 (row) A 38 42 44 37 38 41 40 38 37 39 B 52 53 47 49 46 47 45 43 44 39 C 39 43 44 38 39 52 45 43 46 42 Solving by Stepwise Method: I. II. III. IV. Problem: 1. Is there a significant difference in the performance of the students under three different teachers? 2. Is there a significant difference in the performance of the students under two different methods of teaching? 3. Is there an interaction effect between teachers and methods of teaching factors? Hypotheses: (Ho) 1. There is no significant difference in the performance of the students under three different teachers. 2. There is no significant difference in the performance of the students under two different teachers. 3. There is no significant effect between teachers and methods of teaching. Level of Significance: πΌ = 0.05 Total Degree of Freedom, dft = N-1 = 30-1 = 29 Within Degree of Freedom, dfw = k(ni-1) = 6(5-1) = 24 Column Degree of Freedom, dfc= c-1 = 3-1 = 2 Row Degree of Freedom, dfr= r-1 = 2-1 = 1 Interaction Degree of Freedom, dfcr = (c-1)(r-1) = 2*1=2 Statistics: F-test Two-Way ANOVA With interaction effect. Teacher Factor (Column) Method of Teaching 1 (row) Total Method of Teaching 2 (row) Total Total A 38 42 44 37 38 199 41 40 38 37 39 195 394 B 52 53 47 49 46 247 47 45 43 44 39 218 465 C 39 43 44 38 39 203 52 45 43 46 42 228 431 Total 649 641 `1290 Extend the steps in One-Way ANOVA Steps: Calculate 1. CF= (1290)2 30 = 55470 2. SST = 382+422+442+…+462+422- CF = 56080 - 55470 = 610 3. SSW = 56080 - (1992+2472+2032+1952+2182+2282) /5 = 56080 - 55870.4 = 209.6 4. SSc = (3942+4652+4312)/10 - CF = 55722.2 - 55470 = 252.2 5. SSr = (6492+6412)/15 - CF = 55472.13 – 55470 = 2.13 6. SScr = SST-SSW-SSc-SSr = 610 – 209.6 – 252.2 – 2.13 = 146.07 ANOVA TABLE (TWO-WAY) for Teachers and Methods of Teaching Factors Sources of Variation Between Columns (Teacher) Rows (Methods) Interaction Within Total V. VI. F-value Tabular SS df MS 252.20 2 126.10 14.44 3.40 Significant 2.13 1 2.13 0.24 4.26 Not Significant 146.70 209.60 610.00 2 24 29 73.35 8.73 8.40 3.40 Significant Computed Interpretation Decision Rule: If the F-computed value is greater than the F-tabular value, then reject the null hypothesis. Conclusion: The above table shows ANOVA table for Teachers and Methods of Teaching Factors. It can be deduced from this table, that since the F-computed value of 14.44 for the row (teacher) is greater than the F-tabular value of 3.40 at 0.05 level of significance with 2 and 24 degrees of freedom, hence the null hypothesis is rejected. This means that there is a significant difference in the performance of students under three different teachers. Teacher factor affects the performance of the students. This implies that students under teacher B have performed better than those under teacher A and teacher C. While students under teacher A have the poorest performance. On the other hand, since the F-computed value of 0.24 for row is less than the F-tabular value of 4.26 with 1 and 24 degrees of freedom, the null hypothesis is accepted. This indicates that as far as methods of teaching is concerned, the performance of the students is unaffected. Methods of teaching doesn’t matter. The F-computed value of 8.40 against the F-tabular value of 3.40 indicates that there is an interaction effect between teachers and their methods of teaching. Students under teacher B have better performance under methods of teaching 1 while students under teacher C have better performance under methods of teaching 2. LESSON 11 - THE PEARSON PRODUCT MOMENT COEFFICIENT OF CORRELATION, r The Pearson Product Moment Coefficient r is an index of relationship between two variables. The independent variable can be represented by x while the dependent variable can also be represented by y. The value of r ranges from -1 to +1. If the of value of r is +1 or -1, there is a perfect correlation between x and y. Which means that x influences y or y depends upon the value of x. However, if r equals zero, then x and y are independent of each other. Consider the graph below. High r=+ x Low High If the trend of the line graph is going upward, the value of r is positive. This indicates that as x increases, the value of y also increases. Likewise, if x decreases, the value of y also decreases, the x and y are positively correlated. High r=- x Low High If the trend of the line graph is going downward, the value of r is negative. It indicates that as x increases, the corresponding value of y decreases, x and y are negatively correlated. High Low High If the trend of the line graph cannot be established either upward or downward (i.e., when the plotted points are so scattered), then r=0. This indicates that there is no correlation between the x and y variables. The formula for the Pearson Product Moment Coefficient of Correlation, r is: π= π ∑ π₯π¦−∑ π₯ ∑ π¦ √(π ∑ π₯ 2 −(∑ π₯)2 )(π ∑ π¦ 2 −(∑ π¦)2 ) Where: r = the Pearson Product Moment Coefficient of Correlation n = sample size ∑ π₯π¦ = the sum of the product x and y ∑ π₯ ∑ π¦ = the product of the sum of x and the sum of y ∑ π₯ 2 = sum of squares of x ∑ π¦ 2 = sum of squares of y Example 1. Below are the midterm (x) and final (y) grades. x 75 70 65 90 85 80 70 65 90 88 y 80 77 65 94 88 85 88 76 72 91 Solving by Stepwise Method I. II. III. Problem: Is there a significant relationship between the midterm and the final examinations of 10 students in Statistics. Hypotheses: Ho: There is no significant relationship between the midterm grades and the final examination/grades of 10 students in Statistics. Ha: There is a significant relationship between the midterm grades and the final examination/grades of 10 students in Statistics. Level of Significance: πΌ = 0.05 ππ = π − 2 = 10 − 2 =8 π0.05 = 0.632: IV. Statistics: Pearson Product Moment Coefficient of Correlation Computation: No x 1 75 2 70 3 65 4 90 5 85 6 80 7 70 8 65 9 90 10 88 Total 778 π= y 80 77 65 94 88 85 88 76 72 91 816 x2 5625 4900 4225 8100 7225 6400 4900 4225 8100 7744 61444 y2 6400 5929 4225 8836 7744 7225 7744 5776 5184 8281 67344 xy 6000 5390 4225 8460 7480 6800 6160 4940 6480 8008 63943 10(63943) − (778)(816) √[10(61444) − (778)2 ][10(67344) − (816)2 ] π = 0.55 V. VI. Decision Rule: If the computed r value is greater than the r tabular value, reject Ho. Conclusion/Implication: Since the r-computed value of 0.55 is less than the r-tabular value of 0.632 at 0.05 level of significance with 8 degrees of freedom, hence, the null hypothesis is accepted. This means that there is no significant relationship between the midterm grades of the students and the final grades. It implies that the midterm grade has no say with the final grade. Coefficient of Determination The coefficient of determination (CD) is r2 times 100%. This statistics enables us to explain the extent to which the independent variable x influences the dependent variable y. Example. What is the coefficient of determination when r = 0.55? CD = (0.55)2x (100%) CD = 30.25 % This value of 30.25 % indicates that the final examination grade does not depend on the midterm grade, implying that the final grade is not influenced by the midterm grade. LESSON 12 - THE SIMPLE LINEAR REGRESSION ANALYSIS The simple linear regression analysis is used when there is a significant correlation between x and y variables. The established linear equation is used to predict the value of y given the value of x. The formulas for the simple linear regression analysis are illustrated below. π¦Μ = π + ππ₯ Where: π¦Μ = π‘βπ πππππππππ‘ π£πππππππ ππ π‘βπ πππππππ‘ed value of y x = the independent variable a = the y-intercept b = the slope of the line π= π ∑ π₯π¦ − ∑ π₯ ∑ π¦ π ∑ π₯ 2 − (∑ π₯)2 π = π¦Μ − ππ₯Μ π₯Μ = ππππ ππ π₯, π¦Μ = ππππ ππ π¦ Example. A study was made by SM Bicutan to determine the relationship between weekly sales and advertising expenditures. The following data were recorded. Using r at 0.05 level of significance, determine the prediction equation if x and y have significant relationship. Advertising Cost Sales (in thousand Php) x 4.8 3.2 3.6 3.3 5.2 5.6 3.3 4.3 3.8 4.6 (in thousand Php) y 42.5 38.6 40.2 38.5 45.4 48.8 40.0 38.4 42.4 40.7 Solve by Stepwise Method: I. Problem: Are weekly sales influenced by the advertising expenditures? II. III. IV. Hypotheses: Ho: Weekly sales are not influenced by the advertising expenditures. Ha: Weekly sales are influenced by the advertising expenditures. Level of Significance: πΌ = 0.05 ππ = π − 2 = 10 − 2 = 8 π0.05 = 0.632 Statistics: Pearson Product Moment Coefficient of Correlation Computation: No 1 2 3 4 5 6 7 8 9 10 Total X 4.8 3.2 3.6 3.3 5.2 5.6 3.3 4.3 3.8 4.6 41.7 π= y 42.5 38.6 40.2 38.5 45.4 48.8 40.0 38.4 42.4 40.7 415.5 x2 23.04 10.24 12.96 10.89 27.04 31.36 10.89 18.49 14.44 21.16 180.51 y2 1806.25 1489.96 1616.04 1482.25 2061.16 2381.44 1600 1474.56 1797.76 1656.49 17365.91 xy 204 123.53 144.72 127.05 236.08 273.28 132 165.12 161.12 187.22 1754.11 10(1754.11) − (41.7)(415.5) √[10(180.51) − (41.7)2 ][10(17365.91) − (415.5)2 ] π = 0.827 V. VI. Decision Rule: If the computed value is greater than the tabular value, reject null hypothesis. Conclusion: Since the r-computed value of 0.827 is greater than r-tabular value of 0.632 at 0.05 level of significance with 8 degrees of freedom, hence, the null hypothesis is rejected in favor of the research hypothesis. This means that the advertising cost is related to the sales. The sales are influenced by the advertisement, implying that the higher the cost of advertisement, the higher the sales. And since the variable advertising cost, denoted x, is significantly related to the sales, denoted y, Simple Linear Regression can be used. Solving for: 1. π = π ∑ π₯π¦−∑ π₯ ∑ π¦ π ∑ π₯ 2 −(∑ π₯)2 = 10(1754.11)−(41.7)(415.5) 10(180.51)−(41.7)2 , π = 3.243 2. π = π¦Μ − ππ₯Μ , π₯Μ = 4.17, π¦Μ = 41.55, π = 41.55 − 3.243(4.17) = 28.026 3. Thus, the Linear Regression Equation is π¦Μ = 28.026 + 3.243π₯ 4. Check: by solving values for some y’s: when x = 4.8, y = 43.59, and when x = 3.2, y = 38.40. The values computed for y are closed to the values: 42.5 and 38.6, respectively. This implies that the established Simple Linear Regression Equation, π¦Μ = 28.026 + 3.243π₯, can be used to predict values of y, given values of x. LESSON 13 - THE MULTIPLE REGRESSION ANALYSIS The multiple regression analysis is used for predictions. The dependent variable can be predicted given several independent variables. For instance, we can make better predictions of the performance of newly hired workers if we consider not only their education but also their years of experience, attitudes, and some other variables that may influence performance. Many mathematical formulas can be used to express relationship among two or more variables, but the most commonly used in statistics are linear equations. π¦ = π0 + π1 π₯1 + π2 π₯2 + β― + ππ π₯π Where: y = the dependent variable to be predicted x1, x2, x3, …, xn = the known independent variables that may influence y. b0, b1, b2, …, bn = the arbitrary constants whose values can be determined from the observed data. When there are two independent variables x1 and x2 and we want to fit the equation in the equation model, we use the equation, π¦ = π0 + π1 π₯1 + π2 π₯2 We must solve the three normal equations: ∑ π¦ = ππ0 + ∑ π₯1 π1+∑ π₯2 π2 (1) ∑ π₯1 π¦ = ∑ π₯1 π0 + ∑ π₯12 π1 +∑ π₯1 π₯2 π2 (2) ∑ π₯2 π¦ = ∑ π₯2 π0 +∑ π₯1 π₯2 π1 + ∑ π₯22 π2 (3) Example 1. The following are data on the ages and salaries of a random sample of 6 executives working at BBC corporation and their academic achievements while in college. Income (in thousand Php) y 82.3 75.6 85.4 78.8 73.2 69.3 Age Academic Achievement x2 1.75 2.0 1.5 2.0 2.25 2.5 x1 39 33 42 39 30 29 a) Find the equation of the form π¦ = π0 + π1 π₯1 + π2 π₯2 based on the data shown in the table. b) Use the equation obtained in (a) to estimate the average income of a 36-year old executive with 1.5 academic achievement. Computations: No π¦ π₯1 π₯2 π₯12 π₯22 π₯1 π¦ π₯2 π¦ π₯1 π₯2 1 2 3 4 5 6 Total 82.3 75.6 85.4 78.8 73.2 69.3 464.6 39 33 42 39 30 29 212 1.75 2 1.5 2 2.25 2.5 12 1521 1089 1764 1521 900 841 7636 3.0625 4 2.25 4 5.0625 6.25 24.625 3209.7 2494.8 3586.8 3073.2 2196 2009.7 16570 144.03 151.2 128.1 157.6 164.7 173.25 918.88 68.25 66 63 78 67.5 72.5 415.25 From (1) ∑ π¦ = ππ0 + ∑ π₯1 π1+∑ π₯2 π2 464.6 = 6π0 + 212π1 +12π2 (1) From (2) and from (3) ∑ π₯1 π¦ = ∑ π₯1 π0 + ∑ π₯12 π1 +∑ π₯1 π₯2 π2 16570 = 212π0 + 7636π1 +415.25π2 ∑ π₯2 π¦ = ∑ π₯2 π0+∑ π₯1 π₯2 π1 + ∑ π₯22 π2 918.88 = 12π0+415.25π1 +24.625π2 (2) (3) Solving the 3 equations (1), (2) and (3), you get π0 = 83.68 π1 = 0.423 π2 = −10.593 Take note that 3 linear equations with 3 unknowns can be solved by Elimination, Substitution, or by Extended Matrix (Cramer’s Rule) or you can make use of your scientific calculators with built-in functions dedicated for solving such equations. Knowing the values for π0 , π1 , πππ π2 , the regression equation therefore is a) π¦ = 83.68 + 0.423π₯1 − 10.593π₯2 and b) π€βππ π₯1 = 36 πππ π₯2 = 1.5, π¦ = 83.019 Using the regression equation of π¦ = 83.68 + 0.423π₯1 − 10.593π₯2 as established above, based on the given data, when the age of an executive is 36 years old with academic achievement ππ 1.5, then his predicted average salary will be Php 83,019. Exercise: The following data were obtained for 8 students in Management: midterm grades, average grade in quizzes and their final grades. (Antonio S. Broto, Statistics Made Simple, pp 240, exercise 18). Midterm Grade 3.0 1.5 1.25 1.25 1.75 2.75 2.5 2.0 Average Grade in Quizzes 2.5 1.5 1.5 1.25 2.0 2.75 2.5 2.25 Final Grade 2.75 1.5 1.25 1.25 1.5 2.5 2.5 2.0 a) Use the method of least squares (multiple regression) to fit an equation of the form π¦ = π0 + π1 π₯1 + π2 π₯2 . b) Predict the final grade of student whose midterm grade is 1.75 and the average grade in the quizzes is 1.5. MODULE 3 THE NONPARAMETRIC TEST Nonparametric tests do not require a normal distribution. When the value of skewness is not zero (either positive or negative), and the Kurtosis is greater or less than 0.265, the distribution is said to be abnormal. When the value of the Kurtosis is less than 0.265, the distribution is said to be leptokurtic but it is platykurtic when the value is greater than 0.265. Nonparametric tests also utilize both nominal and ordinal data. Nominal data are expressed in categories, whereas, the ordinal data are expressed in ranking. The most commonly used tests under the nonparametric tests are the ChiSquare test; U-test; H-test; Spearman Rank Order Coefficient of Correlation, rs; Sign Test (Median Test); Mc Nemar’s Test, Friedman Test, Fr; and Kendall’s Coefficient of Concordance, W. Objectives: When you have completed the module, you will be able to: • • • • • Distinguish different types of Chi-Square tests and their uses. Identify parametric and nonparametric tests which are counterparts. Recognize data appropriate for nonparametric tests. Convert data suited for parametric tests into data for nonparametric tests. Apply the Stepwise Method in solving statistical problems into nonparametric tests. LESSON 14 - THE CHI-SQUARE TEST This test is a test of difference between the observed and expected frequencies. The Chi-Square is considered a unique test because it has 3 functions which are as follows: • • • The test of goodness-of-fit The test of homogeneity The test of independence THE CHI-SQUARE TEST OF GOODNESS-OF-FIT This is a test of difference between the observed frequencies and expected frequencies. As an illustration, let’s take for example the throwing of a coin, the coin has only two faces, the head and the toe. If a coin is thrown, it may either fall the head or the toe be on top. Which means that, theoretically, there is a fifty-fifty chance for each of the face being on top. Now if you throw a coin 100 times, the expected frequencies for the head being on top should be 50. And this is true for the other face, the head. But the actual observed frequencies may be different from the expected frequencies. The formula for the Chi-Square Test is: π2 = ∑ (π − πΈ)2 πΈ Where: π 2 = π‘βπ πΆβπ − πππ’πππ π‘ππ π‘ π = π‘βπ πππ πππ£ππ πππππ’ππππππ πΈ = π‘βπ ππ₯ππππ‘ππ πππππ’ππππππ Example. A coin is thrown 100 times. The observed frequency for the head being on top when it falls is 48, while the toe being 52. Using Chi-Square, determine if there is a significant difference between the observed and expected frequencies. Use 0.05 level of significance. Solving by Stepwise Method: I. II. Problem: Is there a significant difference between the observed and expected frequencies? Hypotheses: Ho: There is no significant difference between the observed and expected frequencies. Ha: There is a significant difference between the observed and expected frequencies. III. Level of Significance: IV. πΌ = 0.05 ππ = β − 1 = 2 − 1 = 1 2 π0.05 = 3.841 (ππππ π‘πππ’πππ π£πππ’π) Statistics: Chi-Square Test of Goodness- of-Fit Computation: Face Ratio Head Toe Total 1 1 2 (Actual Result) Observed 48 52 100 π2 = ∑ π2 = (Theory) Expected 50 50 100 (π − πΈ)2 πΈ (48 − 50)2 (52 − 50)2 + 50 50 π 2 = 0.16 V. VI. Decision Rule: If the π 2 − πππππ’π‘ππ π£πππ’π is greater than the π 2 − π‘πππ’πππ π£πππ’π, the null hypothesis is rejected. Conclusion: Since the π 2 − πππππ’π‘ππ π£πππ’π of 0.16 is less than the π 2 − π‘πππ’πππ π£πππ’π of 3.481 at 0.05 level of significance with 1 degree of freedom, the null is accepted. This means that there is no significant difference between the observed and the expected frequencies. This implies that the theory of a fifty-fifty chance for each face of the coin being on top holds true inasmuch as the value of π 2 does not warrant the theory to be rejected. Exercise: A certain machine is supposed to mix peanuts, hazelnuts, cashews, and pecans in the ratio of 4:3:2:1. A can containing 500 of these mixed nuts was found to have 275 peanuts, 105 hazelnuts, 76 cashews and 44 pecans. At 0.05 level of significance, test the hypothesis that the machine is mixing the nuts at this specified ratio. THE CHI-SQUARE TEST OF HOMOGENEITY This test involves two or more samples, with only one criterion variable. This is used to determine if two or more populations are homogeneous. The distribution of data is similar with respect to a particular criterion variable. The formula is: π2 = π(ππ − ππ)2 ππππ Where: π 2 = πβπ − π ππ’πππ π‘ππ π‘ π = πππππ π‘ππ‘ππ π, π, π πππ π = π‘βπ πππππ’ππππππ ππππππππ ππ π‘βπ π‘ππππ ππ πππππ€ ππππ = πππππ’ππ‘ ππ π‘βπ πππ€π πππ ππππ’πππ π‘ππ‘ππ Example. Evaluate the attitude of a sample of LDP and Nationalista parties on the issue of peace and order in Mindanao. To carry out the study, a random sampling of members of each party is drawn from the nationwide population of LDP and Nationalista and each individual in both samples responds to the scale. Scores are then classified into “Favorable” or Unfavorable” categories. The following frequencies were recorded: LDP Nationalista Total Favorable 63 a 52 c 115 m Unfavorable 37 b 48 d 85 n Total 100 k 100 l 200 N Solving by Stepwise Method: I. II. III. Is there a significant difference between the attitudes of the two political parties on the issue of peace and order in Mindanao? Hypotheses: Ho: There is no significant difference between the attitudes of the two political parties on the issue of peace and order in Mindanao. Ha: There is a significant difference between the attitudes of the two political parties on the issue of peace and order in Mindanao. Level of Significance: πΌ = 0.05 ππ = (π − 1)(π − 1) = (2 − 1)(2 − 1) ππ = 1 2 π0.05 = 3.841 IV. Statistics: Chi-Square of Homogeneity Computation: π2 = π2 = π(ππ − ππ)2 ππππ 200((63)(48)−(37)(52))2 (100)(100)(115)(85) π 2 =2.476 V. VI. Decision Rule: If the π 2 − πππππ’π‘ππ value is greater than the π 2 − π‘πππ’πππ value, reject null hypothesis. Conclusion: Since the π 2 − πππππ’π‘ππ value of 2.476 is less than the π 2 − π‘πππ’πππ value of 3.841 at 0.05 level of significance with 1 degree of freedom, hence the null hypothesis is accepted. This means that there is no significant difference between the attitudes of the two political parties on the issue of peace and order in Mindanao. Exercise. Using Chi-Square at 0.05 level of significance, determine the attitude of the TCU community on the issue of charter change. To carry out such study, 100 samples from each teaching and non-teaching personnel were taken to respond to the issue categorized as “YES” or “NO”. The following frequencies were obtained: Teaching Non-Teaching Total YES 34 (a) 55 (c) 89 (m) NO 66 (b) 45 (d) 111 (n) Total 100 (k) 100 (l) 200 (N) THE CHI-SQUARE TEST OF INDEPENDENCE (ONE SAMPLE, TWO CRITERION VARIABLES) The one-sample test of independence is different from the test of homogeneity. This sample consists of members randomly drawn from the same population. This test is used to find out whether measures taken on two criterion variables are either independent or associated with one in a given population using such variables as say level of education and performance, income and years of experience, etc. The calculation of this test is similar to the test of goodness-of-test and the test of homogeneity. The formula is: π2 = ∑ (π − πΈ)2 πΈ Where: π 2 = π‘βπ πΆβπ − πππ’πππ π‘ππ π‘ π = π‘βπ πππ πππ£ππ πππππ’ππππππ πΈ = π‘βπ ππ₯ππππ‘ππ πππππ’ππππππ πΈππ = (π ππ€ πππ‘ππ)π π(πΆπππ’ππ πππ‘ππ)π πΊππππ πππ‘ππ i = row number j= column number Example. 100 individuals, male and female, were given an IQ test and their scores were classified into high and low. Using the π 2 -test of independence at 0,05 level of significance, the table is shown as follows: IQ Sex High Male Female Total O 23 32 55 Low E O 33 12 45 Total E Solving by Stepwise Method: I. II. III. Problem: Is there a significant relationship between sex and IQ? Hypotheses: Ho: There is no significant relationship between sex and IQ. Ha: There is a significant relationship between sex and IQ. Level of Significance: πΌ = 0.05 ππ = (π − 1)(π − 1) = (2 − 1)(2 − 1) =1 2 π0.05 = 3.841 56 44 100 IV. Statistics: Chi-Square of Independence Computation: IQ Sex High O 23 32 55 Male Female Total Low E 30.80 24.20 55 O 33 12 45 πΈ11 = (56)(55) = 30.80 100 πΈ12 = (56)(45) = 25.20 100 πΈ21 = (44)(55) = 24.20 100 πΈ22 = (44)(45) = 19.80 100 π2 = ∑ π2 = (23 − 30.80)2 30.80 + Total E 25.20 19.80 45 56 44 100 (π − πΈ)2 πΈ (33 − 25.2)2 25.2 + (32 − 24.20)2 24.20 + (12 − 19.80)2 19.80 π 2 = 9.976 V. Decision Rule: If the π2 − computed value is greater than the π 2 – tabular VI. value, reject the null hypothesis. Conclusion: Inasmuch as the π 2 − computed value of 9.976 is greater than the π 2 − tabular value of 3.841 at 0.05 level of significance with 1 degree of freedom, the null hypothesis is rejected. This means that there is a significant relationship between sex and IQ. It implies that the female has a better IQ than the male counterpart. Students’ Activity: Two lots of 30 experimental guinea pigs were used in testing the effectiveness of a new serum in combating a certain disease. Both were inoculated with the new organism but only one lot was previously given the preventive serum. Is the serum effective? Use 0.01 level of significance. Recovered Died Total Serum 12 2 14 No Serum 3 13 16 Total 15 15 30 Note: When df is 1 and any expected frequency is small, less than 10, the π 2 − π‘ππ π‘, using the Yate’s correction for lack of continuity will be applied because the distribution of Chi-Square is discrete. Whereas the values obtained by the use of the formula result in a continuous probability model. The formula used is π2 = ∑ (|π−πΈ|−0.5)2 πΈ or π 2 π (|ππ − ππ| − 2 ) π2 = ππππ LESSON 15 - THE WILCOXON RANK-SUM TEST OR WILCOXON TWO-SAMPLE TEST The Wilcoxon Rank-Sum Test is commonly used for the comparison of two groups of nonparametric (interval or not normally distributed) data, such as those which are not measured exactly but rather as falling within certain limits (e.g., how many animals died during each hour of an acute study). π1 = π1 − π1 (π1 + 1) 2 π2 = π2 − π2 (π2 + 1) 2 Where: π1 = ππππππ₯ππ π πππ − ππ’π πππ π‘ π1 = π π’π ππ πππππ ππ ππππ’π 1 π1 = π πππππ π ππ§π ππ ππππ’π 1 π2 = ππππππ₯ππ π πππ − ππ’π πππ π‘ π2 = π π’π ππ πππππ ππ ππππ’π 2 π2 = π πππππ π ππ§π ππ ππππ’π 2 π = min (π1 , π2 ) If U is less than the tabular U, reject Ho. Example. Of the eighteen selected patients who were seriously infected of covid 19, ten were treated with a new serum and eight were not. The number of days the patient recovered were then recorded. Using the Wilcoxon-rank-sum test at 0.05 level of significance, test whether the serum is effective, consider the following data: With Treatment No Treatment 6.5 5.4 15.2 18 5.3 7.2 4.5 8.4 6.8 7.3 13.6 12.2 13.1 12.0 17.6 19.4 6.7 8.2 Solving by the Stepwise Method: I. Problem: Is the new serum effective in treating leukemia? II. Hypotheses: Ho: The new serum is not effective in treating leukemia. Ha: The new serum is effective in treating leukemia. III. Level of Significance: πΌ = 0.05 ππ = π1 πππ π2 = 10 πππ 8 π0.05 = 17 IV. Statistics: U-test. Wilcoxon rank-sum test. Computation: Arrange the data jointly from the lowest to the highest value and rank them. Then indicate the ranks in the table below.The sum of ranks with treatment will be π1 , similarly the sum of ranks without treatment will be π2 . With Treatment 6.5 5.4 5.3 7.2 4.5 8.4 6.8 7.3 6.7 8.2 Total Rank 4 3 2 7 1 10 6 8 5 9 π1 =55 π1 = 55 − No Treatment 15.2 18 13.6 12.2 13.1 12 17.6 19.4 Rank 15 17 14 12 13 11 16 18 π2 =116 10(10 + 1) =0 2 π2 = 116 − 8(8 + 1) = 80 2 π = min(0,80) = 0 V. If the computed U is less than the tabular U, reject Ho. VI. Since the smaller of U1=0 and U2=80, which is 0, is less than the tabular value of 17, at 0.05 level of significance with 10 and 8 degrees of freedom, hence the null hypothesis is rejected in favor of the research hypothesis. This means that the serum is effective in the treatment of Covid 19. It implies that patients with treatment of serum recover more rapidly than those patients without treatment of the serum because it takes them less number of days to recover. LESSON 15 - THE KRUSKAL-WALLIS TEST, ALSO CALLED THE KRUSKALWALLIS H-TEST. This test is used to compare 3 or more independent groups. This is a nonparametric test which does not require normal distribution. This is a counterpart of the F-test (ONE- WAY ANOVA) in parametric tests. It uses the table from ChiSquare test for the tabular values. The formula is: 12 π π2 π»= ∑ − 3(π + 1) π(π + 1) ππ Where: H=Kruskal Wallis test N= number of observations Example. Consider the examination scores of samples of College students taught in English using three different methods: Method 1 (Face-to-Face classroom teaching), Method 2 (On-line teaching), and Method 3 (Modular teaching). Use the H-test at 0.05 level of significance to test the null hypothesis that their means are not equal. Consider the following data: Method 1 96 86 91 93 89 87 Method 2 87 88 89 85 81 86 80 Method 3 88 76 74 64 79 Solving by Stepwise Method: I. II. III. Problem: Are there significant differences in the average scores of these students using the different methods of teaching English? Hypotheses: Ho: There are no significant differences in the average scores of students using 3 methods of teaching English. Ha: There are significant differences in the average scores of students using 3 methods of teaching English. Level of Significance: πΌ = 0.05 ππ = β − 1 = 3 − 1 = 2 2 π0.05 = 5.991 IV. Statistics: Kruskal Wallis, H-test Computation: Method 1 96 86 91 93 89 87 R1 18 8.5 16 17 14.5 10.5 n1=6 84.5 Method 2 87 88 89 85 81 86 80 n2=7 R2 10.5 12.5 14.5 7 6 8.5 5 64 Method 3 88 76 74 64 79 R3 12.5 3 2 1 4 n3=5 22.5 12 π π2 π»= ∑ − 3(π + 1) π(π + 1) ππ 12 84.52 642 22.52 π»= ( + + ) − 3(18 + 1) 18(18 + 1) 6 7 5 π» = 8.840 V. VI. Decision Rule: if the computed H is greater than the tabular value, reject Ho. Conclusion: Since the computed H of 8.840 is greater the tabular value of 5.991 at 0.05 level of significance with 2 degrees of freedom, the null hypothesis is rejected. This means that there are significant differences in the average scores of 18 students using three different Methods of teaching. It implies that Method 1 (face-to-face teaching) is more effective than the other two methods (on-line and modular teaching, respectively). Students’ Exercise: The data on war on rape cases under the Duterte administration committed from January to December in 3 cities in Metro Manila are as follows: Month January February March April May June July August September October November December City B 6 7 8 3 5 7 4 3 4 5 4 2 A 5 6 4 3 4 5 4 3 7 5 6 4 C 6 7 8 9 7 10 9 6 7 9 10 8 Perform Kruskal-Wallis H-test to determine if the hypothesis that the average rape cases in the 3 cities are significantly the same. Use πΌ = 0.05 level of significance. LESSON 16 - THE SPEARMAN RANK ORDER COEFFICIENT OF CORRELATION rs The Spearman Rank Order Coefficient of Correlation is denoted rs. This test of correlation does not require the strict assumption of normality like the Pearson Product Moment Coefficient of Correlation, denoted r. The formula is: ππ = 1 − 6 ∑ π·2 π(π2 − 1) Where: ππ = Spearman Rank Order Coefficient Correlation ∑ π· 2 = sum of the squares of the difference between rank x and rank y π = π πππππ π ππ§π Example. The following are the number of hours spent by 12 students studying for the final examination and the scores they obtained in Calculus. Calculate ππ at 0.05 level of significance. Number of Hours Spent x Midterm Scores y 6 5 10 11 2.75 3 2 2 12 18 15 20 8 9 14 1.75 1.5 1.5 1.25 2.25 2.25 1.25 12 2 Solving by Stepwise Method: I. II. III. IV. Problem: Is there a significant relationship between the number of hours spent for studying Calculus and the scores students obtained in the Final Examination? Hypotheses: Ho: There is no significant relationship between the number of hours spent for studying Calculus and the scores students obtained in the Final Examination. Ha: There is a significant relationship between the number of hours spent for studying Calculus and the scores students obtained in the Final Examination. Level of Significance: πΌ = 0.05 df = π − 1 = 12 − 1 = 11 ππ 0.05 = 0.532 Statistics: Spearman Rank Order Coefficient of Correlation, rs. Number Midterm of Scores Hours y Spent x 6 5 10 11 12 18 15 20 8 9 14 12 2.75 3 2 2 1.75 1.5 1.5 1.25 2.25 2.25 1.25 2 Rx Ry D D2 11 12 8 7 5.5 2 3 1 10 9 4 5.5 11 12 7 7 5 3.5 3.5 1.5 9.5 9.5 1.5 7 0 0 1 0 0.5 -1.5 -0.5 -0.5 0.5 -0.5 2.5 -1.5 0 0 1 0 0.25 2.25 0.25 0.25 0.25 0.25 6.25 2.25 ∑ π·2 = 13 ππ = 1 − 6 ππ = 1 − 6 V. VI. ∑ π·2 π(π2 − 1) 13 = 0.955 12(122 − 1) Decision Rule: If the computed rs is greater than the tabular value, reject Ho. Conclusion: The computed value of rs=0.955 being greater than the tabular value of 0.532 at 0.05 level of significance with 11 degrees of freedom, leads to the rejection of the null hypothesis. This indicates that there is a significant relationship between the number of hours spent for studying and the scores the students obtained in Calculus. It implies that the more students spend their time in studying Calculus, the higher the scores they obtain. Students’ Exercise: The following is the ranking of two judges given to the work of 10 artists. Using rs at 0.05 level of significance, test the hypothesis that the two judges differ most in their opinions about these artists. Judge A 6 Judge B 5 7 9 8 10 5 4 9 8 3 5 2 2 1 1 4 6 10 7 LESSON 17 - A SIGN TEST FOR TWO INDEPENDENT SAMPLES (MEDIAN TEST TWO-SAMPLE CASE) This test is known as the Median Test under nonparametric statistics. This test is used to compare the median of two independent samples. It’s the counter part of the t-test under parametric test, though what is being compared by t-test are the means of two independent groups or samples. The data consist of two independent samples of n1 and n2 observations. The procedure is to get the median of the data jointly. The data above this median are assigned a (+) sign, while those at or below this median a (-) sign. Then the number of + and – signs for each sample is obtained. A Chi-Square test is used to determine whether the observed frequencies of +and – signs differ significantly. π2 = π(ππ − ππ)2 ππππ Where: π 2 = πβπ − π ππ’πππ π‘ππ π‘ π = πππππ π‘ππ‘ππ π πππ π = π‘βπ πππ πππ£ππ (+)πππππ’ππππππ π πππ π = π‘βπ πππ πππ£ππ (−)πππππ’ππππππ π πππ π = π‘βπ πππ€ π‘ππ‘πππ π πππ π = π‘βπ ππππ’ππ π‘ππ‘πππ Example. Consider the IQ test scores of 12 female and 10 male students. Female 75 Male 68 98 87 87 93 65 54 102 85 95 93 110 78 132 106 127 115 96 96 106 118 Solving by Stepwise Method: I. II. Is there a significant difference between the IQ test scores of the male and female students? Hypotheses: Ho: There is no significant difference between the IQ test scores of the male and female students. III. IV. Ha: There is a significant difference between the IQ test scores of the male and female students. Level of Significance: πΌ = 0.05 ππ = (π − 1)(π − 1) ππ = (2 − 1)(2 − 1) = 1 2 π0.05 = 3.841 Statistics: Median Test for Two Independent Samples Computation: • • • Determine the median. Arrange the data from lowest to highest. The median is the middle item if there are odd number data, or the average of the two middle items if there are even number data. The median is 95.5. Mark or assign a plus (+) sign to those data above the median, while those at or below will be marked or assigned a minus (-) sign. Count each plus and minus sign under the female and male columns respectively. The observed frequencies are illustrated and summarized as follows: Female 75 98 87 65 102 95 110 132 127 96 106 118 Female Male Total • Sign + + + + + + + + Above (+) 8 (a) 3 (c) 11 (m) π2 = VI. At or Below (-) 4 (b) 7 (d) 11 (n) Sign + + + Total 12 (k) 10 (l) 22 (N) Use the formula for Chi-Square. π2 = V. Male 68 87 93 54 85 93 78 106 115 96 π(ππ − ππ)2 ππππ 22(8π₯7 − 4π₯3)2 12(10)(11)(11) = 2.933 Decision Rule: If the computed π 2 is greater than the tabular value, reject null hypothesis. Conclusion: Since the computed value π 2 of 2.933 is less than the tabular value π 2 of 3.841 at 0.05 level of significance with 1 degree of freedom, hence the null hypothesis is accepted. This means that there is no significant difference between the IQ scores of the female and male students. Exercise. Consider the test scores of 25 students in spelling. The students are composed of 15 females and 10 males. The following are their scores: Female 13 Male 9 14 16 17 12 24 15 16 8 15 9 12 17 16 11 22 24 18 16 25 23 14 19 20 LESSON 18 - A SIGN TEST FOR TWO CORRELATED SAMPLES (FISHER SIGN TEST) This test is under nonparametric statistics. This is the counterpart of the t-test for correlated sample under the parametric test. The Fisher Sign Test compares two correlated samples and is applicable to data composed of N paired observations. The difference between the paired observations is obtained. This test is based on the assumption that half the difference between the paired observations will be positive and the other half will be negative. The formula is: π= |π·| − 1 √π Where: Z = the Fisher Sign Test D = the difference between the number of + and – signs. Example. The pretest and the posttest results before and after the implementation of the program are presented below: Pretest x Posttest y 16 18 20 24 30 28 34 32 12 12 10 8 18 21 16 14 11 15 15 17 Solving by Stepwise Method: I. II. III. Problem: Is there a significant difference between the pretest and posttest results of the 10 students? Hypothesis: Ho: There is no significant difference between the pretest and posttest results of the 10 students. Ha: There is a significant difference between the pretest and posttest results of the 10 students. Level of Significance: πΌ = 0.05 π0.05 = ±1.96 IV. Statistics: Z-test (The Fisher Sign Test) Computation: Pretest x 16 20 30 34 12 10 18 16 11 15 Posttest y 18 24 28 32 12 8 21 18 15 17 Sign of x-y D + + 0 + - In this example, there are 3 + signs, 6 – signs, and 1 zero. Zero is omitted. Thus, π= π= |π·| − 1 √π |3 − 6| − 1 √9 π = 0.667 V. VI. Decision Rule: If Z-computed value is greater than the Z-tabular value, reject the null hypothesis. Conclusion: Inasmuch as the Z-computed value of 0.667 is less than the Z-tabular value of 1.96 at 0.05 level of significance, thus the null hypothesis is accepted. This means that there is no significant difference between the pretest and posttest results of the 10 students. Exercise: Perform a Fisher Sign Test for the example in t-test for correlated samples then compare the results. LESSON 19 - A SIGN TEST FOR K INDEPENDENT SAMPLES (THE MEDIAN TEST: MULTI-SAMPLE CASE) This test is under the nonparametric tests. This is a forthright extension of the median test for two independent samples. The Chi-Square test formula is used for this test. Example. A sampling of the acidity of rain for 10 randomly selected rainfalls was recorded at three different locations in the province of Albay: Guinubatan, Legaspi City, and Polangui. The pH readings for these 24 rainfalls are shown in the table. (Note: pH readings range from 0 to 14; 0 is acid, 14 is alkaline. Pure water falling through clean air has a pH reading of 5.7). Guinobatan 4.6 4.2 4.1 3.6 2.5 3.9 4.4 3.8 Legaspi City 4.5 4.8 4.5 3.5 2.8 4.3 4.9 3.5 Polangui 4.9 5.4 5.3 3.8 5.6 5.2 5.1 3.4 Use the Median test at 0.05 level of significance to test the hypothesis that there is no significant difference among the pH readings of the 3 different municipalities/cities of Albay. Solving by Stepwise Method: I. II. III. Problem: Is there a significant difference among the pH readings of the 3 different municipalities/cities of Albay. Hypotheses: Ho: There is no significant difference among the pH readings of the 3 different municipalities/cities of Albay. Ha: There is a significant difference among the pH readings of the 3 different municipalities/cities of Albay. Level of Significance: πΌ = 0.05 ππ = (π − 1)(π − 1) ππ = (3 − 1)(2 − 1) = 2 2 π0.05 = 5.991 IV. Statistics: Sign Test for K Independent Samples (Median Test: Multi-Sample Case) Computation: Guinobatan 4.6 4.2 4.1 3.6 2.5 3.9 4.4 3.8 Sign + + - Legaspi City 4.5 4.8 4.5 3.5 2.8 4.3 4.9 3.5 Sign + + + + + Polangui 4.9 5.4 5.3 3.8 5.6 5.2 5.1 3.4 Sign + + + + + + - The median is 4.35 (obtained using MS-excel by the median function). Above Md At or Less Total Guinobatan O E 2 4.33 6 3.67 8 8 Legaspi City O E 5 4.33 3 3.67 8 8 Polangui O E 6 4.33 2 3.67 8 8 Total 13 11 24 (π − πΈ)2 π =∑ πΈ 2 π 2 = 4.36 or p-value = 0.113 V. VI. Decision Rule: If the computed value is greater than the tabular value, reject the null hypothesis. Since the π 2 − πππππ’π‘ππ value of 4.36 is less than the π 2 − π‘πππ’πππ value of 5.991 at 0.05 level of significance with 2 degrees of freedom, the null hypothesis is accepted. This means that there is no significant difference among the pH readings of 3 different municipalities/cities of Albay. Note: Another way to arrive at this decision and conclusion is to use the pvalue. You can obtain this value using MS-Excel. It is compared to the level of significance used, usually, πΌ ππ 0.05 ππ 0.01. π − π£πππ’π =CHISQ.TEST(ofr,efr). Where: ofr = observed frequency range efr = expected frequency range CHISQ.TEST = the name of the function in MS-Excel When using the p-value, the decision rule is: If the p-value is less than or equal to the level of significance, reject the null hypothesis. Since the p-value of 0.113 is greater than 0.05 level of significance, the null hypothesis is accepted. As you can see, you arrived at same decision and interpretation/conclusion. LESSON 20 - THE MC NEMAR’S TEST FOR CORRELATED PROPORTIONS This test belongs to the nonparametric statistics which doesn’t require normal distribution of data. A Chi-Square test for the situations when samples are matched should not be independent. This is a before and after design to test whether there is a significant change between the before and after situations. The formula is: π2 = (π − π)2 π+π Where: π 2 = πΆβπ − πππ’πππ π = π‘βπ ππππ π‘ ππππ ππ π‘βπ 2ππ ππππ’ππ ππ π 2π₯2 π‘ππππ π = π‘βπ ππππ π‘ ππππ ππ π‘βπ 2ππ πππ€ ππ π 2π₯2 π‘ππππ Example. Data on seat belt use before and after involvement in car accidents for a sample of 120 accident victims. Wore seat belt regularly before the accident Wore seat belt regularly after the accident Total Yes a = 74 c = 23 97 86 34 120 Yes No Total No b = 12 d = 11 23 Solving by Stepwise Method: I. II. Problem: Is there a significant difference in the use of seat belt before and after involvement in the car accident? Hypotheses: Ho: There is no significant difference in the use of seat belt before and after involvement in the car accident. III. IV. Ha: There is a significant difference in the use of seat belt before and after involvement in the car accident. Level of Significance: πΌ = 0.05 ππ = (π − 1)(π − 1) = (2 − 1)(2 − 1) = 1 2 π0.05 = 3.841 Statistics: Mc Nemar’s test for correlated proportion Computation: π2 = V. VI. (π−π)2 π+π = (12−23)2 12+23 = 3.457 Decision Rule: If the computed π 2 is greater than the tabular π 2 , reject Ho. Conclusion: Since the computed π 2 of 3.457 is less than the tabular π 2 of 3.841 at 0.05 level of significance and 1 degree of freedom, the null hypothesis is accepted. This means that there is no significant difference in the use of seat belt before and after involvement in the car accident. This implies that, their involvement in the car accident did not change their attitudes towards wearing seat belts. LESSON 21 - THE FRIEDMAN Fr TEST FOR RANDOMIZED BLOCK DESIGN The Friedman test is a nonparametric statistical test developed by Milton Friedman. This test is similar to the parametric repeated measures ANOVA used for comparing the distributions of measurements for k treatments laid out in b blocks using randomized block design. The procedure for conducting the test is similar to that used for the Kruskal-Wallis H-test. When either the number of k treatments or the number of b blocks is larger the five, sampling distribution of Fr test can be approximated by a Chi-Square distribution with (k-1) df. The formula is: πΉπ = 12 ∑ ππ2 − 3π(π + 1) ππ(π + 1) Where: πΉπ = Friedman test π = ππ’ππππ ππ ππππππ π = ππ’ππππ ππ π‘ππππ‘ππππ‘π ππ = ππππ π π’π πππ π‘ππππ‘ππππ‘ π π = 1,2,3, … , π Example. In a study of the probability of a vaccine for children, 6 sample healthy children were used as subjects to assess their reaction to the taste of four vaccines. The children’s response was measured on a 10-point visual digital scale incorporating the use of faces, from sad (low score) to happy (high score). The minimum score was 0 and the maximum was 10. The following data were recorded: Vaccine Children 1 2 3 4 5 6 1 5.6 8.8 5.4 7.6 4.8 4.1 2 2.4 9.2 2.8 9.6 7.8 8.3 3 6.8 6.7 3.6 5.2 2.7 3.4 4 6.5 9.6 6.7 8.9 2.8 3.4 Solving by Stepwise Method: I. II. III. IV. Problem: Is there a significant difference in the reaction of 6 children on the 4 different vaccines? Hypotheses: Ho: There is no significant difference in the reaction of 6 children on the 4 different vaccines. Ha: There is a significant difference in the reaction of 6 children on the 4 different vaccines. Level of Significance: πΌ = 0.05 ππ = π − 1 = 4 − 1 = 3 2 π0.05 = 7.815 Statistics: Friedman Fr Test Computation: Vaccines Children 1 2 3 4 5 6 Rank Sum 1 Reaction (cm) 5.6 8.8 5.4 7.6 4.8 4.1 Rank 2 2 3 2 3 3 15 2 Reaction (cm) 2.4 9.2 2.8 9.6 7.8 8.3 Rank 1 3 1 4 4 4 17 3 Reaction (cm) 6.8 6.7 3.6 5.2 2.7 3.4 Rank 4 1 2 1 1 1.5 10.5 4 Reaction Rank (cm) 6.5 3 9.6 4 6.7 4 8.9 3 2.8 2 3.4 1.5 17.5 πΉπ = πΉπ = 12 ∑ ππ2 − 3π(π + 1) ππ(π + 1) 12 (152 + 172 + 10.52 + 17.52 ) − 3(6)(4 + 1) (6)(4)(4 + 1) πΉπ = 3.05 V. VI. Decision Rule: If the computed Fr is greater than the tabular Fr, reject Ho. Conclusion: Inasmuch as the πΉπ − πππππ’π‘ππ value of 3.05 is less than the πΉπ − π‘πππ’πππ value of 7.815 at 0.05 level of significance with 3 degrees of freedom, the null hypothesis of no significant difference in the reaction of 6 children on the 4 different vaccines was accepted. Exercise: In a study of an antibiotics for children, 5 sample healthy children were used as subjects to assess their reaction to the taste of four antibiotics. The children’s response was measured by an apps developed available in the Google playstore. The minimum score was 0 and the maximum was 10. The following data were recorded: Antibiotics Children 1 2 3 4 5 1 4.3 6.8 5.4 7.2 6.1 2 8.8 9.2 8.5 9.6 9.3 3 5.4 6.7 4.3 5.2 3.9 4 6.5 9.4 8.4 9.3 7.6 Use the Friedman Fr test to assess the children’s reaction on the 4 different antibiotics at 0.01 level of significance. LESSON 22 - THE KENDALL’S COEFFICIENT OF CONCORDANCE, π² This statistic is another nonparametric test used to determine if there is an agreement or concordance among the raters or judges of N objects or individuals. When π² is 1, it is interpreted as ‘High agreement’ and when π² is, ‘No agreement’. The formula is: 12(∑ π·2 ) π²= 2 π (π)(π 2 − 1) Where: π² = the coefficient of concordance π· = the difference between the individual sum of ranks of the raters or judges and the average of the sum of ranks of the object or individuals. ∑ π·2 = the sum of squares of the difference π = number of Judges or raters π = objects or individuals being rated or ranked Example. Below are the recorded ranks of the 4 judges to 10 contestants. Contestants 1 2 3 4 5 6 7 8 9 10 Judges A 1 4 3 5 2 8 7 10 9 6 B 2 3 4 5 1 7 8 10 9 6 C 3 2 5 4 1 6 7 9 10 8 D 2 4 3 5 1 7 6 10 9 8 Solving by Stepwise Method: I. II. III. Problem: Is there an agreement or concordance in the rankings of the 4 judges to the 10 contestants? Hypotheses: Ho: There is no agreement or concordance in the rankings of the 4 judges to the 10 contestants. Ha: There is an agreement or concordance in the rankings of the 4 judges to the 10 contestants. Level of Significance: πΌ = 0.05 ππ = π = 4 πππ π = 10 π²0.05 = 0.44 IV. Statistics: Kendall’s Coefficient of Concordance, π² Computation: Contestants A 1 4 3 5 2 8 7 10 9 6 1 2 3 4 5 6 7 8 9 10 Total Judges B C 2 3 3 2 4 5 5 4 1 1 7 6 8 7 10 9 9 10 6 8 π Μ = π²= V. VI. D 2 4 3 5 1 7 6 10 9 8 Sum of |π Μ -Sum of Ranks| Ranks D 9 13.1 13 9.1 15 9.1 19 3.1 5 17.1 28 5.9 28 5.9 39 16.9 37 14.9 28 5.9 221 D2 171.61 82.81 82.81 9.61 292.1 34.81 34.81 285.61 222.01 34.81 1250.99 221 = 22.1 10 12(1250.99) = 0.948 42 (10)(102 − 1) Decision Rule: If the computed π² is greater than the tabular π², reject Ho. Conclusion: The computed π² of 0.948 being greater than the tabular π² value of 0.44 at 0.05 level of significance with 4 and 10 degrees of freedom, the null hypothesis is accepted. This means that there is a significant agreement or concordance in the rankings of the 4 judges to the 10 contestants. Exercise. Three judges rank a group of 5 contestants in a beauty pageant. The following are the ranks of the judges: Students Judges A B C a 2 1 2 b 2 1 3 c 4 3 2 d 3 4 5 e 5 5 4 Module 1 - Additional Exercises 1.1 The performance ratings of 50 policemen in the capital town of NCR are shown below. Classes Frequency 70-74 5 75-79 11 80-84 13 85-89 15 90-94 6 Determine the value of the following measures: a) b) c) d) e) f) g) h) i) j) Mean π₯Μ Median π₯π Standard deviation SD 1st Quartile π1 3rd Quartile π3 10th Percentile π10 90th Percentile π90 Skewness SK Kurtosis Ku Tell and justify if the distribution of data is normal or abnormal based on the value of SK and Ku 1.2 Consider the performance rating of 65 college professors in TCU. Performance Rating Frequency 65-69 70-74 75-79 80-84 85-89 90-94 95-99 3 8 13 17 13 8 3 A. Determine the value of the following measures: 1. π₯Μ 2. Md 3. Q1 4. Q3 5. P10 6. P90 B. Find the value of the skewness and interpret the result. C. Find the value of the kurtosis and interpret the result. 1.3 In a PE class, 50 students were made to run in a 100-meter dash. The following results are presented in a frequency distribution. Seconds Frequency 10-11 2 12-13 8 14-15 14 16-17 16 18-19 8 20-21 2 a. Find the value of the kurtosis. Then interpret the result. b. Graph the data and tell whether it is leptokurtic, mesokurtic or platykurtic. Module 2 – Additional Exercises 2.1 During the lockdown due to pandemic, a certain livelihood program was given to 15 jeepney drivers to enhance their income. The data were recorded before and after the implementation as shown below. Income Before the Implementation After the Implementation (Php) (Php) 6,000 7,000 8,000 7,600 9,000 10,500 8,500 9,500 8,800 9,000 9,200 10,400 9,500 10,500 11,300 12,000 8,600 9,500 7,200 8,700 9,400 10,500 10,500 10,000 11,600 12,500 12,000 13,500 15,000 15,800 Use the t-test for correlated samples at .05 level of significance to test whether there were significant changes in the income of the jeepney drivers after the implementation of the program. 2.2 The data below represent the number of hours of pain relief provided by two brands of headache syrups administered to 20 individuals. These individuals were randomly divided into two groups and each group was rated with a different brand. Brand X Brand Y 6 6 8 7 9 5 4 3 3 4 7 6 8 5 6 3 5 6 8 6 Use the t-test at 0.05 level of significance to test the null hypothesis that there is no significant difference between the average number of hours of pain relief provided by the two brands of headache syrups. 2.3 The following data show the weight losses (in mg) of certain machine parts due to friction using two different lubricants. Lubricant π΄ Lubricant π΅ 10 7 13 6 12 6 11 5 13 7 14 10 7 9 10 4 8 11 11 6 12 13 Test at 0.01 level of significance whether the differences between the two-sample means are significant. 2.3 The table shows the number of errors made on 10 occasions by two compositors on their technical report. Is there a significant difference in the number of errors made in general by the two compositors? Use the t-test for independent samples at 0.05 level of significance. Composition 1 Composition 2 11 14 12 11 10 13 12 9 11 11 14 13 13 11 11 12 14 11 9 9 2.4 Ten samples were given an attitude test on a controversial issue. They were shown a favorable film regarding the issue and the same attitude test was administered before and after. Make a directional test at 0.05 level of significance. Pretest Posttest 15 19 19 21 15 24 23 27 21 22 24 28 23 24 17 23 16 19 14 15 2.5 A school administrator in a laboratory school claimed that the reading comprehension of grade 12 students in their school have an average of 82.3 with a standard deviation of 7.4. To test if the claim is valid, a researcher conducted a reading comprehension test for 60 randomly selected grade 12 students and found out that the average is 82.6. Use the Z-test at 0.05 level of significance to determine the validity of the claim. 2.6 A law student wants to confirm the claim of RTC Judge that convicted illegal gamblers spend on the average of 12.6 months in jail. A random sample of 50 cases from court files, showed that π₯Μ = 13.2 and the SD = 3.3 months. Use the Z-test to test the null hypothesis the π = 12.6 months against the alternative hypothesis of π ≠ 12.6 months at 0.05 level of significance. 2.7 Is there a significant difference between the average weight of females born from the two different countries. A random sampling yielded the following results: π1 = 100 π₯1 = 64.2 Μ Μ Μ ππ·1 = 2.48 π2 = 120 π₯1 = 63.8 Μ Μ Μ ππ·2 = 2.53 Use the Z-test at 0.05 level of significance. 2.8 A study was made to check whether the male average income is higher than the female average income. Use Z-test at 0.05 level of significance. Sex Number of Workers Male Female Mean 80 75 SD Php 1,218 Php 980 Php 42.30 Php 38.40 2.9 The table below shows the number of minutes that patients had to wait for their appointment with 5 doctors. Doctor A B C D E 21 19 22 31 29 10 12 16 13 19 19 18 17 16 21 10 12 29 30 16 20 31 25 27 26 Use the F-test at 0.05 level of significance to test if there are significant differences among the means of the samples. 2.10 The following are the data on homicide cases committed from January to December in 3 cities in the National Capital Region are as follows: City January A B C February March April May June July August September November December 4 5 3 6 4 7 8 5 3 2 5 7 6 7 3 2 5 3 7 3 6 5 5 7 9 8 10 5 6 8 Perform the one-way analysis of variance to test the null hypothesis that the average homicide cases in the 3 cities are equal at πΌ = 0.05 level of significance. 2.11 A research study was conducted on 4 groups of students. The following number of correct responses were recorded out of 10 trials. Apply the analysis of variance to find out if the groups differed significantly in their performance. Use πΌ = 0.05. Group Trial 1 2 3 4 5 6 7 8 9 10 A B C D 8 6 9 10 11 10 8 10 6 8 7 8 9 10 9 12 14 10 14 9 3 4 6 5 6 6 3 2 4 3 2 7 9 8 7 8 6 4 6 2 2.12 The following are the test scores of 20 students under two teachers and two kinds of modules used in teaching English. Apply two-way-ANOVA, at 0.05 level of significance. TEACHERS Textbooks A B A 5 6 3 2 4 9 8 3 8 9 B 6 9 3 2 1 7 8 8 9 5 Total Sub Total Total 2.13 Two sets of attitudinal scales were administered to two groups of students from private and public schools. Perform the two-way-ANOVA at 0.05 level of significance if there is significant difference between the two groups of students coming from two different schools and two groups of students given Set A and Set B on attitudinal scale? SCHOOLS Attitudinal Scale Set A Private Public 22 18 9 10 8 8 16 17 6 8 5 19 23 19 21 17 19 11 24 25 19 18 17 11 Sub Total Set B Total Total 2.14 The table below shows the percentage of the votes predicted by a poll survey A for 10 candidates for the senate on different cities and the percentage of the votes which they actually received B. Poll Survey A Actually Received B 42 47 32 38 26 35 39 51 19 25 48 50 37 38 25 34 47 53 31 27 Use the Pearson Product Moment Correlation Coefficient at 0.05 level of significance to determine if there is a significant relationship between the poll survey and the actual votes received. 2.15 The following are the scores of 10 students in the final examination in mathematics of investment (MOI) and accounting. MOI (x) Accounting (y) 81 64 65 74 75 76 77 80 83 88 84 69 66 64 77 81 76 84 96 84 Find the value of r and interpret the result at 0.05 level of significance. 2.16 A study was made by the SM Bicutan Hypermarket to determine the relationship between weekly sales and advertising expenditures. The following data were recorded. Use π at 0.05 level of significance. Adverting Cost (in thousand pesos) 5.0 2.5 3.0 2.5 5.5 6.0 2.0 3.5 2.5 3.0 Sales (in thousand pesos) 37.5 40.2 38.4 35.1 45.6 56.7 23.5 42.5 43.8 46.3 2.17 The following midterm grade, average grade in quizzes and final grade of 8 college students in Civil Engineering were obtained by a certain professor in TCU. Midterm Grade π₯1 3.0 1.75 1.5 1.25 2.0 2.75 2.5 1.75 Average Grade in Quizzes π₯2 Final Grade y 2.75 1.5 1.5 1.25 2.25 2.5 2.5 2.0 2.75 1.5 1.5 1.25 2.0 2.5 2.5 1.75 a. Use the method of least squares to fit an equation of the form π¦ = π0 + π1 π₯1 + π2 π₯2 b. Predict the final grade of student whose midterm grade is 1.5 and the average grade in the quizzes is 1.75. 2.18 The table below displays the area of the lot, the number of bedrooms, and the prices at which 10 one-family cottages sold at DMCI subdivision: Area in Square Meters π1 250 200 Number of Bedrooms π2 4 3 Prices y (In Million Php) 1.7 1.5 160 180 240 320 270 300 250 220 2 2 4 5 3 4 3 3 0.9 1.2 1.8 2.6 1.8 2.2 1.6 1.4 Determine a linear equation which will enable us to predict the average sale price of one - family cottage in terms of the area in square meters and the number of bedrooms. Module 3 – Additional Exercises 3.1 In 120 tosses of a coin, 64 heads and 56 tails are recorded. Is this a balanced coin? Use π 2 − π‘ππ π‘ at 0.05 level of significance. 3.2 The grades in Mathematics in the Modern World for a particular semester are as follows: Grades Observed 1.25 1.50 1.75 2.00 2.25 15 19 32 22 17 Use the π 2 − π‘ππ π‘ at 0.05 and test the hypothesis that the distribution of grades is uniform. 3.3 A random sample of 270 voters classified according to the political affiliation were asked if they were in favor of the ongoing peace negotiation in selected cities/towns in Mindanao. Political Affiliation Favor Not in Favor Total NP LP PDP-LABAN 38 53 54 52 37 36 90 90 90 Total 145 125 270 Use Χ 2 − π‘ππ π‘ at 0.05 level of significance to test the hypothesis that the sample belongs to the same population. 3.4 A random sample of 60 adults are grouped according to sex and their opinion regarding the ceasefire between the government and the CPP-NPA at Christmas. Ceasefire Sex Yes No Total Male Female 19 21 11 9 30 30 Total 40 20 60 Use Χ 2 − π‘ππ π‘ at 0.05 level of significance to determine if there is no significant difference between the opinion of male and female adults on the issue of ceasefire between the government and the CPP-NPA. 3.5 A random sample of 400 adults are classified according to their age bracket and drinking habits. Drinking Habits Nondrinkers Moderate drinkers Heavy Drinkers 20-29 30-39 28 33 53 40-above 75 74 63 19 38 17 Test the hypothesis that age bracket is dependent of drinking habit. Use π 2 − π‘ππ π‘ at 0.05 level of significance. 3.6 The following data were taken from 200 individuals in a study to determine the dependence of lung cancer and smoking habits, Nonsmokers Moderate Smokers Heavy Smokers With Cancer No Cancer 26 49 48 21 46 10 Total 75 69 56 Test the hypothesis that the lung cancer is independent of smoking habits. Use 0.05 level of significant. 3.7 From 18 students of Math class, 9 students are selected at random and given additional instruction by the teacher. The rest of the students were given no additional instruction. The results on the final examination were as follows: Grades in the Final Examination With Additional Instruction 86 91 85 82 85 86 80 84 86 No Additional Instruction 76 80 80 84 73 82 84 78 82 Use the Wilcoxon rank-sum test at 0.05 level of significance if the additional instruction affects the average grade. 3.8 Electric cables are being manufactured by two companies. To determine if there is a difference in the mean breaking strength of the cables, 8 pieces from each company are selected at random and tested for breaking strength. The results are: Company A Company B 10.5 11.2 8.9 8.6 10.7 9.6 9.8 7.5 11.4 9.6 11.5 10.8 12.6 11.6 9.7 7.4 Use the Wilcoxon rank-sum test at 0.05 level of significance if there is a difference in the mean breaking strength of the cables manufactured by two companies. 3.9 Random sample of 3 brands of cigarettes were tested for nicotine content. The following figures show the milligrams of nicotine found in the 15 cigarettes tested. Brand A B C 16 15 14 17 13 18 17 19 20 21 12 11 10 9 11 Use the Kruskal-Wallis test, at the 0.05 level of significance, to test whether there is a significant difference in nicotine content among the 3 brands of cigarettes. 3.10 The following data represent the operating time in hours for 4 Brands of midrange cellphones before a recharge is required. Brand Samsung Oppo Vivo Xiaomi 12.8 13.2 13.6 13.4 13.3 13 11.6 12.7 12.4 12.5 12.5 12.4 12.8 12.6 12.2 11.6 12.4 12.6 11.8 12.2 Use the Kruskal-Wallis test, at the 0.01 level of significance to test the hypothesis that the operating times for all 4 brands of cellphones are equal. 3.11 Two judges of a city fiesta parade in NCR ranked 10 floats in the following order: Judge A Judge B 8 5 10 9 3 4 6 7 2 1 5 7 9 5 2 4 8 6 1 3 Use Spearman’s Rank correlation coefficient to test if there is a significant correlation in the ranking of two judges. 3.12 Two groups of experimental dogs are given two brands of vitamins and the following weight gains in grams were obtained. Group 1 Group 2 250 67 206 75 106 138 280 180 212 193 106 48 112 78 142 82 168 67 185 88 Apply the sign test to determine if the two samples come from population with the same median. 3.13 The following are the scores of 10 male and 10 female students in Psychological laboratory examination. Male 22 Female 46 83 94 51 62 67 75 74 56 79 42 46 84 20 92 37 49 69 96 Test the significant difference between the scores of the male and female students using sign test. 3.14 The following are the weights of 15 women enrolled in a 10-week slimming program. Their weights were taken before and after the program. Before After 120 130 140 125 140 115 135 146 141 130 126 114 115 120 136 120 125 115 126 145 100 120 126 134 129 124 112 106 112 121 Use the sign test at 0.05 level of significance to test if there is a significant difference in the weights of women enrolled in the programs before and after. 3.15 The following are the scores obtained by the groups of 6 subjects each given with 3 different methods of teaching in Algebra. Method 1 Method 2 Method 3 38 37 30 40 36 34 40 38 50 46 32 48 70 40 70 69 50 39 Apply a median test at 0.05 level of significance. 3.16 The following data were obtained before and after a televised debate on charter change for a sample of 60 registered voters. After the Debate Before the Debate Yes No Total Yes 20 15 35 No 12 13 25 Total 32 28 60 Apply the Mc Nemar’s test at 0.05 level of significance. 3.17 The following data were recorded for 6 subjects exposed to 4 different treatments. Treatments Subjects T1 T2 T3 T4 1 2 3 4 5 6 10 9 8 10 5 9 10 5 7 9 6 9 4 6 9 8 3 8 6 3 10 7 4 7 Apply the Friedman rank test at 0.05 level of significance. 3.18 The following data were obtained for 5 subjects repeatedly measured under 3 different conditions. Condition Subject 1 2 3 1 2 3 4 5 11 29 22 3 13 14 34 30 5 11 13 41 18 17 16 Use the Friedman rank test at 0.05 level of significance. 3.19 Three judges ranked 5 contestants in a beauty contest. The following were obtained: Students Judge A B C D E X 2 3 1 4 5 Y 2 1 4 3 5 Z 1 3 2 5 4 Compute the coefficient of concordance π at 0.05 level of significance to determine if there is an agreement among the ranks of the three judges on the 5 students . Works Cited (n.d.). Retrieved from https://www.sciencedirect.com>topics Bayanito, M. R. (2015). The Contemporary World. Taguid City: National Book Store. Bibliography (n.d.). Retrieved from https://www.sciencedirect.com>topics Broto, A. S. (2006). Statistics Made Simple. Mandaluyong City: National Book Store. Garcia, G. A. (2003). Fundamental Concepts and Methods in STATISTICS (Part 1). Manila, Philippines: University of Santo Tomas Publishing House.