HYPOTHESIS TESTING
INFERENTIAL STATISTICS
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enables us to make estimates of population values called parameters and to make statements about
computed statistics acceptable to some degree of confidence
Two areas of inferential statistics
1. estimation
2. hypothesis testing
Research Problem:
e.g. How effective is Minoxidil in treating male pattern baldness?
Specific Objectives:
1. To estimate the population proportion of patients who will show new hair growth after being treated with
Minoxidil. (Can be answered by estimation)
2.To determine whether treatment using Minoxidil is better than the existing treatment that is known to
stimulate hair growth among 40% of patients with male pattern baldness. (Answered by hypothesis testing)
Hypothesis testing is a procedure on sample evidence and probability, used to test claims regarding a
characteristic of one or more populations.
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Hypothesis testing evaluates a conjecture about some characteristic of the parent population based
upon the information contained in the random sample.
Usually, the conjecture concerns one of the unknown parameters of the population, μ (mean), σ2
(variance), σ (standard deviation).
Examples of parameters are population mean and population proportion
The parameter must be identified before analysis
HYPOTHESIS

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A statement or claim regarding a characteristic of one or more populations.
A preconceived idea, assumed to be true but has to be tested for its truth or falsity
Procedures for Testing Hypothesis
1. State the null and alternative hypothesis.
2. Set the level of significance or alpha level (α).
3. Determine the test distribution to use.
4. Calculate test statistic or p - value.
5. Make statistical Decision 6. Draw Conclusion
COMPONENTS OF A FORMAL HYPOTHESIS TEST
Null Hypothesis (none, empty)
 denoted by Ho
 the statement being tested
 it represents what the experimenter doubts to be true
 must contain the condition of equality and must be written with the symbol =, ≤ or ≥.
 Mean of the population is = to the mean of sample.
Example of NULL HYPOTHESIS:
Ho: μ = some value
The null hypothesis corresponding to the common belief that the mean body temperature is 37˚C is expressed
as
Ho:
μ
=
37˚C,
The
mean
body
temperature
is
equal
to
37˚C.
Note: The hypothesized value can be obtained from the previous studies or from knowledge of the population.
Alternative Hypothesis
 denoted by Ha
 Is the statement that must be true if the null hypothesis is false
 the operational statement of the theory that the experimenter believes to be true and wishes to prove
 Is sometimes referred to as the research hypothesis
 Must contain the condition of equality and must be written with the symbol ≠, < or >.
For the mean, the alternative hypothesis will be stated in only one of three possible forms:
 Ha: μ ≠ some value (significant difference)
 Ha: μ > some value (there is a significant increased)
 Ha: μ < some value (there is a significant decrease)
Note: Ha is the opposite of Ho.
For example, if Ho is given as µ= 37.0, then it follows that the alternative hypothesis is given by any of the ff:
 Ha: µ ≠ 37.0
 Ha: µ > 37.0
 Ha: µ < 37.0
WHAT IS A TEST OF SIGNIFICANCE?
 A test of significance is a problem of deciding between the null and the alternative hypotheses on the
basis of the information contained in a random sample.
 The goal will be to reject Ho in favor of Ha , because the alternative is the hypothesis that the
researcher believes to be true. If we are successful in rejecting Ho, we then declare the results to be
“significant”
 Test of difference
 Test of effect
 Test of relationship
TWO TYPES OF ERRORS
Type I Error
 The mistake (error) of rejecting the null hypothesis when it is true. (Sinabing may significant difference
kahit wala naman)
 It is not a miscalculation or procedural misstep; it is an actual error that can occur when a rare event
happens by chance.
 The probability of rejecting the null hypothesis when it is true is called the significance level (α).
 The value of α is typically predetermined, and very common choices are α = 0.05 and α = 0.01
Example of Type I Error
The mistake of rejecting the null hypothesis that the mean body temperature is 37.0 when that mean is really
37.0.
Example of situation resulting from Type I Error
 BFaD allows the release of an ineffective medicine
Type II Error
 The mistake of failing to reject the null hypothesis when it is false. (Sinabing walang significant
difference kahit meron).
 The symbol (beta) is used to represent the probability of a type II error
Example Type II Error
The mistake of failing to reject the null hypothesis (μ = 37.0) when it is actually false (that is, the mean is not
37.0).
Example of situation resulting from Type II Error
 BFaD does not allow the release of an effective drug
Decision
REJECT Ho
RETAIN Ho
Ho is actually
TRUE
Type I Error (False Positive)
Correct
FALSE
Correct
Type II Error (False Negative)
Ho: The defendant is innocent.
Ha: The defendant is not innocent.
What happen to the defendant if the injury made type I and Type II error?
Answer:
A type I error is putting an innocent person in jail.
A type II error is like letting a guilty person go free.
TYPES OF TESTS
1. Two-tailed test
If we are primarily concerned with deciding whether the true value of a population parameter is
different from a specified value, then the test should be two-tailed.
 Alternative hypothesis represents as NOT EQUAL, both sides will be test (positive or
negative)
2. Left-tailed test
 If we are primarily concerned with deciding whether the true value of a parameter is less than
a specified value, then the test should be left-tailed.
 Alternative hypothesis represents as LESS THAN, left side will be test (with lower value)
3. Right-tailed test
 If we are primarily concerned with deciding whether the true value of a parameter is greater
than a specified value, then we should use the right-tailed test.
 Alternative hypothesis represents as GREATER THAN, right side will be test (with greater
value)
P-VALUE (T-TABLE)
 This is the smallest level of significance at which Ho will be rejected based on the information
contained in the sample (Point of rejection)
 It is commonly generated by the statistical software.
Decision rule: Reject Ho if the p-value is less than or equal to the level of significance (α)
EXAMPLE OF MAKING DECISIONS USING THE P-VALUE
 If the level of significance α = 0.05 and p-value is 0.01, reject Ho.
 If the level of significance α = 0.05 and p-value is 0.05, reject Ho.
 If the level of significance α = 0.05 and p-value is 0.10, do not reject Ho
SUMMARY OF THE STEPS IN HYPOTHESIS TESTING
1. State the null and alternative hypotheses.
2. Decide on a level of significance, Determine the testing procedure and methods of analysis (responsibility
of the statistician).
3. Decide on the type of data to be collected and choose an appropriate test statistic and testing procedure.
4. State the decision rule. Reject Ho if p-value is less than or equal to level of significance (alpha).
5. Collect the data and do the test of significance.
6. Determine the p-value. If p-value is less than or equal to alpha, reject Ho.
7. Write the conclusion and interpret the results.
ONE SAMPLE Z-TEST
 A one sample z-test is used to test whether a population parameter is significantly different from some
hypothesized value assuming that the entries are normally distributed and variable is known.
Z = (x̅ – μ0) / σ /√n
x̅= mean of population,
μ0= mean of sample
n= sample size
It depends on the number of sample size.
*If hindi nakalagay sa Df ang sample size; z-test
gagamitin
*LNF ang gagamitin na numbers
1. Ho: The average grade of Professor J’s student is equal to 80%.
Ho: µ=80%
Ha: The average grade of Professor J’s student is greater than 80%.
Ha: µ > 80% *atleast
2. α = 0.05, right tailed test
Z = (x̅ – μ0) / σ /√n,
3. 𝑧0.05 = 1.645, n=100, one tailed test (0.05, LMF= 1.645)
4. Reject Ho if 𝑧𝑐 >𝑧𝛼 . Otherwise accept the null hypothesis
5. Zc = (85 – 80) / 16 /√100
6. 𝑧𝑐 = 3.125 > 𝑧0.05 = 1.645 reject the null hypothesis in favor of the
alternative hypothesis
7. We conclude that the claim of professor j the average grade of his
students is greter than at 80% at 5% level of significance.
1. Ho: The average BMI of the female high school students is equal to 19
points.
Ho: µ = 19 points
Ha: The average grade of Professor J’s student is greater than 80%.
Ha: µ < 19 points
2. α = 0.05, left tailed test,
Z = (x̅ – μ0) / σ /√n,
3. 𝑧0.05 = −1.711, n=25-1, one tailed test
4. Reject Ho if 𝑡𝑐 < 𝑧𝛼 . Otherwise accept the null hypothesis
5. Tc = (18 – 19) / 4.5 /√25 = -1.11
6. 𝑇𝑐 = −1.11 > 𝑇0.05,24 = −1.711 accept the null hypothesis
7. We conclude that the average BMI of the female high school students
is about 19 points.
*sample size is in the degree of
freedom,
1. Ho: The mean weight of the sample of 100 persons from the Honolulu Heart Study is equal to 60kg.
Ho: µ = 60kg
2. Ha: The mean weight of the sample of 100 persons from the Honolulu Heart Study is greater than 60kg
Ha: µ > 60kg
3. α = 0.05, right tailed test,
Z = (x̅ – μ0) / σ /√n,
4. 𝑧0.05 = 1.645
5. Reject Ho if 𝑍𝑐 > 𝑧𝛼 . Otherwise accept the null hypothesis
6. Zc = (63 – 60) / 10 /√100 = 3
7. 𝑍𝑐 = 3 > 𝑇𝑍0.05, = 1.645 reject the null hypothesis
8. We conclude that the mean weight of the sample of 100 persons from the Honolulu Heart Study is greater than
60kg at 5% level of significance.