Sample Test 1 Solutions - Math 1342 Each question is worth 7 points. Your results, plots and comments will be typed into Microsoft Word and then either submitted to the D2L dropbox as a Word File or as a pdf file. The following data shows the number of grams per serving of 30 different cakes 32 47 51 41 46 30 46 38 34 34 52 48 48 38 43 41 21 24 25 29 33 45 51 32 32 27 23 23 34 35 1. Enter the data into StatCrunch and create a summary statistics table. Remember your data must go down column 1 (var1) Paste your table into your Word document. You can use the default selection that StatCrunch uses but also include the IQR as well. ANS. Summary statistics: Column n var1 Mean Variance Std. dev. Median Range Min Max Q1 Q3 IQR 30 36.766667 87.909195 9.3759904 34.5 31 21 52 30 46 16 2. Use StatCrunch to create a Stem and Leaf plot and paste it into your Word document. ANS. Decimal point is 1 digit(s) to the right of the colon. 2 : 1334 2 : 579 3 : 02223444 3 : 588 4 : 113 4 : 566788 5 : 112 Or shortened 2 : 1334579 3 : 02223444588 4 : 113566788 5 : 112 1|P age 3. Looking at the Summary Statistics and the Stem and Leaf plot why does the median have a fractional part and not a whole number when all of the numbers in the data are whole numbers? What are the two numbers on each side of the median? ANS. The data has an even number of data items and therefore the middle 2 numbers were averaged. The two numbers that were averaged are 34 and 35 4. Use StatCrunch to create a Box Plot which goes side to side (not up and down) and paste it into your Word document. Is the median equally spaced between Q1 and Q3? If not which one is it closer to? ANS. The Box Plot is shown below and the median is shifted toward Q1. 2|P age 5. Use your IQR to determine if the data has medium or extreme outliers. See page 176 question 18 in the textbook. ANS. From StatCrunch Min Max Q1 Q3 IQR 21 52 30 46 16 1.5 * IQR = 1.5 * 16 = 24 Q1 – (1.5 * IQR) = 30 – 24 = 6 Q3 + (1.5 * IQR) = 46 + 24 = 70 So if there were any outliers they would be outside of the range of 6 to 70. Our Min = 21 which is not lower than 6 and Our Max = 52 which is not higher than 70 so our data has NO outliers. 6. Create a Grouped Frequency Table from the data using 5 classes. Make sure the width is an odd number and complete the following table. Examples -> 4.3 rounds to w = 5, 7.3 rounds to w = 9 Column n Mean Variance Std. Dev. Std. Err. var1 30 36.766666 87.909195 9.37599 1.7118138 Median Range Min Max Q1 Q3 IQR 34.5 31 21 52 30 46 16 ANS. Range = 31 Width = 31/5 = 6.2 and round up to the closest odd number which is 7 Class 1 ->>> 21 up to 21 + (7 - 1) = 21 + 6 = 27 Class 2 up by 1 ->>> 28 up to 28 + 6 = 34 Class Limits w = 6 Class Boundaries True w = 7 21 to 27 28 to 34 35 to 41 42 to 48 49 to 55 20.5 to 27.5 27.5 to 34.5 34.5 to 41.5 41.5 to 48.5 48.5 to 55.5 3|P age 2 : 1334579 3 : 02223444588 4 : 113566788 5 : 112 Class 1 2 3 4 5 Class Limits Class Boundaries Frequency Midpoints Freq * Midpoints Freq *Mid^2 21 to 27 20.5 to 27.5 6 24 144 28 to 34 35 to 41 27.5 to 34.5 34.5 to 41.5 9 5 31 38 279 190 42 to 48 41.5 to 48.5 7 45 315 49 to 55 48.5 to 55.5 3 52 156 Sums 30 3456 8649 7220 14175 8112 41612 1084 7. Use the Grouped Frequency Table to calculate the grouped mean and compare this number with the mean from your StatCrunch summary of the ungrouped data. Are the results the same or are they different? Which answer do you think is more accurate and why? ANS. Grouped Mean = Sum of Column of Freq * Midpoints / Sample Size = 1084 / 30 = 36.13 The mean from the original ungrouped data is 36.76 and the grouped mean is 36.13. The numbers are not the same and the reason that they are not is that the grouped table has lost some of the precision due to the midpoints being used to represent the data. 8. Use your Grouped Frequency Table to calculate the grouped variance and the grouped standard deviation. ANS. pg 139 n * sum(freq * midpoints^2) = 30 * 41612 = 1248360 ( sum( freq * midpoints) )^2 = 1084^2 = 1175056 n * (n - 1) = 30 * 29 = 870 s^2 = grouped variance = (1248360 – 1175056) / 870 = 84.257 s = grouped std = sqrt(84.257) = 9.179 4|P age 9. Use StatCrunch and your Grouped Frequency Table to make a Histogram. Paste the graph into your Word document. ANS. 5|P age 10. Use the data shown below and StatCrunch to create a Pie Chart and also a Pareto Chart. Paste the graphs into your Word document. Favorite Colors Red Blue Green Yellow Orange Frequency 5 10 14 3 12 ANS. 6|P age 11. Use the data: 5, 8, 12, 15, 9, 11, 13, 20, 8, 10, 9, 8, 6, 20, 8 a. Calculate the z Score for 11? Show your work. Hint: Use StatCrunch to help. b. Calculate the z Score for 8? Show your work. Hint: Use StatCrunch to help. ANS. First use StatCrunch with the data to get the Summary Summary statistics: Column n Mean Variance Std. Dev. Median Range Min Max Q1 Q3 var1 15 10.8 20.6 4.5387225 9 15 5 20 8 13 a. z score for 11 = (value – mean)/std = (11 – 10.8)/4.538 = 0.0441 b. z score for 8 = ( 8 – 10.8)/4.538 = -0.6170 12. Use the data: 5, 8, 12, 15, 9, 11, 13, 20, 8, 10, 9, 8, 6, 20, 8 a. Calculate the percentile rank for 12? Show your work. b. Determine the value that corresponds to the 30th percentile? Show your work. ANS. We need the data sorted Data Position 5 1 6 2 8 3 8 4 8 5 8 6 9 7 9 8 10 9 11 10 12 11 13 12 15 13 20 14 20 15 a. Percentile Rank (page 155) How many numbers are under 12? 10 of them Percentile = (10 + 0.5) / 15 * 100 = 10.5/15 * 100 = 70 th 7|P age b. 30th Percentile (page 157) c = (15 * 30) / 100 = 4.5 Since this is not a whole number round up to next whole number = 5 5 is NOT the answer, 5 is the position where the 30 th percentile is at. Position 5 number is 8. Data Position 5 1 6 2 8 3 8 4 8 5 8 6 9 7 9 8 10 9 11 10 12 11 13 12 15 13 20 14 20 15 13. Chebyshev’s Theorem - If the mean is 10 and the standard deviation is 4 what is the predicted percentage of values that will live in the interval (1, 19)? ANS. From the theory we know the interval is always ( mean – k * standard deviation, mean + k * standard deviation ) So matching mean = 10, standard deviation = 4, k = unknown ( 10 – 4k, 10 + 4k ) = (1 , 19) So 10 – 4k = 1 and solve for k -4k = -9 k = 9/4 = 2.25 The proportion is 1 - 1/k^2 = 1 – 1/(2.25^2) = 1 - 0.1975 = 0.8025 So about 80.25% of the data will live in the interval ( 1, 19) 8|P age 14. For the following data use StatCrunch to make a Scatter Plot. Paste the graph into your Word document. Age, X 34 22 46 56 62 Hours, y 5.5 7 3.5 3 1 ANS. 9|P age