AP Statistics Page 1 of 4 Practice: t Intervals and Hypothesis Tests With Matched Pairs Data Answers

1. A. Answer: Yes, this is an appropriate situation for a matched pairs analysis. The observations are paired on the students' grade point averages, class (freshman, sophomore, junior, senior), and after-school activities. B. Answer: No, this is not an appropriate situation for a matched pairs analysis. There is no natural pairing of individual measurements. C. Answer: No, this is not an appropriate situation for a matched pairs analysis. There is no natural pairing of individual measurements. D. Answer: Yes, this is an appropriate situation for a matched pairs analysis. The outcome of the first sample is related to the outcome of the second (the after weight is going to depend on what the before weight was). The data can be analyzed as a single sample representing the differences within the pairs. 2. A. Answer:

( )ABH −µ:0 = 0

(There is no difference between the final exam scores between students in curriculum A and students in curriculum B, where µ is the true population mean difference.)

( )ABaH −µ: ≠ 0 (There is a difference between the final exam scores between students in

curriculum A and students in curriculum B, where µ is the true population mean difference.)

AP Statistics Page 2 of 4 Practice: t Intervals and Hypothesis Tests With Matched Pairs Data Answers

B. Answer: The next step in analyzing these data is to compute the differences between the two scores.

Student Curriculum

Pair A B B � A 1 83 90 7 2 90 94 4 3 78 75 �3 4 82 78 �4 5 80 72 �8 6 93 94 1 7 90 96 6 8 84 80 �4 9 83 84 1 10 86 88 2 11 79 82 3 12 96 96 0 13 93 90 �3 14 86 84 �2 15 79 83 4 16 83 84 1 17 83 84 1 18 85 87 2 19 80 85 5 20 76 79 3

Now, determine if you can use a t procedure by seeing if the data are severely skewed or have outliers. A box-and-whisker plot of the differences shows no outliers and no severe skewness. Also, the mean (.8) and the median (1) of the differences are close to one another. So you can assume that the data are close enough to normal to use a t procedure. C. Answer: To test for a difference between the two curricula, construct a confidence interval for the true difference between the final scores in the population.

x = .8, s = 3.82, n = 20, t = = 20/82.3

08. −=

8542.8.

= .9366.

Since this is a two sided test, P = 2*tcdf(.9366,E99,19) = .3607. This P-value is greater than alpha = .05 and we cannot reject the null hypothesis that there is no difference between the means.

AP Statistics Page 3 of 4 Practice: t Intervals and Hypothesis Tests With Matched Pairs Data Answers

D. Answer: We cannot reject the null hypothesis that there is no difference between the curricula because 0 is contained in the 95% confidence interval for the true population difference. 3. A. Answer:

( )BAH −µ:0 = 0 (There is no difference between the before and after weights, where µ

is the true population mean difference.) ( )BAaH −µ: < 0 (The difference between the before and after weights is less than zero

(the after weight is less than the before weight), where µ is the true population mean

difference.) B. Answer:

Weight (in lbs.) Before After (After � Before) 185 180 �5 187 183 �4 195 192 �3 176 179 3 193 193 0 206 208 2 200 190 �10 189 182 �7 179 183 4 193 187 �6 196 196 0 220 218 �2 215 212 �3 220 212 �8 196 191 �5 179 171 �8 188 182 �6 192 194 2 190 181 �9 225 215 �10

Now, determine if you can use a t procedure by seeing if the data are severely skewed or have outliers. A box-and whisker plot of the differences shows no outliers and no severe skewness. Also, the mean (�3.75) and the median (�4.5) are relatively close to one another. So you can assume that the data are close enough to normal to use a t procedure. C. Answer: We would use a t statistic because we do not know the true population standard deviation and the sample does not show any outliers or skewness so that we assume that the data are close enough to normal.

AP Statistics Page 4 of 4 Practice: t Intervals and Hypothesis Tests With Matched Pairs Data Answers

D. Answer: t = �3.769 p = .00065 E. Answer: We have enough evidence to reject the null hypothesis that there is no difference between before and after weights because p = .0006 and α = .05. Therefore, there is a difference between before and after weights.