The following table contains information on matched sample v

The following table contains information on matched sample values whose differences are normally distributed. Number: 1 2 3 4 5 6 7 8 9 Sample 1: 103 143 132 118 132 162 142 128 172 Sample 2:121 132 157 127 156 172 117 143 179. a. Enter the data in Excel and calculate the differences and then calculate the sample mean and the sample standard deviation of the differences. Your calculation table is similar to the table on page 157 of the course packet or Table 10.4 on page 330 of the textbook. b. Construct a 95% confidence interval for the mean difference. c. Specify the competing hypotheses in order to test whether the mean difference differs from zero. Use the results in part a to conduct the test at the 5% significance level with the critical value approach. d. Use Excel’s Data Analysis to get the test results for part c. In Data Analysis, you use “tTest: Paired Two Sample for Means” to get the results. e. Use the printout from part d to conduct the test in part c with the pvalue approach at the 5% significance level.

Solution

PART A)

SAMPLE 1   SAMPLE 2   DIFFERENCE
103 121   -18
143 132   11
132 157   -25
118 127   -9
132 156   -24
162 172   -10
142 117   25
128 143   -15
172 179   -7

Variable Mean StDev
DIFFERENCE -8.00 16.39

PART B)

95% CI for difference: (-29.7, 13.7)

PART C)

T-Test of H0: difference = 0 vs H1: not =0

T-Value = -0.78

CRITICAL VALUE=  2.11991

so at 5% level of significance we fail to reject H0