Suppose you want to determine whether the average values for populations 1 and 2

Question

Suppose you want to determine whether the average values for populations 1 and 2 are different, and you randomly gather the following data. Use these data to construct a 98% confidence interval for the difference in the two population means. Sample 1: 2, 10, 7, 8, 2, 5, 9, 1, 8, 0, 2, 8, 11, 2, 4, 5, 3, 9 Sample 2: 10, 12, 8, 7, 9, 11, 9, 8, 9, 10, 11, 10, 11, 10, 7, 8, 10, 10

Choose the correct answer. a. negative 6.2799 less-than-or-equal-to mu 1 minus mu 2 less-than-or-equal-to negative 1.9421

b. negative 3.6347 less-than-or-equal-to mu 1 minus mu 2 less-than-or-equal-to negative 2.8093

c. negative 4.5367 less-than-or-equal-to mu 1 minus mu 2 less-than-or-equal-to negative 3.9073

d. negative 7.164 less-than-or-equal-to mu 1 minus mu 2 less-than-or-equal-to negative 1.280

e. negative 4.3749 less-than-or-equal-to mu 1 minus mu 2 less-than-or-equal-to negative 2.0691

Explanation / Answer

Two-Sample T-Test and CI: sample1, sample2

Two-sample T for sample1 vs sample2

N Mean StDev SE Mean
sample1 17 5.18 3.50 0.85
sample2 18 9.44 1.42 0.34

Difference = mu (sample1) - mu (sample2)
Estimate for difference: -4.268
98% CI for difference: (-6.578, -1.958)
T-Test of difference = 0 (vs not =): T-Value = -4.67 P-Value = 0.000 DF = 20

Please check your data. Same data I have used below

Answere approximately matched with a. negative 6.2799 less-than-or-equal-to mu 1 minus mu 2 less-than-or-equal-to negative 1.9421

sample1 sample2 2 10 10 12 7 8 8 7 2 9 5 11 9 9 1 8 0 9 2 10 8 11 11 10 2 11 4 10 5 7 3 8 9 10 10

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