Question

To perform a test of the null and alternative hypotheses shown below, random samples were selected from the two normally distributed populations with equal variances. The data are shown below. Test the null hypothesis using an alpha level equal to 0.10.

Sample from Population 1: 38,28,28,39,39,33,29,37,43,38

Sample from Population 2: 45,53,37,47,44,38,43,46,46,41

H0: ?1 ? ?2 = 0

HA: ?1 – ? ? 0

Determine the rejection region for the test statistic t. Select the correct choice below and fill in the answer box(es) to complete your choice.

A. t < ____ or t > ____

B. t > ___

C. t < ___

Calculate the value of the test statistic. t = ___

Since the test statistic ____ (is not, or is) in the rejection region, _____ (reject, or do not reject) the null hypothesis. There is ____ (sufficient, or insufficient) evidence to conclude that the two population means are different.

show formulas in excel

Answer 1

go to

data -> data analysis -> t-test with equal variance

df = n1 +n2-2 = 18

alpha = 0.10

critical values ==T.INV.2T(0.1,18)

= 1.73406 and -1.73406

A) t > 1.73406 or t < -1.73406

Two-sample T for C9 vs C10

      N   Mean StDev SE Mean
C9   10 35.20   5.33      1.7
C10 10 44.00   4.64      1.5

Difference = ? (C9) - ? (C10)
Estimate for difference: -8.80
95% CI for difference: (-13.50, -4.10)
T-Test of difference = 0 (vs ?): T-Value = -3.94 P-Value = 0.001 DF = 18
Both use Pooled StDev = 4.9978

t = -3.94

Since the test statistic is in the rejection region, reject the null hypothesis. There is sufficient evidence to conclude that the two population means are different.