# 1. 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

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

.
 Sample from Population 1 Sample from Population 2 H0​: μ1−μ2=0 HA​: μ1−μ2≠0 32 28 26 34 39 32 41 38 36 32 38 29 32 40 31 26 35 36 39 31
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. (Round to four decimal places as needed.)

Difference Scores Calculations

Treatment 1

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N1: 10
df1 = N – 1 = 10 – 1 = 9
M1: 32.9
SS1: 206.9
s21 = SS1/(N – 1) = 206.9/(10-1) = 22.99

Treatment 2

N2: 10
df2 = N – 1 = 10 – 1 = 9
M2: 34.6
SS2: 176.4
s22 = SS2/(N – 1) = 176.4/(10-1) = 19.6

T-value Calculation

s2p = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22) = ((9/18) * 22.99) + ((9/18) * 19.6) = 21.29

s2M1 = s2p/N1 = 21.29/10 = 2.13
s2M2 = s2p/N2 = 21.29/10 = 2.13

t = (M1 – M2)/√(s2M1 + s2M2) = -1.7/√4.26 = -0.82

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