Question

Consider the following competing hypotheses and accompanying sample data. (You may find it useful to reference the appropriate table: z table or t table)

H0: μ1 – μ2 = 9

HA: μ1 – μ2 ≠ 9

x−1 = 54 , s1 = 21.6 , n1 = 22

x−2 = 32 , s2 = 15.3, n2 = 18

Assume that the populations are normally distributed with equal variances.

a-1. Calculate the value of the test statistic. (Round intermediate calculations to at least 4 decimal places and final answer to 3 decimal places.)

a-2. Find the p-value. 0.05 ≤ p-value < 0.10, p-value ≥ 0.10, p-value < 0.01, 0.01≤ p-value < 0.02, 0.02 ≤ p-value < 0.05

b. At the 1% significance level, can you conclude that the difference between the two means differs from 9?

		H0. At the 1% significance level,	we
conclude that the difference between the means differs from 9.

Answer 1

using miita>stat>basic stat>two sample t

we have

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 22 54.0 21.6 4.6
2 18 32.0 15.3 3.6

Difference = μ (1) - μ (2)
Estimate for difference: 22.00
99% CI for difference: (5.59, 38.41)
T-Test of difference = 0 (vs ≠): T-Value = 3.64 P-Value = 0.001 DF = 38
Both use Pooled StDev = 19.0410

a-1) the value of the test statistic is 3.635

a-2 ) p value is 0.001

p-value < 0.01

a-3 ) since 99% cofidence interval contain 9

so accept Ho

At the 1% significance level,accept Ho	we
conclude that the difference between the means differs from 9.