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 = 5
H_A: μ₁ – μ₂ ≠ 5

x−1x−1 = 57	x−2x−2 = 43
s1= 21.5	s2= 15.2
n1 = 22	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.

• p-value < 0.01

• 0.01 ≤ p-value < 0.02

• 0.02 ≤ p-value < 0.05

• 0.05 ≤ p-value < 0.10

• p-value ≥ 0.10

b. At the 5% significance level, can you conclude that the difference between the two means differs from 5?

Answer 1

Solution:

Given:

Hypothesis:

the populations are normally distributed with equal variances.

Part a-1:  Calculate the value of the test statistic.

where

Thus

.

Part a-2. Find the p-value.

df = n1 + n2 - 2 = 22 + 18 - 2 = 38

Since in t-table, we do not have df = 38 , so we look in t-table for its previous df= 30 and find the interval in which t = 1.495 fall. Then we look for corresponding two tail area ( Two tail area , since this is two tailed test)

.

t = 1.495 fall between 1.310 and 1.697

corresponding two tail area is between 0.10 and 0.20

Thus we have:

0.10 < p-value < 0.20

Thus from given options, p-value ≥ 0.10 is the correct option.

Part b. At the 5% significance level, can you conclude that the difference between the two means differs from 5?

Decision rule:

If p-value < 0.05 significance level, we reject H0 , otherwise we fail to reject H0.

Since p-value ≥ 0.10 > 0.05, thus we fail to reject H0 and hence at the 5% significance level,we can not conclude that the difference between the two means differs from 5