Question

A sample of 66 observations is selected from one population with a population standard deviation of 0.68. The sample mean is 2.67. A sample of 45 observations is selected from a second population with a population standard deviation of 0.68. The sample mean is 2.59. Conduct the following test of hypothesis using the 0.1 significance level:

H₀: μ₁ – μ₂≤ 0

H₁: μ₁ – μ₂ > 0

a. Is this a one-tailed or a two-tailed test?

This is a  (Click to select)  one  two  -tailed test.

b. State the decision rule. (Round the final answer to 2 decimal places.)

The decision rule is to reject H₀ if z is   (Click to select)  less than  greater than  equal to   .

c. Compute the value of the test statistic. (Negative answer should be indicated by a minus sign. Round the final answer to 2 decimal places.)

The test statistic is z  

d. What is your decision regarding H₀?

H₀ is  (Click to select)  not rejected  rejected  .

e. What is the p-value? (Round the final answer to 4 decimal places.)

The p-value is  .

Answer 1

a)

This is a one tailed test

b)

Rejection Region
This is right tailed test, for α = 0.1
Critical value of z is 1.28.
Hence reject H0 if z > 1.28

The decision rule is to reject H0 if z is greater than equal to 1.28

c)

Pooled Variance
sp = sqrt(s1^2/n1 + s2^2/n2)
sp = sqrt(0.4624/66 + 0.4624/45)
sp = 0.1315

Test statistic,
z = (x1bar - x2bar)/sp
z = (2.67 - 2.59)/0.1315
z = 0.61

d)

H0 is not rejected

e)

P-value Approach
P-value = 0.2709

p value is calculated By using excel:

= 1 - NORM.DIST(0.61,0,1,TRUE)