Question

A sample of 65 observations is selected from one population with a population standard deviation of 0.75. The sample mean is 2.67. A sample of 50 observations is selected from a second population with a population standard deviation of 0.66. The sample mean is 2.59/ conduct the following test of hypothesis using the 0.08 significance level:

H_o: m₁ £ m₂     H₁: m₁ > m₂

1. Is this a one-tailed or a tow-tailed test?
2. State the decision rule.
3. Compute the value of the test statisti
4. What is your decision regarding Ho?
5. What is the p-value?

Answer 1

Solution:

Given:

Population 1:

Population standard deviation =   = 0.75

Sample Size = n1 = 65

Sample Mean =

Population 2:

Population standard deviation =   = 0.66

Sample Size = n2 = 50

Sample Mean =

significance level = 0.08

H_o: m₁ =  m₂     H₁: m₁ > m₂

Part a) Is this a one-tailed or a tow-tailed test?

Since   H₁: m₁ > m_{2 ,} it is > type this is one tailed ( right tailed) test.

Part b) State the decision rule.

Find z critical value:

significance level = 0.08 ,

thus find Area = 1 - 0.08 = 0.92

Look in z  table for Area = 0.9200 or its closest area and find corresponding z value.

Area 0.9207 is closest to 0.9200 and it corresponds to 1.4 and 0.01

thus z critical value = 1.41

Thus decision rule is:
Reject null hypothesis H0, if z test statistic value > z critical value = 1.41 , otherwise we fail to reject H0.

Part c) Compute the value of the test statistic

Part d) What is your decision regarding Ho?

Since z test statistic value = 0.61 < z critical value = 1.41 , we fail to reject H0.

Part e) What is the p-value?

p-value = P( Z > z test statistic value)

p-value = P( Z > 0.61 )

p-value = 1 - P( Z < 0.61 )

Look in z table for z = 0.6 and 0.01 and find corresponding area.

P( Z< 0.61 ) = 0.7291

Thus

p-value = 1 - P( Z < 0.61 )

p-value = 1 - 0.7291

p-value = 0.2709