Question

A sample of 33 observations is selected from a normal population. The sample mean is 53, and the population standard deviation is 6. Conduct the following test of hypothesis using the 0.05 significance level. H0: μ = 57 H1: μ ≠ 57

What is the decision rule? Reject H0 if −1.960 < z < 1.960 Reject H0 if z < −1.960 or z > 1.960

What is the value of the test statistic? (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

What is your decision regarding H0? Fail to reject H0 Reject H0

e-1. What is the p-value? (Round your z value to 2 decimal places and final answer to 4 decimal places.)

e-2. Interpret the p-value? (Round your z value to 2 decimal places and final answer to 2 decimal places.)

there is a ......% chance of finding a z value this large by "sampling error" when H0 is true.

Answer 1

a) As we are given the population standard deviation here, we use the Z test here to test the hypothesis. For 0.05 level of significance, we have from the standard normal tables:
P( -1.96 < Z < 1.96) = 0.95

Therefore the decision here is given as:
Reject H₀ when Z < -1.96 or Z > 1.96

b) The test statistic here is computed as:

Therefore -3.83 is the required test statistic value here.

As the test statistic value is -3.83 < -1.96, therefore it lies in the rejection region and we can reject the null hypothesis here.

e-1) As this is a two tailed test, the p-value here is computed from the standard normal tables as:
p = 2P(Z < -3.83) = 2*0.0001 = 0.0002

Therefore 0.0002 is the required p-value here.

e-2) As the p-value here is 0.0002, therefore the interpretation of this is that there is a 0.02% change of finding a z value this large by sampling error when H0 is true.