Question

A sample of 31 observations is selected from a normal population. The sample mean is 69, and the population standard deviation is 8. Conduct the following test of hypothesis using the 0.01 significance level.

H₀: μ = 72

H₁: μ ≠ 72

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

• One-tailed test

• Two-tailed test

2. What is the decision rule?

• Reject H₀ if −2.576 < z < 2.576

• Reject H₀ if z < −2.576 or z > 2.576

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

value of the test statistic __________

4. What is your decision regarding H₀?

• Reject H₀

• Fail to reject H₀

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

p. value = ___________

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

Solution :

Given that ,

= 72

= 69

= 8

n = 31  

The null and alternative hypothesis is ,

H0 :   = 72

Ha :    72

This is the two tailed test .

= 0.05  

Z = Z0.01 = +/-2.576

Reject H0 if z < −2.576 or z > 2.576

Test statistic = z

= ( - ) / / n

= ( 69 - 72 ) / 8 / 31

= -2.09

The test statistic = -2.09

-2.09 < +/-2.576

Test statistic < Critical value

Reject Ho

P - value = 2 * P ( z < -2.09 )

= 2 * 0.0183

= 0.0366

e-1. P-value = 0.0366

e-2. Interpret the p-value = 03.66%

1. There is a 03.66 % chance of finding a z value this large by “sampling error” when H0 is true.