Question

10. In each part, we have given the value obtained for the test statistic z in a one-mean z-test. We have also specified whether the test is two tailed, left tailed, or right tailed. Determine the P-value in each case and decide whether, at the 5% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis. EXPLAIN AND SHOW ALL WORK

a. z = −1.25; left-tailed test

b. z = 2.36; right-tailed test

c. z = 1.83; two-tailed test

Answer 1

Solution,

= 0.05

a) test statistic z = -1.25

This is left tailed test,

P( z < -1.25)

= P-valuee = 0.1056

P-value >

Fail to reject the null hypothesis

b) test statistic z = 2.36

This is right tailed test,

P( z > 2.36) = 1 - P( z < 2.36) = 1 - 0.9909

= P-valuee = 0.0091

P-value <

Reject the null hypothesis

C) test statistic z = 1.83

This is two tailed test,

P( z > 1.83) = 1 - P( z < 1.83) = 1 - 0.9664 = 0.0336

= P-valuee = 2 * P( z > 1.83)

= P-valuee = 2 * 0.0336

= P-valuee = 0.0672

P-value >

Fail to reject the null hypothesis