Question

(d) For the hypothesis test in part (c)("If a hypothesis test was conducted using a 5% significance level, and the P-value turned out to be 0.075, should the null hypothesis be rejected, or not? Justify your answer."), the alternate hypothesis was H1: p ≠ 0.8 and the sampling distribution of the test statistic was standard Normal. What value(s) could the test statistic have been?

ANSWER GIVEN: (d) Left-tail area = .075 / 2 = .0375. [1] Z = ± 1.78

can someone please explain how they got this answer? as in what tables they went to to find the Z and stuff?

Thanks

Answer 1

Here test is two tailed test.

So we know that P value for two tailed test is

P value = 2 * P(z > test statistic)

or P value = 2 * P(z < - test statistic)

P value = 0.075

Left tail area = 0.075 / 2 = 0.0375

Z for left tail area 0.0375 = -1.78                 (From excel using command: =NORMSINV(0.0375))

Right tail area = 0.075 /2 = 0.0375

left area = 1- right area = 1 - 0.0375 = 0.9625     

z for left area 0.9625 is 1.78       (From excel using command: =NORMSINV(0.9625))

Here P value = 0.075 > level of significance = 0.05

So we do not reject null hypothesis.