Question

(a) Find the P - value for the test statistic z=1.36 for the following null and alternative hypotheses:

Answer 1

Step-by-step explanation:

we need to look up the p-value in the z-table.

for z = 1.36 we get 0.91309.

this is p(the population mean is less than 9).

left-tailed test (less) : p-value = z-table(1.36)

right-tailed test (larger) : p-value = 1 - z-table(1.36)

two-tailed test (not equal to) :

p-value = 2×minimum(z-table(1.36), 1 - z-table(1.36))

(a) is clearly the left-tailed test.

p-value (as above) = 0.91309

(b) is clearly the two-tailed test.

z-table(1.36) = 0.91309

1 - z-table(1.36) = 1 - 0.91309 = 0.08691

the minimum of these 2 values is 0.08691.

so,

p- value = 2×0.08691 = 0.17382

FYI - in both cases that p- value is not small enough (typically it should be <= 0.05) to drop H0.