Question

(10 pts) In an advertisement, a pizza shop claims that its mean delivery time is less than 30 minutes. A random sample of 36 delivery imes has a mean of 28.5 minutes and a standard deviation of 3.5 minutes. Is there enough evidence to support the claim at α-.05 ? (a) Set up the null and alternative hypotheses (b) Find the test statistic. (c) Find the rejection region and state your conclusion. (d) What is the P-value for this test?

Answer 1

Solution :

This is the left tailed test .

The null and alternative hypothesis is ,

H₀ :  $\mu$ = 30

H_a : $\mu$   $\neq$ 30

$\bar x$ = 28.5

$\mu$ = 30

$\sigma$ = 3.5

n = 36

Test statistic = z

= ($\bar x$ - $\mu$ ) / $\sigma$ / $\sqrt$ n

= (28.5 - 30) / 3.5 / $\sqrt$ 36

= -2.57

Test statistic = -2.57

P(z < -2.57) = 0.0051

P-value = 0.0051

$\alpha$ = 0.05

Z_0.05 = -2.576

P-value > $\alpha$

Fail to reject the null hypothesis .

There is not enough evidence to suggest that the claim at $\alpha$ = 0.05 .