Question

A national grocer’s magazine reports the typical shopper spends 10 minutes in line waiting to check out. A sample of 20 shoppers at the local Farmer Jack’s showed a mean of 9.6 minutes with a standard deviation of 3.7 minutes. Is the waiting time at the local Farmer Jack’s less than that reported in the national magazine? Use the 0.025 significance level. What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.) What is your decision regarding H0? Reject H0 Do not reject H0

Answer 1

Solution :

= 10

= 9.6

s = 3.7

n = 20

This is the left tailed test .

The null and alternative hypothesis is ,

H₀ :   = 10

H_a : < 10

The significance level is α=0.025

and the critical value for a left-tailed test is t_c​=−2.093.

Test statistic = t

= ( - ) / s / n

= (9.6 - 10) /3.7 / 20

= -0.48

P (Z < -0.48) = 0.3171

P-value = 0.3171

= 0.025  

p= 0.3171 ≥ 0.025, it is concluded that the null hypothesis is not rejected.

Do not reject H0

There is not enough evidence to claim that the population mean μ is less than 10, at the 0.025 significance level.