Question

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line 8.2 is minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from this standard. a. Formulate the hypotheses for this application. b. A sample of 110 shoppers showed a sample mean waiting time of 9.0 minutes. Assume a population standard deviation of 3.2 minutes. What is the z-value? (to 2 decimals) what is the p-value (to 4 decimals) c. At 0.05, what is your conclusion? .Reject null hypothesis,can conclude that the population mean waiting time differs from minutes. d. Compute a 95% confidence interval for the population mean. Does it support your conclusion? , (to 2 decimals) ; in the interval.

Answer 1

a)

H0: = 8.2

Ha: 8.2

b)

Test statistics

z = - / / sqrt(n)

= 9 - 8.2 / 3.2 / sqrt(110)

= 2.62

This is test statistics value.

p-value = 2 * P( Z > z)

= 2 * P( Z > 2.62)

= 2 * 0.0044

= 0.0088

c)

Since p-value < 0.05 level we have sufficient evidence to reject H0.

We conclude that,

Reject the null hypothesis, we can conclude that the population mean waiting time differs from 8.2 minutes.

d)

95% confidence interval for is

- Z* / sqrt(n) < ​​​​​​ < + Z* / sqrt(n)

9 - 1.96 * 3.2 / sqrt(110) < < 9 + 1.96 * 3.2 / sqrt(110)

8.40 < < 9.60

95% CI is (8.40 , 9.60)

Since Claimed mean 8.2 outside the confidence interval, we have sufficient evidence to reject H0.

We have sufficient evidence to support the claim.