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 is 8.3 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. Ho 8.3 Ha 8.3 b. A sample of 100 shoppers showed a sample mean waiting time of 8.9 minutes. Assume a population standard deviation of o 3.2 minutes. What is the p-value? (to 2 decimals) z value 1.875 (to 4 decimals) p-value 1.9392 c. At a 0.05, what is your conclusion? Select your answer conclude that the population mean waiting time differs from 8.3 minutes. -Select your answer- d. Compute a 95% confidence interval for the population mean. Does it support your conclusion? ) (to 2 decimals) in the interval. -Select your answer- 8.3 Select your answer

Answer 1

b)

p value = 0.0601

c)

As P-value >= 0.05, fail to reject null hypothesis.

we cannot conclude

d)

sample mean, xbar = 8.9
sample standard deviation, σ = 3.2
sample size, n = 100

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 3.2/sqrt(100)
ME = 0.63

CI = (xbar - Zc * s/sqrt(n) , xbar + Zc * s/sqrt(n))
CI = (8.9 - 1.96 * 3.2/sqrt(100) , 8.9 + 1.96 * 3.2/sqrt(100))
CI = (8.27 , 9.53)

Fail to reject H0 mu = 8.3 lies in the interval