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 7.7 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.

	- Select your answer -<>≤≥≠=Item 1
	- Select your answer -<>≤≥=≠Item 2

b. A sample of 150 shoppers showed a sample mean waiting time of 8 minutes. Assume a population standard deviation of 2.8 minutes. What is the p -value?

z value	(to 2 decimals)
p-value	(to 4 decimals)

c. At a=.01 what is your conclusion?

- Select your answer -Reject null hypothesisDo not reject null hypothesisItem 5 .

- Select your answer -Cannot CanItem 6 conclude that the population mean waiting time differs from  minutes.

d. Compute a 99 % confidence interval for the population mean. Does it support your conclusion?

  ,   (to 2 decimals)

- Select your answer -YesNoItem 9 ;  - Select your answer -isis notItem 10 in the in

Answer 1

a)

null hypothesis:Ho μ		=	7.7
Alternate Hypothesis:Ha μ		≠	7.7

b)

test stat z = '(x̄-μ)*√n/σ=	1.31
p value        =	0.1902

c)

Do not reject null hypothesis

Cannot

d)

for 99 % CI value of z=		2.576
margin of error E=z*std error =		0.59
lower bound=sample mean-E=		7.4111
Upper bound=sample mean+E=		8.5889
from above 99% confidence interval for population mean =(7.41,8.59)

Yes 7.7 is in the interval

answered by: anonymous