Question

A group of researchers estimates the mean length of time (in minutes) the average U.S. adult spends watching television using digital recorders and other forms of time-shifted television each day. To do so, the group takes a random sample of 40 U.S. adults and obtains the times (in minutes) listed below. Assume the population standard deviation to be 6.5 minutes. construct a 90% and 99% confidence interval for the population mean.

29, 27, 28, 26, 14,28,32, 12, 16, 21, 17, 23, 24, 13, 24, 25, 33, 17, 29, 24, 24, 12, 13, 19, 28,21, 20,30, 31, 32,13, 33, 18, 26, 21,15, 27,16, 27, 28

Answer 1

	Values ( X )
	29	37.21
	27	16.81
	28	26.01
	26	9.61
	14	79.21
	28	26.01
	32	82.81
	12	118.81
	16	47.61
	21	3.61
	17	34.81
	23	0.01
	24	1.21
	13	98.01
	24	1.21
	25	4.41
	33	102.01
	17	34.81
	29	37.21
	24	1.21
	24	1.21
	12	118.81
	13	98.01
	19	15.21
	28	26.01
	21	3.61
	20	8.41
	30	50.41
	31	65.61
	32	82.81
	13	98.01
	33	102.01
	18	24.01
	26	9.61
	21	3.61
	15	62.41
	27	16.81
	16	47.61
	27	16.81
	28	26.01
Total	916	1639.6

Mean

Standard deviation

Confidence Interval :-



Lower Limit =
Lower Limit = 21.2095
Upper Limit =
Upper Limit = 24.5905
90% Confidence interval is ( 21.2095 , 24.5905 )

Confidence Interval :-



Lower Limit =
Lower Limit = 20.2527
Upper Limit =
Upper Limit = 25.5473
99% Confidence interval is ( 20.2527 , 25.5473 )