Question

The table below shows the lengths of some randomly chosen CDs in John’s very large collection. Give the 91% confidence interval for the population mean, assuming that the population is approximately normally distributed. Round the endpoints to two decimal places.

48.19	63.29	41.40	70.15	55.33	59.09	52.72	48.57	43.20	58.33
40.36	53.42	39.17	45.22	63.60	46.21	47.13	43.84	65.37	35.94
44.64	73.43	73.76	62.24	52.90	44.25	53.48	42.56	41.58	58.46
44.50	42.77	60.32	50.24	54.37

Confidence interval: (_ ,_ )

Answer 1

Solution:

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 52.00085714

S = 10.04650404

n = 35

df = n l- 1 = 34

Confidence level = 91%

Critical t value = 1.7451

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 52.00085714 ± 1.7451*10.04650404/sqrt(35)

Confidence interval = 52.00085714 ± 2.9635

Lower limit = 52.00085714 - 2.9635 = 49.04

Upper limit = 52.00085714 + 2.9635 = 54.96

Confidence interval = (49.04, 54.96)