Question

A simple random sample with n=50 provided a sample mean of 2.3 and a sample standard deviation of 4.7.

a. Develop a 90% confidence interval for the population mean (to 2 decimals). ,

b. Develop a 95% confidence interval for the population mean (to 2 decimals). ,

c. Develop a 99% confidence interval for the population mean (to 2 decimals). ,

Answer 1

Solution:

Part a

Confidence interval for Population mean is given as below:

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

We are given

Xbar = 2.3

S = 4.7

n = 50

df = n – 1 = 50 – 1 = 49

Confidence level = 90%

Critical t value = 1.6766

(by using t-table)

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

Confidence interval = 4.7 ± 1.6766*4.7/sqrt(50)

Confidence interval = 4.7 ± 1.6766*0.664680374

Confidence interval = 4.7 ± 1.1144

Lower limit = 4.7 - 1.1144 = 1.19

Upper limit = 4.7 + 1.1144 = 3.41

Confidence interval = (1.19, 3.41)

Part b

Confidence interval for Population mean is given as below:

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

We are given

Xbar = 2.3

S = 4.7

n = 50

df = n – 1 = 50 – 1 = 49

Confidence level = 95%

Critical t value = 2.0096

(by using t-table)

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

Confidence interval = 4.7 ± 2.0096*4.7/sqrt(50)

Confidence interval = 4.7 ± 2.0096*0.664680374

Confidence interval = 4.7 ± 1.3357

Lower limit = 4.7 - 1.3357= 0.96

Upper limit = 4.7 + 1.3357= 3.64

Confidence interval = (0.96, 3.64)

Part c

Confidence interval for Population mean is given as below:

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

We are given

Xbar = 2.3

S = 4.7

n = 50

df = n – 1 = 50 – 1 = 49

Confidence level = 99%

Critical t value = 2.68

(by using t-table)

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

Confidence interval = 4.7 ± 2.68*4.7/sqrt(50)

Confidence interval = 4.7 ± 2.68*0.664680374

Confidence interval = 4.7 ± 1.7813

Lower limit = 4.7 - 1.7813= 0.52

Upper limit = 4.7 + 1.7813= 4.08

Confidence interval = (0.52, 4.08)

AS confidence level increases, the width of the confidence interval also increases.