Question

A statistics practitioner took a random sample of 52 observations from a population whose standard deviation is 27 and computed the sample mean to be 100.

Note: For each confidence interval, enter your answer in the form (LCL, UCL). You must include the parentheses and the comma between the confidence limits.

A. Estimate the population mean with 95% confidence.

Confidence Interval =

B. Estimate the population mean with 95% confidence, changing the population standard deviation to 45;

Confidence Interval =

C. Estimate the population mean with 95% confidence, changing the population standard deviation to 11;

Confidence Interval =

Answer 1

Solution :

Given that,

= 100

n = 52

A)

= 27

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (27 / 52)

= 7.34

At 95% confidence interval estimate of the population mean is,

- E < < + E

100 - 7.34 < < 100 + 7.34

92.66 < < 107.34

(92.66 , 107.34)

(B)

= 45

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (45 / 52)

= 12.23

At 95% confidence interval estimate of the population mean is,

- E < < + E

100 - 12.23 < < 100 + 12.23

87.77 < < 112.23

(87.77 , 112.33)

(C)

= 11

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (11 / 52)

= 2.99

At 95% confidence interval estimate of the population mean is,

- E < < + E

100 - 2.99 < < 100 + 2.99

97.01 < < 102.99

(97.01 , 102.99)