Question

A statistics practitioner took a random sample of 55 observations from a population whose standard deviation is 35 and computed the sample mean to be 101.

Note: For each confidence interval, enter your answer in the form (Lower limit, Upper limit). 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 90% confidence.

Confidence Interval =

C. Estimate the population mean with 99% confidence.

Confidence Interval =

Answer 1

Solution :

Given that,

= 101

= 35

n = 55

a ) 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 = 1.960

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

= 1.960 * ( 35 / 55 )

= 9.25

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

- E < < + E

101 - 9.25 < < 101 + 9.25

91.75 < < 110.25

Lower limit, = 91.75

Upper limit = 110.25

B ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

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

= 1.645 * ( 35 / 55 )

= 7.76

At 90% confidence interval estimate of the population mean is,

- E < < + E

101 - 7.76 < < 101 + 7.76

93.24 < < 108.76

Lower limit, = 93.24

Upper limit = 108.76

C ) At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (35 / 55 )

= 12.16

At 99% confidence interval estimate of the population mean is,

- E < < + E

101 - 12.16 < < 101 + 12.16

88.84 < < 113.16

Lower limit, = .24

Upper limit = 113.16