Question

A statistics practitioner took a random sample of 41 observations from a population whose standard deviation is 29 and computed the sample mean to be 96.

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 52;

Confidence Interval =

C. Estimate the population mean with 95% confidence, changing the population standard deviation to 12;

Confidence Interval =

Answer 1

Solution :


Point estimate = sample mean = = 96  

Sample size = n =41

a) Population standard deviation = = 29

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z0.025 = 1.960

Z/2 = 1.960

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

= 1.960 * (29 /  41 )

=8.88

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

- E < < + E

96 - 8.88 <   < 96 + 8.88

87.12<   < 104.88

( 87.12,104.88 )

Confidence Interval =(87.12 and 104.88)

b)

Population standard deviation = = 52

Z/2 = 1.960

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

= 1.960 * (52 /  41 )

= 15.92

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

- E < < + E

96 - 15.92 <   < 96 + 15.92

80.08 <   < 111.92
( 80.08 ,111.92)

Confidence Interval =( 80.08 and 111.92)

c)

Population standard deviation = = 12

Z/2 = 1.960

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

= 1.960 * (12  /  41 )

= 3.67

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

- E < < + E

96 - 3.67 <   < 96 + 3.67

92.33 <   < 99.67

( 92.33 ,99.67)

Confidence Interval =( 92.33 and 99.67)