Question

A simple random sample with n=54   provided a sample mean of 21.0 and a sample standard deviation of 4.7 .

a. Develop a 90% confidence interval for the population mean (to 1 decimal).

,

b. Develop a 95% confidence interval for the population mean (to 1 decimal).

,

c. Develop a 99% confidence interval for the population mean (to 1 decimal).

,

d. What happens to the margin of error and the confidence interval as the confidence level is increased?

- Select your answer -They increaseThey decreaseThey stay the sameIt cannot be determined from the given dataThe margin of error increases and the confidence interval becomes narrowerThe margin of error decreases and the confidence interval becomes widerItem 7 .

Answer 1

Solution :

Given that,

a)

Point estimate = sample mean = = 21

sample standard deviation = s = 4.7

sample size = n = 54

Degrees of freedom = df = n - 1 = 54 - 1 = 53

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_0.05,21 = 1.674

Margin of error = E = t_/2,df * (s /n)

= 1.674* ( 4.7/ 54)

= 1.07

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

- E < < + E

21 - 1.07 < < 21 + 1.07

19.9 < < 22.1

(19.9 , 22.1)

b)

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,53 = 2.006

Margin of error = E = t_/2,df * (s /n)

= 2.006 * ( 4.7/ 54)

= 1.3

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

- E < < + E

21 - 1.3 < < 21 + 1.3

19.7 < < 22.3

(19.7 , 22.3)

c)

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,53 = 2.672

Margin of error = E = t_/2,df * (s /n)

= 2.672* ( 4.7/ 54)

= 1.7

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

- E < < + E

21 - 1.7 < < 21 + 1.7

19.3 < < 22.7

(19.3 , 22.7)

The Confidence level is increased then the margine of error of increased.