Question

In a random sample of 35 ​refrigerators, the mean repair cost was ​$141.00 and the population standard deviation is ​$19.10. A 95​% confidence interval for the population mean repair cost is left parenthesis 134.67 comma 147.33 right parenthesis. Change the sample size to nequals70. Construct a 95​% confidence interval for the population mean repair cost. Which confidence interval is​ wider? Explain. Construct a 95​% confidence interval for the population mean repair cost. The 95​% confidence interval is ​( ​, ​). ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 141

Population standard deviation = = 19.10

Sample size = n = 70

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 * (19.10 / 70)

= 4.47

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

- E < < + E

141 - 4.47 < < 141 + 4.47

136.53 < < 145.47

(136.53 , 145.47)

n = 35 is the wider confidence interval .