Question

A simple random sample with

n = 57

provided a sample mean of 22.5 and a sample standard deviation of 4.4. (Round your answers to one decimal place.)

(a)

Develop a 90% confidence interval for the population mean.

   to  

(b)

Develop a 95% confidence interval for the population mean.

  to  

(c)

Develop a 99% confidence interval for the population mean.

   to  

Answer 1

Solution :

Given that n = 57 , mean x-bar = 22.5 , standard deviation s = 4.4

=> df = n - 1 = 56

(a)
=> for 90% confidence interval, t = 1.673

=> A 90% confidence interval for the population mean is

=> x-bar +/- t*s/sqrt(n)

=> 22.5 +/- 1.673*4.4/sqrt(57)

=> 21.5250 to 23.4750

=> 21.5 to 23.5 (rounded)

(b)
=> for 95% confidence interval, t = 2.003

=> A 95% confidence interval for the population mean is

=> x-bar +/- t*s/sqrt(n)

=> 22.5 +/- 2.003*4.4/sqrt(57)

=> 21.3327 to 23.6673

=> 21.33 to 23.67 (rounded)

(c)
=> for 99% confidence interval, t = 2.667

=> A 99% confidence interval for the population mean is

=> x-bar +/- t*s/sqrt(n)

=> 22.5 +/- 2.667*4.4/sqrt(57)

=> 20.9457 to 24.0543

=> 20.95 to 24.05 (rounded)