Question

Sixty items are randomly selected from a population of 660 items. The sample mean is 38, and the sample standard deviation 3. Develop a 98% confidence interval for the population mean. (Round the final answers to 2 decimal places.)            

    

The confidence interval is between       and     .

Answer 1

Solution :

Given that,

= 38

s =3

n = 660

Degrees of freedom = df = n - 1 = 660- 1 = 659

a ) At 98% confidence level the t is

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02/ 2 = 0.01

t /2,df = t0.01,659 = 2.332 ( using student t table)

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

= 2.332 * ( 3/ 660)

= 0.27

The 98% confidence interval estimate of the population mean is,

- E < < + E

38-0.27 < < 38+0.27

37.73 < < 38.27