Question

Marty Funkhouser has a sample of 64. He finds a sample average of 34 and a sample standard deviation of 9. What is the 95% confidence interval?

A

​[33, 35]

B

​[32, 36]

C

​[31, 37]

D

​[30, 38]

Answer 1

Solution :

Given that,

= 34

s =9

n =64

Degrees of freedom = df = n - 1 =64 - 1 = 63

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,63 = 1.998 ( using student t table)

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

= 1.998* ( 9/ 64)

= 2

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

- E < < + E

34-2 < < 34+2

​[32, 36]