Question

1. A study of 35 golfers showed that their average score on a particular course was 92. The standard deviation of the sample is 5.

a. Find the best point estimate of the mean.

b. Find the 95% confidence interval of the mean score for all golfers.

c. Find the 99% confidence interval of the mean score for all golfers.

Answer 1

Solution :

Given that,

= 92

s =5

n = 35

Degrees of freedom = df = n - 1 =35 - 1 = 34

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,34 =2.032 ( using student t table)

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

= 2.032* ( 5/ 35)

= 1.7174

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

- E < < + E

92 - 1.7174 < < 92+ 1.7174

9.2826 < < 93.7174

( 9.2826 ,93.7174)

(B)

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2  df = t0.005,34 = 2.728    ( using student t table)

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

= 2.728* ( 5/ 35) =2.3056

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

- E < < + E

92 - 2.3056 < <92 + 2.3056

89.6944 < < 94.3056

( 89.6944 , 94.3056 )