Question

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

a. Find the best point estimate of the mean

b. What is the margin of Error?

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

d. Find the minimum number of golfers for 99 percent confidence when the score when the margin of error is 3.

Answer 1

Solution :

Given that,

Point estimate = sample mean =     =92

Population standard deviation =    = 5

Sample size n =35

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ( /n)

= 1.96 * ( 5 / 35 )

E= 1.6565
At 95% confidence interval estimate of the population mean
is,

- E < < + E

92 - 1.6565 <   < 92+ 1.6565

90.3435 <   < 93.6565

( 90.3435, 93.6565 )

(B)

Given that,

standard deviation =   =5

Margin of error = E = 3

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

sample size = n = [Z/2* / E] 2

n = ( 2.58* 5 / 3)2

n =18.49

Sample size = n =19