Question

two hundred auditors each took their own simple random sample of fifteen values from a large data set of account balances, and each constructed a 90% confidence interval for the mean balance of the entire data set. About how many of the auditors had the true mean for the entire set in their personal confidence interval?

Answer 1

Solution :

Given that,

n = 200

x = 15

Point estimate = sample proportion = = x / n = 15 / 200 = 0.075

1 - = 1 - 0.075 = 0.925

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05  = 1.645

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.645 (((0.075 * 0.925) / 200)

= 0.031

A 90% confidence interval for population proportion p is ,

± E  

= 0.075  ± 0.031

= ( 0.044, 0.106 )