Question

A survey of 25 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $59.00 per hour. The sample standard deviation was $6.05 per hour. (Use t Distribution Table.)

1. What is the best estimate of the population mean?

2. Develop a 99% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)

3. How large a sample is needed to assess the population mean with an allowable error of $2.00 at 99% confidence? (Round up your answer to the next whole number.)

Answer 1

Solution :

Given that,

a) Point estimate = sample mean = = 59.00

sample standard deviation = s = 6.05

sample size = n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

b) At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,24 = 2.797

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

= 2.797 * (6.05 / 25)

Margin of error = E = 3.38

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

  ± E  

= 59.00 ± 3.38

= ( 55.62, 62.38 )

c) margin of error = E = 2.00

sample size = n = [t/2,df* s / E]2

n = [2.797 * 6.05 / 2 ]2

n = 71.58

Sample size = n = 72