A survey of 27 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $63.00 per hour. The sample standard deviation was $6.11 per hour Ụse t Distribution Table) a. What is the best estimate of the population mean?
Estimated population mean
b. Develop a 95% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)
Confidence interval for the population mean wage is and between
c. How large a sample is needed to assess the population mean with an allowable error of $2.00 at 90% confidence? (Round up your answer to the next whole number.)
Sample size

a) sample mean = 63

b) 95% confidence interval for the population mean wage (including benefits) for these employees is ( 60.58, 65.42 )

c) sample size is 26

Step-by-step explanation:

given the data in the question

a)  estimate of the population mean

sample mean = ($63.00 per hour) = 63

b)

given that; standard deviation σ = 6.11, sample size n = 27

df = n-1 = 27 - 1 = 26

now at 95% CI, t will be;

∝ = 1 - 95% = 0.05, = 0.05/2 = 0.025

= = 2.056

Margin of Error E = × (σ/√n)

Margin of Error E = 2.056 × (6.11/√27)

Margin of Error E = 2.056 × 1.17587

Margin of Error E = 2.4175 ≈ 2.42

CI estimate of the population mean will be; ( 95% )

- E < μ < + E

we substitute

63 - 2.42 < μ < 63 + 2.42

60.58 < μ < 65.42

Therefore, 95% confidence interval for the population mean wage (including benefits) for these employees is ( 60.58, 65.42 )

c)

At 90% confidence level and Margin of Error of 2

sample size n = ?

∝ = 1-90% = 0.10, ∝/2 = 0.10/2 = 0.05

= = 1.645

Sample size n = ( × σ/ / E )²

we substitute

Sample size n = (1.645×6.11 / 2)²

n = (10.05095 / 2)²

n = ( 5.025475)²

n = 25.255

number of employed judges cant be decimal,

Therefore, sample size is 26