Question

A survey of 24 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $57.00 per hour. The sample standard deviation was $6.02 per hour. (Use t Distribution Table.)

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

2. Develop a 98% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = $57.00

sample standard deviation = s = $6.02

sample size = n = 24

Degrees of freedom = df = n - 1 = 24 - 1 = 23

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,23 = 2.500

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

= 2.500 * ( 6.02/ 24)

Margin of error = E = 3.07

The 98% confidence interval estimate of the population mean is,

- E < < + E

57.00 - 3.07 < < 57.00 + 3.07

53.93 < < 60.07

A 98% confidence interval for the population mean: 53.93 , 60.07