Question

A random sample of 28 people employed by the Florida state authority established they earned an average wage (including benefits) of $59.00 per hour. The sample standard deviation was $5.84 per hour. (Use  t Distribution Table.)

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

Estimated population mean

$

b.	Develop a 90% 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 between  and  .

c.	How large a sample is needed to assess the population mean with an allowable error of $1.00 at 90% confidence? (Round up your answer to the next whole number.)

Sample size

Answer 1

Solution :

Given that,

= 59.00

s = 5.84

n = 28

a) The best estimate of the population mean =59.00

Degrees of freedom = df = n - 1 = 28 - 1 = 27

b) At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2,df = t_{0.05,27 = 1.70}

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

= 1.70 * (5.84 / 28)

= 1.88

Margin of error = 1.88

The 90% confidence interval estimate of the population mean is,

- E < < + E

59- 1.88 < < 59 + 1.88

57.12 < < 60.88

(57.12, 60.88)

C) E =1

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

t _/2 = t_{0.05, = 1.645}

Sample size = n = (Z_/2 * ) / E)²

= (1.645 * 5.84) / 1)²

= 92.29 = 92

Sample size = 92.