Question

1. A study was conducted of recent high school graduates who began full-time jobs rather than going to college. We wish to find the average starting income µ of such workers. A random sample of size 60 gave a sample mean of $24,500 with a population standard deviation of $2350.Round answers to 2 decimal places.

1. Find the standard error.

2. Identify the upper and lower z score for a 95% confidence interval for µ.

3. Calculate the upper and lower bounds of the confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 24500

Population standard deviation = = 2350

Sample size = n = 60

a)

Standard error

= / n = 2350 / 60 = 303.38

b)

P( -z < Z < z ) = 0.95

P( Z < z ) - P( Z < -z ) = 0.95

2*P(Z < z ) - 1 = 0.95

2*P(Z < z ) = 1 + 0.95

P( Z < z ) = 1.95 / 2

P( Z < z ) = 0.975

P( Z < 1.96 ) = 0.975

z = 1.96

and

z = -1.96

c)

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

Margin of error = E = Z_/2* ( /n)

= 1.96 * (2350 / 60 )

= 594.63

At 95% confidence interval estimate of the population mean is,

- E < < + E

24500 - 594.63 < < 24500 + 594.63

23905.37 < < 25094.63

( 23905.37 , 25094.63 )