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A study was conducted of recent high school graduates who began full-time jobs rather than going...

  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.
  1. Identify the upper and lower z score for a 95% confidence interval for µ.
  1. Calculate the upper and lower bounds of the confidence interval.

Solutions

Expert Solution

Solution :

Given that,

Sample size = n = 60

a.

Standard error =   / n = 2350 / 60 = 303.38

b.

Z/2 = 1.96

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

= 1.96 * (2350 / 60)

Margin of error = E = 594.63

c.

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

- E < < + E

24500 - 594.63 < < 24500 + 594.63

23905.37< < 25094.63

lower bounds = 23905.37

Upper bounds = 25094.63

milcah answered 2 years ago
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