Question

In a random sample of 14 senior-level chemical engineers, the mean annual earnings was 138850 and the standard deviation was 35800. Assume the annual earnings are normally distributed and construct a 95% confidence interval for the population mean annual earnings for senior-level chemical engineers.

1. The critical value:

2. The standard error of the sample mean:

3. The margin of error:

4. The lower limit of the interval:

5. The upper limit of the interval:

Answer 1

Solution :

Given that,

= 138850

s =35800

n = Degrees of freedom = df = n - 1 = 14- 1 = 13

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

critical value t /2,df = t0.025,13 = 1.812 ( using student t table)

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

= 1.812 * ( 35800/ 14)

= 17337.13

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

- E < < + E

138850 - 17337.13 < <138850 + 17337.13

121512.87 < < 156187.13

4. The lower limit of the interval:121512.87

5. The upper limit of the interval:156187.13