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,

Point estimate = sample mean = = 138850

sample standard deviation = s = 35800

sample size = n = 14

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

1) At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

  t/2,df = 2.160

2) Standard of error = SE =  (s /n)

SE = ( 35800 / 14)

SE = 9568

3) Margin of error = E = t/2,df * SE

= 2.160 * 9568

Margin of error = E = 20667

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

  ± E

138850 ± 20667

4. The lower limit of the interval: 118183

5. The upper limit of the interval: 159517