Question

1. A random sample of 20 starting salaries in a certain field found an average of $23 per hour, and a standard deviation of $3. Construct a 95% confidence interval for the true starting salary in this field.

Answer 1

Solution :

degrees of freedom = n - 1 = 20 - 1 = 19

t/2,df =  t_{0.025 , 19 = 2.093}

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

= 2.093 * ( 3 / 20)

Margin of error = E = 1.40

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

  ± E

23  ± 1.40

(21.6 , 24.4)