Question

In: Statistics and Probability

A university surveyed recent graduates of the English department for their starting salaries. Three hundred graduates...

A university surveyed recent graduates of the English department for their starting salaries. Three hundred graduates returned the survey. The average salary was $25,000. The sample standard deviation was $1,500. What is the 95% confidence interval for the mean salary of all graduates from the English department?

Solutions

Expert Solution

Solution :

Given that,

= $25000

s =$1500

n =300

Degrees of freedom = df = n - 1 = 300- 1 = 299

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,299 = 1.968 ( using student t table)

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

= 1.968 * (1500 / 300)

= 170.4338

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

- E < < + E

25000- 170.4338 < < 25000+ 170.4338

24829.5662 < < 25170.4338

( 24829.5662 ,25170.4338)


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