Question

The American Economic Association, which surveyed economics students finishing their PhDs, found that academic job seekers were offered an average of 5.99 interviews. The sample standard deviation was 6.62 interviews, and there was 211 job candidates in the study.

a. Construct a 95% confidence interval for the mean number of interviews among those seeking academic jobs.

b. Provide a concise interpretation of this interval.

Answer 1

Solution :

Given that,

$\bar x$ = 5.99

s =6.62

n = Degrees of freedom = df = n - 1 =211 - 1 = 210

a ) At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

  $\alpha$/ 2$\alpha$ = 0.05 / 2 = 0.025

t_{$\alpha$ /2,df} = t_0.025,210 =

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.971* ( 6.62/ $\sqrt$ 211)

=0.8983

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

$\bar x$ - E < $\mu$ < $\bar x$ + E

5.99 - 0.8983< $\mu$ < 5.99+ 0.8983

5.0917 < $\mu$ < 6.8883

( 5.0917 , 6.8883)

The 95% confidence interval estimate of the mean is,( 5.0917 , 6.8883)