Question

2. A random sample of 12 recent college graduates reported an average starting salary of $54,000 with a standard deviation of $6,000.

To construct a 95% confidence interval for the mean starting salary of college graduates, what will be the margin of error? Round to whole dollars.

Answer #2

Solution :

Given that,

= 54000

s =6000

n =12

Degrees of freedom = df = n - 1 =12 - 1 = 11

a ) At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2= 0.05 / 2 = 0.025

t /2,df = t0.025,11 = 2.201 ( using student t table)

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

= 2.201* (6000 / 12)

=3812

answered by: anonymous

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