Starting salaries of 130 college graduates who have taken a statistics course have a mean of $43,917 and a standard deviation of $9,456. Using 99% confidence, find both of the following:
A. The margin of error E E
B. The confidence interval for the mean μ : < μ<
df = n - 1= 130 - 1 = 129
t critical value at 0.01 level with 129 df = 2.614
A)
Margin of error = t * S / sqrt(n)
= 2.614 * 9456 / sqrt(130)
= 2167.91
B)
99% Confidence interval is
- E < < + E
43917 - 2167.91 < < 43917 + 2167.91
41749.09 < < 46084.91
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