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earnstyle [38]
3 years ago
7

Starting salaries of 90 college graduates who have taken a statistics course have a mean of $43,993 and a standard deviation of

$9,144. Using 99% confidence level, find the following: A. The margin of error E: _______________________________________ B. The confidence interval for the mean μ: _______________
Mathematics
1 answer:
inysia [295]3 years ago
3 0

Answer:

a) E = 2482.90

b) 99% Confidence interval: (41510.1,46475.9)

Step-by-step explanation:

We are given the following in the question:

Sample mean, \bar{x} =$43,993

Sample size, n = 90

Alpha, α = 0.051

Sample standard deviation, s = $9,144

a) The margin of error

Formula:

z_{critical}\text{ at}~\alpha_{0.01} = 2.576

E =z_{critical}\times \displaystyle\frac{s}{\sqrt{n}}\\\\E = 2.576\times \frac{9144}{\sqrt{90}}\\\\E = 2482.90

b) 99% Confidence interval:  

\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

43993 \pm 2482.90 = (41510.1,46475.9)

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