Question

A random sample of 15 recent college graduates found that starting salaries for attorneys in New York City had a mean of $102,342 and a standard deviation of $21,756. There are no outliers in the sample data set. Construct a 95% confidence interval for the average starting salary of all attorneys in the city. Group of answer choices (89869.82, 114814.18) (90292.73, 114391.27) (91331.94, 113352.06) (90371.37, 114312.63)

Answer 1

Answer:

(91331.94, 113352.06)

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{21756}{\sqrt{15}} = 11010.06$

The lower end of the interval is the sample mean subtracted by M. So it is 102342 - 11010.06 = 91331.94.

The upper end of the interval is the sample mean added to M. So it is 102342 + 11010.06 = 113.352.06.

So the correct answer is:

(91331.94, 113352.06)