Question

A random sample of 10 recent college graduates found that starting salaries for accountants in New York City had a mean of $47,589 and a standard deviation of $11,364. There are no outliers in the sample data set. Construct a 95% confidence interval for the average starting salary of all accountants in the city.

A. (40545.52, 54632.48)

B. (39460.25, 55717.75)

C. (39582.43, 55595.57)

D. (39020.54, 56157.46)

Answer 1

Answer:

B. (39460.25, 55717.75)

Step-by-step explanation:

Our sample size is 10.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

$df = 10-1 = 9$

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

$\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025$

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 9 and 0.025 in the t-distribution table, we have $T = 2.26$.

Now, we need to find the standard deviation of the sample. That is:

$s = \frac{11.364}{\sqrt{10}} = 3593.6$

Now, we multiply T and s

$M = T*s = 2.262*3593.61 = 8128.75$

For the lower end of the interval, we subtract the mean by M. So 47589 - 8128.75 = 39460.25

For the upper end of the interval, we add the mean to M. So 47589 + 8128.75 = 55717.75.

The correct answer is:

B. (39460.25, 55717.75)