Q:

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)

Accepted Solution

A:
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[tex]df = 10-1 = 9[/tex]Then, we need to subtract one by the confidence level \alpha and divide by 2. So:[tex]\frac{1-0.95}{2} = \frac{0.05}{2} = 0.025[/tex]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 [tex]T = 2.26[/tex].Now, we need to find the standard deviation of the sample. That is:[tex]s = \frac{11.364}{\sqrt{10}} = 3593.6[/tex]Now, we multiply T and s[tex]M = T*s = 2.262*3593.61 = 8128.75[/tex]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)