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Pavel [41]
3 years ago
5

A random sample of 10 recent college graduates found that starting salaries for accountants in New York City had a mean of $47,5

89 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)
Mathematics
1 answer:
MAXImum [283]3 years ago
7 0

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)

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Step-by-step explanation:

Hi, to answer this question we have to write each inequality:

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