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Arturiano [62]
2 years ago
13

A random sample of 12 recent college graduates reported an average starting salary of $54,000 with a standard deviation of $6,00

0.
a. Construct a 95% confidence interval for the mean starting salary of college graduates.

a. 50188 to 57812$
b. 45,0000 to 50,000$
c. 48,000 to 60,000$
d. 50,000 to 54000$

b. The National Association of Colleges and Employers reports that the mean starting salary for 2017 college graduates is $50,516. Does your confidence interval (part a) indicate that starting salaries have significantly changed since 2017? Why or why not?
Mathematics
1 answer:
Marianna [84]2 years ago
4 0

Answer: a.) $50188 to $57812

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> (CI) is an interval of values in which we are confident the true mean is in.

The interval is calculated as

x ± z\frac{s}{\sqrt{n} }

a. For a 95% CI, z-value is 1.96.

Solving:

54,000 ± 1.96.\frac{6000}{\sqrt{12} }

54,000 ± 1.96\frac{6000}{3.464}

54,000 ± 1.96*1732.102

54,000 ± 3395

This means the interval is

50605 < μ < 57395

<u>With a 95% confidence interval, the mean starting salary of college graduates is between 50605 and 57395 or </u><u>from 50188 to 57812$.</u>

<u />

b. The mean starting salary for college students in 2017 is $50,516, which is in the confidence interval. Therefore, since we 95% sure the real mean is between 50188 and 57812, there was no significant change since 2017.

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