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

a.) $50188 to $57812

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

The interval is calculated as

x ±

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

Solving:

54,000 ±

54,000 ±

54,000 ± 1.96*1732.102

54,000 ± 3395

This means the interval is

50605 < μ < 57395

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

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.

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3

- both graphs increase on their domain.

- the graphs of both functions have the same x-intercept.

- neither graph has a y-intercept.