N a survey of twelve Harbor Business School graduates, the mean starting salary was $93,000, with a standard deviation of $17,000. What is the 95% confidence interval for the average starting salary among all Harbor graduates?

The confidence interval would be (82198.71, 103801.28).

Explanation:

Since we have given that

Mean = $93,000

Standard deviation = $17,000

n = 12 < 30

So, we will use 't-test' to find the 95% confidence interval,

Here ,

and

So, according to t table, we get that

So, interval would be

Hence, the confidence interval would be (82198.71, 103801.28).

its c i google it wish u luck

it should be d or b but i'm sure it's d