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## The starting salary of business students in a university is known to be normally distributed. A random sample of 18 business students result

Question

The starting salary of business students in a university is known to be normally distributed. A random sample of 18 business students results in a mean salary of $46,500 with a standard deviation of $10,200. Construct the 90% confidence interval for the mean starting salary of business students in this university. Multiple Choice 46,500 ± 1.740 ( 10,200 / 18−−√ ) 46,500 ± 1.330 ( 10,200 / 18−−√ ) 46,500 ± 1.734 ( 10,200 / 18−−√ ) 46,500 ± 1.645 ( 10,200 / 18−−√ )

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Mathematics
3 years
2021-08-22T01:40:23+00:00
2021-08-22T01:40:23+00:00 1 Answers
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## Answers ( )

Answer:Step-by-step explanation:We have the standard deviation for the sample, which means that the

t-distributionis used to solve this question.The

first stepto solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. Sodf = 18 – 1 = 17

90% confidence intervalNow, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.74

The margin of error is:

Confidence interval:Sample mean plus/minus margin of error. So