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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 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−−√ )

Sagot :

Answer:

[tex]46500 \pm 1.74\frac{10200}{\sqrt{18}}[/tex]

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 18 - 1 = 17

90% confidence interval

Now, 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 [tex]1 - \frac{1 - 0.9}{2} = 0.95[/tex]. So we have T = 1.74

The margin of error is:

[tex]M = T\frac{s}{\sqrt{n}} = 1.74\frac{10200}{\sqrt{18}}[/tex]

Confidence interval:

Sample mean plus/minus margin of error. So

[tex]46500 \pm 1.74\frac{10200}{\sqrt{18}}[/tex]

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