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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?

Answer :

Answer: 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 , [tex]\alpha =1-0.095=0.05[/tex]

and [tex]n-1=12-1=11[/tex]

So, according to t table, we get that

[tex]t_{n-1,\frac{\alpha }{2}}=2.200985[/tex]

So, interval would be

[tex]\bar{x}\pm t_{n-1,\frac{\alpha {2}}\dfrac{\sigma}{\sqrt{n}}\\\\=93000\pm 2.200985\times \dfrac{17000}{\sqrt{12}}\\\\=93000\pm 10801.2839\\\\=(93000-10801.2839,93000+10801.2839)\\\\=(82198.71,103801.28)[/tex]

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

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