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#8A study of the consultants in a particular industry has determined that the standard deviation of the hourly fee of the consultants is $24. A random sample of 90 consultants in the industry has a mean hourly fee of $122. Find a 90% confidence interval for the true mean hourly fee of all consultants in the industry. Then give its lower limit and upper limit.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)Lower limit:?Upper limit:?

Sagot :

Given the following information from the question

[tex]\begin{gathered} \bar{x}=\text{ \$}122 \\ s=\text{ \$}24 \\ n=90 \end{gathered}[/tex]

Where

x(bar) is the mean, s is the standard deviation and n is the sample size

The z-score for the 90% confidence interval can be obtained using a calculator and the value is given below

[tex]z_{\frac{\alpha}{2}}=1.645[/tex]

The confidence interval formula is given as

[tex]CI=\bar{x}\pm z_{\frac{\alpha}{2}}\times\frac{s}{\sqrt{n}}[/tex][tex]CI=122\pm1.645\times\frac{24}{\sqrt{90}}[/tex][tex]\begin{gathered} CI=122\pm4.1616 \\ lower\text{ limit}\rightarrow122-4.1616=117.8384 \\ upper\text{ limit}\rightarrow122+4.1616=126.1616 \end{gathered}[/tex]

Hence, the Lower limit is $117.838 and the Upper limit is $126.162

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