Q2-7BSC - Elementary Statistics

Question

Using Correct Distribution. In Exercises 5–8, assume that we want to construct a confidence interval. Do one of the following, as appropriate: (a) Find the critical value   \({t_{{\alpha  \mathord{\left/

 {\vphantom {\alpha  2}} \right.

 \kern-\nulldelimiterspace} 2}}}\) ,(b) find the critical value \({z_{{\alpha  \mathord{\left/

 {\vphantom {\alpha  2}} \right.

 \kern-\nulldelimiterspace} 2}}}\) ,or (c) state that neither the normal distribution nor the  t distribution applies.

 Denver Bronco Salaries confidence level is 99%,\(\sigma  = 3342\)thousand dollars, and the histogram of 61 player salaries (thousands of dollars) is shown in Exercise 6.

Step-by-Step Solution

Answer

In this case, the normal distribution is applying.

 

The critical value\({z_{\frac{\alpha }{2}}}\) is 2.575 .

            

Step 1: Given information

The histogram shows salaries (in thousands of dollars) of 61 players. That means the sample size n is 61.Denver Bronco Salaries confidence level is 99% with known \(\sigma  = 3342\).

Step 2: Describe the critical value

The critical value is the border value which separates the sample statistics that are significantly high or low from those sample statistics that are not significant.

Step3: Find the appropriate distribution

When the samples are randomly selected, the condition for t-distribution and normal distribution is as,

 

If \(\sigma \) is not known and \(n > 30\) then t-distribution is suitable to find the confidence interval.

 

If \(\sigma \) is known and \(n > 30\) then normal distribution is suitable to find the confidence interval.

 

In this case,\(\sigma \)is known and \({\rm{n}} = 61\), which means n>30.So, normal distribution applies here.

Step 4: Find the critical value

The \({z_{\frac{\alpha }{2}}}\)is a z score which separates an area of \(\frac{\alpha }{2}\) in the right tail of the standard normal distribution.

 

The confidence level 99% corresponds to\(\alpha  = 0.01\,\,\,\,{\rm{and}}\,\,\,\,\frac{\alpha }{2} = 0.005\). 

For critical value\({z_{\frac{\alpha }{2}}}\), the cumulative area to its left is\(1 - \frac{\alpha }{2}\).

Mathematically,

 

\(\begin{array}{c}P\left( {z < {z_{\frac{\alpha }{2}}}} \right) = 1 - \frac{\alpha }{2}\\ = 0.995\end{array}\)

 

From the standard normal table, the area of 0.995 is observed corresponding intersection of the row value 2.5 and column value 0.08, which implies\({z_{\frac{\alpha }{2}}}\)is 2.575.