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 $t_{\frac{\alpha }{2}}$,(b) find the critical value $z_{\frac{\alpha }{2}}$,or (c) state that neither the normal distribution nor the t distribution applies.

Denver Bronco Salaries confidence level is 90%, $\sigma$is not known, and the histogram of 61 player salaries (thousands of dollars) is as shown.

The histogram shows salaries (in thousands of dollars) of 61 players, that means sample size n is equal to 61.Denver Bronco Salaries confidence level is 90% with unknown$\sigma$.

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.

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

If$\sigma$is not known androle="math" localid="1647515348481" $n>30$(or distribution is known to be normal) then t-distribution is suitable to find the confidence interval.

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

The histogram is non-normal but the sample size 61 is greater than 30. In this case,is unknown.

Thus, t-distribution applies.

The critical value $t_{\frac{\alpha }{2}}$requires a value for the degrees of freedom.

The degree of freedom is calculated as,

$$\begin{gathered} \text{degree} \text{of} \text{freedom} =n-1 \\ =61-1 \\ =60 \end{gathered}$$

The 90% confidence level corresponds to $\alpha =0.10$, so there is an area of 0.05 in each of the two tails of the t-distribution.

Thus,

$$\begin{gathered} P\left(t>t_{\frac{\alpha }{2}}\right)=\frac{\alpha }{2} \\ P\left(t>t_{0.05}\right)=0.05 \end{gathered}$$

Referring to Table A-3 critical value of t-distribution, the critical value of$t_{\frac{\alpha }{2}}=t_{0.05}$is obtained from the intersection of column with 0.10 for the “Area in Two Tails” (or use the same column with 0.05 for the “Area in One Tail”)and the row value number of degrees of freedom is 60, which is 1.67.

Thus, $P\left(t>1.67\right)=0.05$.