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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$蟽$.

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$蟽$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 $蟽$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 $t_{\frac{伪}{2}}$requires a value for the degrees of freedom.

The degree of freedom is calculated as,

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

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

Thus,

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

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

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