Finding P-values. In Exercises 5–8, either use technology to find the P-value or use Table A-3 to find a range of values for the P-value.

8. Tornadoes. The claim is that for the widths (yd) of tornadoes, the mean is $\mu <140$ yd. The sample size is n = 21 and the test statistic is t = -0.024.

The claim states that the mean widths of tornadoes are less than 140 yd.

The sample size is $n=21$, and the test-statistics is $t=-0.024$

The claim states a non-equality statement. So, it will be the alternate hypothesis, and the null hypothesis will be that the mean width (yd) of tornadoes is equal to (yd).

Thus, the hypotheses are stated as follows.

$$\begin{gathered} H_0:\mu =140 \\ {H}_1:\mu <140 \end{gathered}$$

Here, is the population mean width of the tornadoes.

The test is one-tailed.

The formula for the t-statistic is given below.

$t=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Here,

$$\begin{gathered} \overline{x}:\text{sample}\text{mean} \\ s:\text{sample}\text{stadard}\text{deviation} \\ \mu :\text{population}\text{mean} \\ n:\text{sample}\text{size} \end{gathered}$$

The decision rule is stated below for $\alpha$.

If $\text{P - value}<\alpha$, reject the null hypothesis.

If $\text{P - value}>\alpha$, fail to reject the null hypothesis.

In the given problem, the test statistic is $-0.024$. The sample size is $n=21$, and the degree of freedom of the t-distribution is

$$\begin{gathered} df=n-1 \\ =21-1 \\ =20 \end{gathered}$$

In the t-distribution table (Table A-3), look for the range where the t-statistic lies.

In the table, look for the closest bounds of the absolute test statistic value in the row with a degree of freedom of 20 for the one-tailed test.

In row 20, the test statistic value is 1.325 greater than 0.024 (corresponding to 0.10 level for the one-tailed test).

Thus, the P-value will be greater than 0.10 and hence expressed as $\text{P - value}>0.10$.