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\[140\](yd).

Thus, the hypotheses are stated as follows.

\[\begin{array}{l}{H_{0\;}}:\mu = 140\;\\{H_1}\;:\;\mu \; < 140\end{array}\]

Here,\[\mu \]is the population mean width of the tornadoes.

The test is one-tailed.

The formula for the t-statistic is given below.

\[t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\] .

Here,

\[\begin{array}{l}\bar x\;:\;{\rm{sample}}\;{\rm{mean}}\\s\;:\;{\rm{sample}}\;{\rm{stadard}}\;{\rm{deviation}}\\\mu \;:\;{\rm{population}}\;{\rm{mean}}\\n\;:{\rm{sample}}\;{\rm{size}}\end{array}\]

The decision rule is stated below for\[\alpha \].

If \[{\rm{P - value}}\; < \alpha \;\], reject the null hypothesis.

If \[{\rm{P - value}}\; > \alpha \;\], 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

\[\]

\[\begin{array}{c}df\; = \;n - 1\\ = 21 - 1\\ = 20\end{array}\].

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 \({\rm{P - value}} > 0.10\).