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-value7. Old Faithful. The claim is that for the duration times (sec) of eruptions of the Old Faithful geyser, the mean is $\mu =240$sec. The sample size is n = 6 and the test statistic is t = 1.340.

The claim states that the mean duration time of eruptions is 240 sec. The sample is $n=6$, and the test statistic is $t=1.340$.

The claim has an equality statement that will be the null hypothesis, and the alternate hypothesis will be the opposite of this.

Thus, the hypotheses are as follows.

$$\begin{gathered} H_0:\mu =240\sec \\ {H}_1:\mu \ne 240\sec \end{gathered}$$

Here, $\mu$ is the population mean time (sec) of eruptions of the old faithful geyser.

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 the level of significance $\alpha$.

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

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

In the given problem, the test-statistic is 1.340. The sample size is , and the degree of freedom of the t-distribution is

$$\begin{gathered} df=n-1 \\ =6-1 \\ =5 \end{gathered}$$

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

In the table, look for the closest bounds of the test statistic value in the row with a degree of freedom 5 for the two-tailed test.

In row 5, the test statistic value 1.340 is less than 1.476 (corresponding to 0.20 level for the two-tailed test).

Thus, the P-value will be greater than 0.20 and hence expressed as $\text{P - value}>0.20$.