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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.

\(\begin{array}{l}{H_{0\;}}:\mu = 240\sec \\{H_1}\;:\mu \; \ne \;240\;\sec \end{array}\)

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

The formula for the t-statistic is given below.

\(t = \frac{{\bar x - \mu }}{{\frac{s}{{\sqrt n }}}}\) .

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 the level of significance \(\alpha \).

\({\rm{If}}\;{\rm{P - value}}\; < \alpha \;\), reject the null hypothesis

If \({\rm{P - value}}\; > \alpha \), fail to reject the null hypothesis.

In the given problem, the test-statistic is 1.340. The sample size is \(n = 6\), and the degree of freedom of the t-distribution is

\(\)

\(\begin{array}{c}df\; = \;n - 1\\ = 6 - 1\\ = 5\end{array}\).

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 \({\rm{P - value}} > 0.20\).