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 Body Temperatures The claim is that for 12 am body temperatures, the mean is $\mu <98.6°F$.The sample size is n = 4 and the test statistic is t = -2.503.

The claim states that the mean body temperature is less than . The sample size is $n=4$, and the test statistic is $t=-2.503$.

The claim does not have any equality statement. So, it will be stated as the alternate hypothesis, and the null hypothesis will be that the mean body temperature is equal to $98.6°F$.

$$\begin{gathered} H_0:\mu =98.6°F \\ {H}_1:\mu <98.6°F \end{gathered}$$

Here, $\mu$ is the population mean body temperature at 12 am.

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 test statistic is -2.503.

The decision rule is stated as follows for $\alpha$level of significance.

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

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

In the given problem, the test-statistic is $-2.503$. The sample size is $n=4$, and the degree of freedom of t distribution is

$$\begin{gathered} df=n-1 \\ =4-1 \\ =3 \end{gathered}$$

In the t-distribution table, look for the range where the absolute value of t-statistic lies.

In the row with the degree of freedom 3, the values closest to 2.503 are 2.353 and 3.182, corresponding to 0.025 and 0.05 levels, respectively, for a one-tailed test.

Thus, the range for the P-value for t-statistic $t=-2.503$ is $0.025<\text{P - value}<0.05$