Chapter 9.4, Problem 26E

a.

To determine

Find the p-value and state the conclusion.

Answer to Problem 26E

The p-value is 0.039.

The conclusion is “reject the null hypothesis”.

Explanation of Solution

Calculation:

The given information is the sample size is $n=65$, the sample mean is $\overline{x}=103$, and the sample standard deviation is $s=11.5$.

The hypotheses are given below:

Null hypothesis: $H_0:\mu =100$

Alternative hypothesis: $H_a:\mu \ne 100$

Test statistic:

The formula for finding test statistic is as follows:

$t=\frac{\overline{x}-{\mu }_0}{\frac{s}{\sqrt{n}}}$

Here, $\overline{x}$ represents the sample mean, ${\mu }_0$ represents the hypothesized value of the population mean, s represents the sample standard deviation, and n represents the sample size.

Substitute $\overline{x}=103$, ${\mu }_0=100$, $s=11.5$ and $n=65$ in t formula.

$\begin{array}{c}t=\frac{103-100}{\frac{11.5}{\sqrt{65}}}\\ =\frac{3}{\frac{11.5}{8.0623}}\\ =\frac{3}{1.4264}\\ =2.10\end{array}$

Thus, the value of the test statistic is 2.10.

The degrees of freedom is,

$\begin{array}{c}\text{df}=n-1\\ =65-1\\ =64\end{array}$

Software Procedure:

Step-by-step procedure to obtain the probability value using Excel:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =T.DIST.2T(2.10,64) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

From the output, the exact p-value is approximately 0.040.

Rejection rule:

If $p\text{-value}\le \alpha$, reject the null hypothesis.

If $p\text{-value}>\alpha$, do not reject the null hypothesis.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.040\right)<\alpha \left(=0.05\right)$.

By the rejection rule, the null hypothesis is rejected.

b.

To determine

Calculate the p-value and draw the conclusion.

Answer to Problem 26E

The p-value is 0.013.

The conclusion is “reject the null hypothesis”.

Explanation of Solution

Calculation:

The given information is that the sample size is $n=65$, the sample mean is $\overline{x}=96.5$, and the sample standard deviation is $s=11$.

Test statistic:

Substitute $\overline{x}=96.5$, ${\mu }_0=100$, $s=11$, and $n=65$ in t formula.

$\begin{array}{c}t=\frac{96.5-100}{\frac{11}{\sqrt{65}}}\\ =\frac{-3.5}{\frac{11}{8.0623}}\\ =-\frac{3.5}{1.3644}\\ =-2.57\end{array}$

Thus, the value of the test statistic is –2.57.

From part (a), the degrees of freedom is 64.

Software Procedure:

Step-by-step procedure to obtain the probability value using Excel:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =T.DIST.2T(2.57,64) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

From the output, the exact p-value is approximately 0.0125.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.0125\right)<\alpha \left(=0.05\right)$.

By the rejection rule, the null hypothesis is rejected.

c.

To determine

Obtain the p-value and claim the conclusion.

Answer to Problem 26E

The p-value is 0.130.

The conclusion is “do not reject the null hypothesis”.

Explanation of Solution

Calculation:

The given information is that the sample size is $n=65$, the sample mean is $\overline{x}=102$, and the sample standard deviation is $s=10.5$.

Test statistic:

Substitute $\overline{x}=102$, ${\mu }_0=100$, $s=10.5$ and $n=65$ in t formula.

$\begin{array}{c}t=\frac{102-100}{\frac{10.5}{\sqrt{65}}}\\ =\frac{2}{\frac{10.5}{8.0623}}\\ =\frac{2}{1.3024}\\ =1.54\end{array}$

Thus, the value of the test statistic is 1.54.

From part (a), the degrees of freedom is 64.

Software Procedure:

Step-by-step procedure to obtain the probability value using Excel:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =T.DIST.2T(1.54,64) in the cell A1.
• Press Enter.

Output using the EXCEL software is given below:

From the output, the exact p-value is approximately 0.01285.

Conclusion:

Here, the p-value is greater than the level of significance.

That is, $p\text{-value}\left(=0.1285\right)>\alpha \left(=0.05\right)$.

By the rejection rule, the null hypothesis is not rejected.