Chapter 9.3, Problem 14E

a.

To determine

Find the p-value for $\overline{x}=23$ and state the conclusion.

Answer to Problem 14E

The p-value is 0.3844.

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

Explanation of Solution

Calculation:

It is given that the sample size is $n=75$, the sample mean is $\overline{x}=23$, and the population standard deviation is $\sigma =10$.

The hypotheses are given below:

Null hypothesis: $H_0:\mu =22$

Alternative hypothesis: $H_a:\mu \ne 22$

Test statistic:

The formula for finding test statistic is as follows:

$z=\frac{\overline{x}-{\mu }_0}{\frac{\sigma }{\sqrt{n}}}$

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

Substitute 23 for $\overline{x}$, 22 for ${\mu }_0$, 10 for $\sigma$, and 75 for n in $z$ formula.

$\begin{array}{c}z=\frac{23-22}{\frac{10}{\sqrt{75}}}\\ =\frac{1}{\frac{10}{8.6602}}\\ =\frac{1}{1.1547}\\ =0.87\end{array}$

Thus, the value of the test statistic is 0.87.

For p-value:

In this case, z is greater than 0. Therefore, the p-value is two times the upper tail area.

That is,

$\begin{array}{c}p\text{-value}=2\left(\text{Upper tail area}\right)\\ =2\left(1-\text{Lower tail area}\right)\end{array}$

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value 0.8 in the first column.
• Locate the value 0.07 in the first row.
• The intersecting value that corresponds to 0.87 is 0.8078.

$\begin{array}{c}p\text{-value}=2\left(1-0.8078\right)\\ =2\left(0.1922\right)\\ =0.3844\end{array}$

Thus, the p-value is 0.3844.

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 greater than the level of significance.

That is, $p\text{-value}\left(=0.3844\right)>\alpha \left(=0.01\right)$.

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

b.

To determine

Find the p-value for $\overline{x}=25.1$ and state the conclusion.

Answer to Problem 14E

The p-value is 0.0074.

The conclusion is “reject the null hypothesis”.

Explanation of Solution

Calculation:

Test statistic:

Substitute 25.1 for $\overline{x}$, 22 for ${\mu }_0$, 10 for $\sigma$, and 75 for n in $z$ formula.

$\begin{array}{c}z=\frac{25.1-22}{\frac{10}{\sqrt{75}}}\\ =\frac{3.1}{\frac{10}{8.6602}}\\ =\frac{3.1}{1.1547}\\ =2.68\end{array}$

Thus, the value of the test statistic is 2.68.

For p-value:

In this case, z is greater than 0. Therefore, the p-value is two times the upper tail area.

That is,

$\begin{array}{c}p\text{-value}=2\left(\text{Upper tail area}\right)\\ =2\left(1-\text{Lower tail area}\right)\end{array}$

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value 2.6 in the first column.
• Locate the value 0.08 in the first row.
• The intersecting value that corresponds to 2.68 is 0.9963.

$\begin{array}{c}p\text{-value}=2\left(1-0.9963\right)\\ =2\left(0.0037\right)\\ =0.0074\end{array}$

Thus, the p-value is 0.0074.

Conclusion:

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

That is, $p\text{-value}\left(=0.0074\right)<\alpha \left(=0.01\right)$.

By the rejection rule, the null hypothesis is rejected.

c.

To determine

Find the p-value for $\overline{x}=20$ and state the conclusion.

Answer to Problem 14E

The p-value is 0.083.

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

Explanation of Solution

Calculation:

Test statistic:

Substitute 20 for $\overline{x}$, 22 for ${\mu }_0$, 10 for $\sigma$, and 75 for n in $z$ formula.

$\begin{array}{c}z=\frac{20-22}{\frac{10}{\sqrt{75}}}\\ =\frac{-2}{\frac{10}{8.6602}}\\ =\frac{-2}{1.1547}\\ =-1.73\end{array}$

Thus, the value of the test statistic is –1.73.

For p-value:

In this case, z is less than 0. Therefore, the p-value is two times the lower tail area.

That is,

$p\text{-value}=2\left(\text{Lower tail area}\right)$

Use Table 1: Cumulative probabilities for the standard normal distribution to find probability.

• Locate the value –1.7 in the first column.
• Locate the value 0.03 in the first row.
• The intersecting value that corresponds to–1.73 is 0.0418.

$\begin{array}{c}p\text{-value}=2\left(0.0418\right)\\ =0.0836\end{array}$

Thus, the p-value is 0.0836.

Conclusion:

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

That is, $p\text{-value}\left(=0.0836\right)>\alpha \left(=0.01\right)$.

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