Chapter 9.5, Problem 36E

a.

To determine

Find the p-value and state the conclusion.

Answer to Problem 36E

The p-value is 0.003.

The conclusion is “reject the null hypothesis”.

Explanation of Solution

Calculation:

The given information is that $n=300$ and $\overline{p}=0.68$.

The hypotheses are given below:

Null hypothesis: $H_0:p\ge 0.75$

Alternative hypothesis: $H_a:p<0.75$

Test statistic:

The formula for finding test statistic is as follows:

$z=\frac{\overline{p}-p_0}{\sqrt{\frac{p_0\left(1-p_0\right)}{n}}}$

Here, $\overline{p}$ represents the sample proportion, $p_0$ represents the hypothesized value of the population proportion, and n represents the sample size.

Substitute $\overline{p}=0.68$, $p_0=0.75$, and $n=300$ in z formula.

$\begin{array}{c}z=\frac{0.68-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{300}}}\\ =\frac{-0.07}{\sqrt{\frac{0.1875}{300}}}\\ =-\frac{0.07}{0.025}\\ =-2.80\end{array}$

Thus, the value of the test statistic is −2.80.

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

• Locate the value –2.8 in the first column.
• Locate the value 0.00 in the first row.
• The intersecting value that corresponds to–2.80 is 0.0026.

Thus, the p-value is 0.0026.

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, $p\text{-value}\left(=0.0026\right)<\alpha \left(=0.05\right)$.

By the rejection rule, the null hypothesis is rejected.

b.

To determine

Find the p-value and state the conclusion.

Answer to Problem 36E

The p-value is 0.1151.

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

Explanation of Solution

Calculation:

The given information is that $n=300$ and $\overline{p}=0.72$.

Test statistic:

Substitute $\overline{p}=0.72$, $p_0=0.75$, and $n=300$ in the z formula.

$\begin{array}{c}z=\frac{0.72-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{300}}}\\ =\frac{-0.03}{\sqrt{\frac{0.1875}{300}}}\\ =-\frac{0.03}{0.025}\\ =-1.20\end{array}$

Thus, the value of the test statistic is −1.20.

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

• Locate the value –1.2 in the first column.
• Locate the value 0.00 in the first row.
• The intersecting value that corresponds to the –1.20 is 0.1151.

Thus, the p-value is 0.1151.

Conclusion:

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

That is, $p\text{-value}\left(=0.1151\right)>\alpha \left(=0.05\right)$.

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

c.

To determine

Find the p-value and state the conclusion.

Answer to Problem 36E

The p-value is 0.0228.

The conclusion is “reject the null hypothesis”.

Explanation of Solution

Calculation:

The given information is that $n=300$ and $\overline{p}=0.70$.

Test statistic:

Substitute $\overline{p}=0.70$, $p_0=0.75$, and $n=300$ in the z formula.

$\begin{array}{c}z=\frac{0.70-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{300}}}\\ =\frac{-0.05}{\sqrt{\frac{0.1875}{300}}}\\ =-\frac{0.05}{0.025}\\ =-2.00\end{array}$

Thus, the value of the test statistic is −2.00.

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

• Locate the value –2.0 in the first column.
• Locate the value 0.00 in the first row.
• The intersecting value that corresponds to the –2.00 is 0.0228.

Thus, the p-value is 0.0228.

From the output, the test statistic is −2.00 and the p-value is 0.0228.

Conclusion:

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

That is, $p\text{-value}\left(=0.0228\right)<\alpha \left(=0.05\right)$.

By the rejection rule, the null hypothesis is rejected.

d.

To determine

Find the p-value and state the conclusion.

Answer to Problem 36E

The p-value is 0.7881.

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

Explanation of Solution

Calculation:

The given information is that $n=300$ and $\overline{p}=0.77$.

Test statistic:

Substitute $\overline{p}=0.77$, $p_0=0.75$, and $n=300$ in the z formula.

$\begin{array}{c}z=\frac{0.77-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{300}}}\\ =\frac{0.02}{\sqrt{\frac{0.1875}{300}}}\\ =\frac{0.02}{0.025}\\ =0.80\end{array}$

Thus, the value of the test statistic is 0.80.

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 0.00 in the first row.
• The intersecting value that corresponds to the 0.80 is 0.7881.

Thus, the p-value is 0.7881.

Conclusion:

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

That is, $p\text{-value}\left(=0.7881\right)>\alpha \left(=0.05\right)$.

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