Chapter 6.3, Problem 12E

The following MINITAB output presents the results of a hypothesis test for a population proportion p.

Test and CI for One Proportion : X
Test of p = 0.4 vs p < 0.4
Variable X	X 73	N 240	Sample p 0.304167	95% Upper Bound 0.353013	Z-Value –3.03	P-Value 0.001

1. a. Is this a one-tailed or two-tailed test?
2. b. What is the null hypothesis?
3. c. Can H₀ be rejected at the 2% level? How can you tell?
4. d. Someone asks you whether the null hypothesis H₀ : p ≥ 0.45 versus H₁ : p > 0.45 can be rejected at the 2% level. Can you answer without doing any calculations? How?
5. e. Use the output and an appropriate table to compute the P-value for the test of H₀ : p ≤ 0.25 versus H₁: p > 0.25.
6. f. Use the output and an appropriate table to compute a 90% confidence interval for p.

To determine

Identify whether the hypotheses test is a one tailed or two tailed test.

Answer to Problem 12E

The test is a one tailed test.

Explanation of Solution

Given info:

The MINITAB output represents the results of a hypothesis test for a population proportion.

Justification:

From the given MINITAB output, the alternative hypothesis represents the less than symbol. Therefore, the form of alternative hypothesis is $p<p_0$. Hence, the test is a one tailed test.

To determine

Find the null hypothesis.

Explanation of Solution

Justification:

From the MINITAB output, the form of alternative hypothesis is $p<p_0$. Therefore, the form of null hypothesis is $p\ge p_0$.

Thus, the null hypothesis is $H_0:p\ge 0.4$

To determine

Check whether the null hypothesis is rejected at the 2% level or not.

Answer to Problem 12E

Yes, the null hypothesis can be rejected at the 2% level.

Explanation of Solution

Calculation:

From the given MINITAB output, the test statistic is –3.03 and the P-value is 0.001.

Decision rule:

If $P\text{-value}\le \alpha \text{,}$ then reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}>\alpha \text{,}$ then fail to reject the null hypothesis $\left(H_0\right)$.

Conclusion:

Here, the P-value is less than the level of significance 0.02.

That is $P\text{-value}\left(=0.001\right)<\alpha \left(=0.02\right)$.

Therefore, the null hypothesis is rejected.

To determine

Check whether the null hypothesis $H_0:p\ge 0.45$ versus $H_1:p<0.45$ can be rejected at the 2% level or not.

Answer to Problem 12E

Yes, the null hypothesis can be rejected at the 2% level.

Explanation of Solution

Calculation:

From the given MINITAB output, the sample proportion is 0.304167.

Thus, the P-value for $H_0:p\ge 0.45$ is smaller than the P-value for $H_0:p\ge 0.40$. That is, the P-value for $H_0:p\ge 0.45$ is smaller than 0.001. Therefore, the null hypothesis can be rejected at the 2% level.

To determine

Find the P-value for the test of $H_0:p\le 0.25$ versus $H_1:p>0.25$.

Answer to Problem 12E

The P-value is 0.026.

Explanation of Solution

Calculation:

Test statistic and P-value:

Software Procedure:

Step-by-step procedure to obtain the z-score using the MINITAB software:

• Choose Stat > Basic Statistics > 1 Proportion.
• Choose Summarized data.
• In Number of events, enter 73. In Number of trials, enter 240.
• Check Perform hypothesis test. In Hypothesized proportion, enter 0.25.
• Click Options. Under Alternative, and choose less than.
• Click OK in each dialog box.

Output using the MINITAB software is given below:

From the MINITAB output, the z-score is 1.94 and the P-value is 0.026.

To determine

Find the 90% confidence interval for p.

Answer to Problem 12E

The 90% confidence interval is (0.2588, 0.3560).

Explanation of Solution

Calculation:

The formula for confidence interval is as follows:

$\tilde{p}\pm z_{\frac{\alpha }{2}}\sqrt{\frac{\tilde{p}\left(1-\tilde{p}\right)}{\tilde{n}}}$

Where, $\tilde{n}=n+4$ and $\tilde{p}=\frac{X+2}{\tilde{n}}$.

The value of $\tilde{p}$ is as follows:

$\begin{array}{c}\tilde{p}=\frac{73+2}{240+4}\\ =\frac{75}{244}\\ =\text{0}\text{.307377}\end{array}$

Critical value:

Software procedure:

Step-by-step procedure to obtain the critical value using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose Probability Value and Both Tail for the region of the curve to shade.
• Enter the Probability value as 0.10.
• Click OK.

Output using the MINITAB software is given below:

Thus, $z_{\frac{\alpha }{2}}=\pm 1.645$.

The 90% confidence interval is as follows:

$\begin{array}{c}0.307377\pm 1.645\sqrt{\frac{0.307377\left(1-0.307377\right)}{244}}=0.307377\pm 1.645\left(\text{0}\text{.0295386}\right)\\ =0.307377\pm \text{0}\text{.048591}\\ \text{=}\left(0.307377-\text{0}\text{.048591,}0.307377+\text{0}\text{.048591}\right)\\ =\left(0.2588,0.3560\right)\end{array}$

Thus, the 90% confidence interval is (0.2588, 0.3560).