Six Steps for One-Sample Proportion Hypothesis Test
Steps 1-3 Section
Let's apply the general steps for hypothesis testing to the specific case of testing a one-sample proportion.
- Step 1: Set up the hypotheses and check conditions.
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\( np_0\ge 5 \) and \(n(1−p_0)≥5 \)
One Proportion Z-test Hypotheses
Left-Tailed- \( H_0\colon p=p_0 \)
- \( H_a\colon p<p_0\)
Right-Tailed- \( H_0\colon p=p_0 \)
- \( H_a\colon p>p_0 \)
Two-Tailed- \( H_0\colon p=p_0 \)
- \( H_a\colon p\ne p_0 \)
- Step 2: Decide on the level of significance \(\boldsymbol{(\alpha)}\).
- Step 3: Calculate the test statistic.
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One Proportion Z-test: \(z^*=\dfrac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \)
Rejection Region Approach
Steps 4-6 Section
- Step 4: Find the appropriate critical values for the tests. Write down clearly the rejection region for the problem.
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View the critical values and regions with an \(\alpha=.05\).
- Step 5: Make a decision about the null hypothesis.
- Check to see if the value of the test statistic falls in the rejection region. If it does, then reject \(H_0 \) (and conclude \(H_a \)). If it does not fall in the rejection region, do not reject \(H_0 \).
- Step 6: State an overall conclusion.
P-Value Approach
Steps 4-6 Section
- Step 4: Compute the appropriate p-value based on our alternative hypothesis.
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Left-Tailed
- \(P(Z \le z^*)\)
Right-Tailed- \(P(Z\ge z^*)\)
Two-Tailed- \(2\) x \(P(Z \ge |z^*|)\)
- Step 5: Make a decision about the null hypotheses.
- If the p-value is less than the significance level, then reject the null hypothesis. If the p-value is greater than the significance level, fail to reject the null hypothesis.
- Step 6: State an overall conclusion.
Example 6-5: Penn State Students from Pennsylvania Section
Referring back to example 6-4. Say we take a random sample of 500 Penn State students and find that 278 are from Pennsylvania. Can we conclude that the proportion is larger than 0.5 at a 5% level of significance?
Conduct the test using both the rejection region and p-value approach.
- Step 1: Set up the hypotheses and check conditions.
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Set up the hypotheses. Since the research hypothesis is to check whether the proportion is greater than 0.5 we set it up as a one (right)-tailed test:
\( H_0\colon p=0.5 \) vs \(H_a\colon p>0.5 \)
Can we use the z-test statistic? The answer is yes since the hypothesized value \(p_0 \) is \(0.5\) and we can check that: \(np_0=500(0.5)=250 \ge 5 \) and \(n(1-p_0)=500(1-0.5)=250 \ge 5 \)
- Step 2: Decide on the significance level, \(\alpha \).
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According to the question, \(\alpha= 0.05 \).
- Step 3: Calculate the test statistic:
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\begin{align} z^*&= \dfrac{0.556-0.5}{\sqrt{\frac{0.5(1-0.5)}{500}}}\\z^*&=2.504 \end{align}
Rejection Region Approach
- Step 4: Find the appropriate critical values for the test using the z-table. Write down clearly the rejection region for the problem.
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We can use the standard normal table to find the value of \(Z_{0.05} \). From the table, \(Z_{0.05} \) is found to be \(1.645\) and thus the critical value is \(1.645\). The rejection region for the right-tailed test is given by:
\( z^*>1.645 \)
- Step 5: Make a decision about the null hypothesis.
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The test statistic or the observed Z-value is \(2.504\). Since \(z^*\) falls within the rejection region, we reject \(H_0 \).
- Step 6: State an overall conclusion.
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With a test statistic of \(2.504\) and critical value of \(1.645\) at a 5% level of significance, we have enough statistical evidence to reject the null hypothesis. We conclude that a majority of the students are from Pennsylvania.
P-Value Approach
- Step 4: Compute the appropriate p-value based on our alternative hypothesis:
- \(\text{p-value}=P(Z\ge z^*)=P(Z \ge 2.504)=0.0062\)
- Step 5: Make a decision about the null hypothesis.
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Since \(\text{p-value} = 0.0062 \le 0.05\) (the \(\alpha \) value), we reject the null hypothesis.
- Step 6: State an overall conclusion.
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With a test statistic of \(2.504\) and p-value of \(0.0062\), we reject the null hypothesis at a 5% level of significance. We conclude that a majority of the students are from Pennsylvania.
Try it!
Online Purchases Section
An e-commerce research company claims that 60% or more graduate students have bought merchandise online. A consumer group is suspicious of the claim and thinks that the proportion is lower than 60%. A random sample of 80 graduate students shows that only 22 students have ever done so. Is there enough evidence to show that the true proportion is lower than 60%?
Conduct the test at 10% Type I error rate and use the p-value and rejection region approaches.
- Step 1: Set up the hypotheses and check conditions.
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Set up the hypotheses. Since the research hypothesis is to check whether the proportion is less than 0.6 we set it up as a one (left)-tailed test:
\( H_0\colon p=0.6 \) vs \(H_a\colon p<0.6 \)
Can we use the z-test statistic? The answer is yes since the hypothesized value \(p_0 \) is 0.6 and we can check that: \(np_0=80(0.6)=48 \ge 5 \) and \(n(1-p_0)=80(1-0.6)=32 \ge 5 \)
- Step 2: Decide on the significance level, \(\alpha \).
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According to the question, \(\alpha= 0.1 \).
- Step 3: Calculate the test statistic:
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\begin{align} z^* &=\frac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\\&=\frac{.275-0.6}{\sqrt{\frac{0.6(1-0.6)}{80}}}\\&=-5.93 \end{align}
Rejection Region Approach
- Step 4: Find the appropriate critical values for the test using the z-table. Write down clearly the rejection region for the problem.
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The critical value is the value of the standard normal where 10% fall below it. Using the standard normal table, we can see that the value is -1.28.
- Step 5: Make a decision about the null hypothesis.
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The rejection region is any \(z^* \) such that \(z^*<-1.28 \) . Since our test statistic, -5.93, is inside the rejection region, we reject the null hypothesis.
- Step 6: State an overall conclusion.
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There is enough evidence in the data provided to suggest, at 10% level of significance, that the true proportion of students who made purchases online was less than 60%.
P-Value Approach
- Step 4: Compute the appropriate p-value based on our alternative hypothesis:
- \( \text{p-value}=P(Z \le -5.93) = 0.0000000003 \)
- Step 5: Make a decision about the null hypothesis.
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Since our p-value is very small and less than our significance level of 10%, we reject the null hypothesis.
- Step 6: State an overall conclusion.
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There is enough evidence in the data provided to suggest, at 10% level of significance, that the true proportion of students who made purchases online was less than 60%.