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.\( np_0\ge 5 \) 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.
One Proportion Z-test: \(z^*=\dfrac{\hat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \)
The first few steps (Step 1 - Step 3) are exactly the same as the rejection region or p-value approach. The next part will discuss steps 4 - 6 for both approaches.
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.
- Left-Tailed Test
- Right-Tailed Test
- Two-Tailed Test
𝛼z𝛼Normal curve with a left tailed test shaded.
Reject \(H_0\) if \(z^* \le z_\alpha\)
𝛼1-𝛼z1-𝛼Normal curve with a right tailed test shaded.
Reject \(H_0\) if \(z^* \ge z_{1-\alpha}\)
𝛼/21-𝛼z1-𝛼/2𝛼/2z𝛼/2Normal curve with a two-tailed test shaded
Reject \(H_0\) if \(|z^*| \ge |z_{\alpha/2}|\)
View the critical values and regions with an \(\alpha=.05\).
Critical Values for \(\alpha=.05\)
These graphs show the various z-critical values for tests at an \(\alpha=.05\). *The graphs are not to scale.
Left-Tailed Test
.05 -1.65 Normal curve with a left tailed test shaded.Reject \(H_0\) if \(z^* \le -1.65\)
Right-Tailed Test
.05 .95 1.65Reject \(H_0\) if \(z^* \ge 1.65\)
Two-Tailed Test
.025 .95 .025 1.96 -1.96Reject \(H_0\) if \(|z^*| \ge |-1.96|\)
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.
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.Note! Recall that the P-value is a probability of obtaining a value of the test statistic or a more extreme value of the test statistic assuming that the null hypothesis is true.
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.
Answer
- Steps 1-3
- Steps 4-6: Rejection Region
- Steps 4-6: P-Value
Step 1: Set up the hypotheses and check conditions.
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, \(\alpha \).According to the question, \(\alpha= 0.05 \).
Step 3: Calculate the test statistic:\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.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.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.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.
Since \(\text{p-value} = 0.0062 \le 0.05\) (the \(\alpha \) value), we reject the null hypothesis.
Step 6: State an overall conclusion.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.
Answer
- Steps 1-3
- Steps 4-6: Rejection Region
- Steps 4-6: P-Value
Step 1: Set up the hypotheses and check conditions.
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, \(\alpha \).According to the question, \(\alpha= 0.1 \).
Step 3: Calculate the test statistic:\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.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.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.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.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.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%.
- Previous6a.4.2 - More on the P-Value and Rejection Region Approach
- Next6a.5 - Relating the CI to a Two-Tailed Test