Two-Sample Test for Proportions: Apply It 1

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Two-Sample Test for Proportions: Apply It 1

Recognize when a one-sample [latex]z[/latex]-test or a two-sample [latex]z[/latex]-test is needed to answer a research question.
Complete a two-sample [latex]z[/latex]-test for proportions from hypotheses to conclusions.

Two-Sample Hypothesis Test for Proportions

As mentioned before, hypothesis testing is part of inference. We previously stated that the purpose of a hypothesis test is to use sample data to test a claim about a population parameter. However, studies often compare two groups. Comparing two proportions is quite common. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.

two-sample [latex]z[/latex]-test of proportions

Write out the null and alternative hypotheses.
- Null hypothesis: [latex]H_0: p_1=p_2[/latex] or [latex]H_0: p_1-p_2=0[/latex]
- Alternative hypothesis:
  - [latex]H_A: p_1\lt p_2[/latex] or [latex]H_A: p_1-p_2\lt 0[/latex]
  - [latex]H_A: p_1>p_2[/latex] or [latex]H_A: p_1-p_2>0[/latex]
  - [latex]H_A: p_1\ne p_2[/latex] or [latex]H_A: p_1-p_2\ne0[/latex]
Check the conditions for the hypothesis test. For testing a two-sample [latex]z[/latex]-test for proportions, we require:
- Large Counts: For [latex]\hat{p}_c = \frac{x_1+x_2}{n_1+n_2}[/latex], check that: [latex]n_1\hat{p}_c \ge 10[/latex], [latex]n_2\hat{p}_c \ge 10[/latex], [latex]n_1(1-\hat{p}_c) \ge 10[/latex], and [latex]n_2(1-\hat{p}_c) \ge 10[/latex].
- Random Samples/Assignment: Check that the two samples are independent and random samples or that they come from randomly assigned groups in an experiment.
- 10%: Check that [latex]n_1<0.10(N_1)[/latex] and [latex]n_2<0.10(N_2)[/latex].
Calculate a test statistic.
Calculate a P-value.

Compare the P-value to the significance level, [latex]\alpha[/latex], to make a decision.

Decision	Conclusion
If P-value [latex]\le\alpha[/latex], there is enough evidence to reject the null hypothesis.	At the [latex]\alpha\times[/latex]100% significance level, the data provide convincing evidence in support of the alternative hypothesis.
If P-value [latex]\gt\alpha[/latex], there is not enough evidence to reject the null hypothesis.	At the [latex]\alpha\times[/latex]100% significance level, the data do not provide convincing evidence in support of the alternative hypothesis.

Write a conclusion in context (e.g., we do/do not have convincing evidence…).

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Vaccine Efficiency^[1]In 2020, after a trial showed that Pfizer’s coronavirus vaccine had an efficacy rate of [latex]95\%[/latex], the Food and Drug Administration approved the vaccine.

The [latex]43,660[/latex] subjects were split evenly between the placebo and vaccine groups (about [latex]21,830[/latex] subjects per group). In the placebo group — the group that got a “fake” vaccine — [latex]162[/latex] became infected with the coronavirus and showed symptoms. In the vaccine group — the group that got the real vaccine — that number was only eight. At the [latex]5\%[/latex] significance level, is the vaccine group’s infection risk that different from the placebo group’s?

Step 1: Write out the hypotheses.

[latex]H_0: p_p = p_v[/latex] or [latex]H_0: p_p - p_v=0[/latex]
[latex]H_A: p_p \ne p_v[/latex] or [latex]H_A: p_p - p_v\ne0[/latex]

where [latex]p_p[/latex] is the proportion of the placebo subjects that became infected and [latex]p_v[/latex] is the proportion of the vaccine subjects that became infected.

Step 2: Check the conditions.

The sample is random, so there is no problem there. Again, we want to determine whether the normal model is a good fit.

[latex]\hat{p}_c = \frac{162+8}{21830+21830} = \frac{170}{43660}[/latex]

[latex]n_p\hat{p}_c = 21830(\frac{170}{43660})= 85\ge 10[/latex],
[latex]n_v\hat{p}_c = 21830(\frac{170}{43660})= 85\ge 10[/latex],
[latex]n_p(1-\hat{p}_c) = 21830(1-\frac{170}{43660})= 21745\ge 10[/latex], and
[latex]n_v(1-\hat{p}_c) =21830(1-\frac{170}{43660})= 21745\ge 10[/latex].

We can use the normal model to find the P-value.

Step 3: Calculate the test statistic.

[latex]p_p = \frac{162}{21830} \approx 0.74\%[/latex]
[latex]p_v =\frac{8}{21830} \approx 0.04\%[/latex]

We can use our statistical tool to calculate our test statistic.

Appropriate alternative text can be found in the description above.

[latex]z=11.8[/latex]

The difference in the proportion is about [latex]11.8[/latex] standard errors above the null hypothesis value ([latex]0[/latex]).

Step 4: Calculate the P-value.

Using our statistical tool, the P-value is [latex]0[/latex].

Step 5: Compare the P-value with the significance level.

Now we compare the P-value to the level of significance, [latex]\alpha = 0.05[/latex]. In this case, the P-value of [latex]0[/latex] is less than [latex]0.05[/latex], which means we do have enough evidence to reject the null hypothesis.

Step 6: Write the conclusion in the context of the problem.

The data from this study does provide convincing evidence that the vaccine group’s infection risk is different from the placebo group’s.

Fun Fact: How did Pfizer get an efficacy rate of [latex]95\%[/latex]?

Efficacy Rate = [latex]\frac{p_p-p_v}{p_p} =\frac{0.74\%-0.04\%}{0.74\%}=\frac{0.7\%}{0.74\%} \approx 95\%[/latex]

https://www.nytimes.com/2020/12/13/learning/what-does-95-effective-mean-teaching-the-math-of-vaccine-efficacy.html ↵