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When constructing a two sample proportions hypothesis test the assumptions and conditions must be met in order to use the normal model.
Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populationsor it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions $({\hat{p}}_{A}-{\hat{p}}_{B})$ reflects a difference in the population proportions.
The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, ${H}_{o}:{p}_{A}={p}_{B}$ . To conduct the test, we use a pooled proportion, ${\hat{p}}_{\mathrm{pooled}}$ .
Two types of medication for hives are being tested to determine if there is a medication A is more effective than medication B. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.
This is a test of 2 population proportions.
How do you know?
The problem asks for a difference in proportions.
Let $A$ and $B$ be the subscripts for medication A and medication B. Then ${p}_{A}$ and ${p}_{B}$ are the desired population proportions.
${H}_{o}:{p}_{A}={p}_{B}\phantom{\rule{25pt}{0ex}}$ or $\phantom{\rule{25pt}{0ex}}{p}_{A}-{p}_{B}=0$
${H}_{a}:{p}_{A}>{p}_{B}\phantom{\rule{25pt}{0ex}}$ or $\phantom{\rule{25pt}{0ex}}{p}_{A}-{p}_{B}>0$
The words "is more effective" tell you the test is one-tailed.
Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: ${p}_{c}={\hat{p}}_{\mathrm{pooled}}$
${\hat{p}}_{\mathrm{pooled}}=\frac{{x}_{A}+{x}_{B}}{{n}_{A}+{n}_{B}}=\frac{20+12}{200+200}=0.08\phantom{\rule{12pt}{0ex}}1-{\hat{p}}_{\mathrm{pooled}}=0.92$
Therefore, $\phantom{\rule{10pt}{0ex}}{\hat{p}}_{A}-{\hat{p}}_{B}~N[0,\sqrt{\left(0.08\right)\cdot \left(0.92\right)\cdot (\frac{1}{200}+\frac{1}{200})}]$
${\hat{p}}_{A}-{\hat{p}}_{B}$ follows an approximate normal distribution.
Calculate the p-value using the normal distribution: p-value = 0.1404.
Estimated proportion for group A: $\phantom{\rule{12pt}{0ex}}{\hat{p}}_{A}=\frac{{x}_{A}}{{n}_{A}}=\frac{20}{200}=0.1$
Estimated proportion for group B: $\phantom{\rule{12pt}{0ex}}{\hat{p}}_{B}=\frac{{x}_{B}}{{n}_{B}}=\frac{12}{200}=0.06$
${\hat{p}}_{A}-{\hat{p}}_{B}=0.1-0.06=0.04$ .
The p-value is above 0.1404.
Compare $\alpha $ and the p-value: $\alpha =0.01$ and the $\text{p-value}=0.1404$ . $\alpha <$ p-value.
Make a decision: Since $\alpha <\text{p-value}$ , do not reject ${H}_{o}$ .
Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adultpatients who did not react after 30 minutes to medication A and medication B.
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