# 10.3 Hypothesis testing: two population means and two population

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## Assumptions and conditions: two sample proportions test

When constructing a two sample proportions hypothesis test the assumptions and conditions must be met in order to use the normal model.

• Randomization Condition: The data must be sampled randomly. Is one of the good sampling methodologies discussed in the Sampling and Data chapter being used?
• Independence Assumption: The sample values must be independent of each other. This means that the occurrence of one event has no influence on the next event. Usually, if we know that people or items were selected randomly we can assume that the independence assumption is met.
• 10% Condition: When the sample is drawn without replacement (usually the case), the sample size, n, should be no more than 10% of the population. This must be true for both groups.
• Success/ failure Condition: must be met for both groups. When working with proportions we need to be especially concerned about sample size when the proportion is close to zero or one. To check that the sample size is large enough calculate the successes by multiplying the sample percentages by the sample sizes and calculate failure by multiplying one minus the sample percentages by the sample sizes. If all of these products are larger than ten then the condition is met.
• ${\stackrel{̂}{p}}_{1}{\left(n}_{1}\right)$ >10 and ${\left(1-\stackrel{̂}{p}}_{1}\right){\left(n}_{1}\right)$ >10 and
• ${\stackrel{̂}{p}}_{2}{\left(n}_{2}\right)$ >10 and ${\left(1-\stackrel{̂}{p}}_{2}\right){\left(n}_{2}\right)$ >10
• Independent Groups: The two groups you are working with must be independent of one another. Is there reason to believe that one group would have influence on the other group? A hypothesis test that is comparing a group of respondent’s pre-test and post-test scores would not be independent. Likewise, groups made by splitting married couples into husband and wife groups would not be independent.

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 $\left({\stackrel{̂}{p}}_{A}-{\stackrel{̂}{p}}_{B}\right)$ 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, ${\stackrel{̂}{p}}_{\mathrm{pooled}}$ .

## The pooled proportion is calculated as follows:

${\stackrel{̂}{p}}_{\mathrm{pooled}}=\frac{{x}_{A}+{x}_{B}}{{n}_{A}+{n}_{B}}$

## The distribution for the differences is:

${\stackrel{̂}{p}}_{A}-{\stackrel{̂}{p}}_{B}~N\left[0,\sqrt{{\stackrel{̂}{p}}_{\mathrm{pooled}}·\left(1-{\stackrel{̂}{p}}_{\mathrm{pooled}}\right)·\left(\frac{1}{{n}_{A}}+\frac{1}{{n}_{B}}\right)}\right]$

## The test statistic (z-score) is:

$z=\frac{\left({\stackrel{̂}{p}}_{A}-{\stackrel{̂}{p}}_{A}\right)-\left({p}_{A}-{p}_{B}\right)}{\sqrt{{\stackrel{̂}{p}}_{\mathrm{pooled}}·\left(1-{\stackrel{̂}{p}}_{\mathrm{pooled}}\right)·\left(\frac{1}{{n}_{A}}+\frac{1}{{n}_{B}}\right)}}$

## Two population proportions

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.

## Determining the solution

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.

## Random variable:

${\stackrel{̂}{p}}_{A}-{\stackrel{̂}{p}}_{B}=$ difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B.

${H}_{o}:{p}_{A}={p}_{B}\phantom{\rule{25pt}{0ex}}$ 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: ${p}_{c}={\stackrel{̂}{p}}_{\mathrm{pooled}}$

${\stackrel{̂}{p}}_{\mathrm{pooled}}=\frac{{x}_{A}+{x}_{B}}{{n}_{A}+{n}_{B}}=\frac{20+12}{200+200}=0.08\phantom{\rule{12pt}{0ex}}1-{\stackrel{̂}{p}}_{\mathrm{pooled}}=0.92$

Therefore, $\phantom{\rule{10pt}{0ex}}{\stackrel{̂}{p}}_{A}-{\stackrel{̂}{p}}_{B}~N\left[0,\sqrt{\left(0.08\right)\cdot \left(0.92\right)\cdot \left(\frac{1}{200}+\frac{1}{200}\right)}\right]$

${\stackrel{̂}{p}}_{A}-{\stackrel{̂}{p}}_{B}$ follows an approximate normal distribution.

Calculate the p-value using the normal distribution: p-value = 0.1404.

Estimated proportion for group A: $\phantom{\rule{12pt}{0ex}}{\stackrel{̂}{p}}_{A}=\frac{{x}_{A}}{{n}_{A}}=\frac{20}{200}=0.1$

Estimated proportion for group B: $\phantom{\rule{12pt}{0ex}}{\stackrel{̂}{p}}_{B}=\frac{{x}_{B}}{{n}_{B}}=\frac{12}{200}=0.06$

## Graph:

${\stackrel{̂}{p}}_{A}-{\stackrel{̂}{p}}_{B}=0.1-0.06=0.04$ .

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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