Comparing Two Comparing Two Population ParametersPopulation Parameters

Comparing two-population proportions

Test-statistic

Cholesterol and Heart attackHigh levels of cholesterol in the blood are associated

with higher risk of heart attacks. Will using a drug to lower blood cholesterol reduce heart attacks? The Helsinki Heart Study looked at this question. Middle-aged men were assigned at random to one of two treatments: 2051 men took the drug gemfibrozil to reduce their cholesterol levels, and a control group of 2030 men took a placebo. During the next five years, 56 men in the gemfibrozil group and 84 men in the placebo group had heart attacks.

gemfibrozil

X1 56n1 2051

p1 0.0273

placebo

X2 84n2 2030

p20.041

4

Step 1: Hypotheses

p1 the proportion of middle-aged men who would suffer heart attacks after taking gemfibrozil, and p2, the proportion of middle-aged men who would suffer heart attacks if they only took a placebo

Step 2: Hypotheses

SRS The two samples can be viewed as SRSs from their respective populations or are the two groups in a randomized experiment.

Normality The estimated counts of “successes” and “failures” n1 c, n1(q), n2c, and n2(q) are all greater than 5.

Independence The samples are independent. When sampling without replacement, the two populations must be at least 10 times as large as the corresponding samples.

Step 3: Test Statistic

P = 0.0068.

Step 4: Interpretation Since P < 0.01, the results are statistically significant at the α = 0.01 level. There is strong evidence that gemfibrozil reduced the rate of heart attacks.

Step 3: Test Statistic

P = 0.0068.

Practice:Don't drink the water!Two-proportion z test for an observational study

The movie A Civil Action tells the story of a major legal battle that took place in the small town of Woburn, Massachusetts. A town well that supplied water to East Woburn residents was contaminated by industrial chemicals. During the period that residents drank water from this well, a sample of 414 births showed 16 birth defects. On the west side of Woburn, a sample of 228 babies born during the same time period revealed 3 with birth defects. The plaintiffs suing the companies responsible for the contamination claimed that these data show that the rate of birth defects was significantly higher in East Woburn, where the contaminated well water was in use. How strong is the evidence supporting this claim? What should the judge for this case conclude?

Step 3: Calculations

•Test statistic From the computer output, the test statistic is z = 1.82.

•P-value The computer output shows that the P-value is 0.0341.

Step 4: Interpretation The P-value, 0.0341, tells us that it is unlikely that we would obtain a difference in sample proportions as large as we did if the null hypothesis is true. Judges have generally adopted a 5% significance level as their standard for convincing evidence. More than likely, the judge in this case would conclude that the companies who contaminated the well water were responsible for the higher proportion of birth defects in East Woburn.

n1c = (414) (0.0296) = 12.25n2c = (228) (0.0296) = 6.75n1(1 − c) = (414) (0.9704) = 401.75n2(1 − c ) = (228) (0.9704) = 221.25Since all four of these values are larger than 5, we are safe using a Normal approximation.Independence We must consider that both populations are at least 10 times as large as the samples of babies.

Normality

Step 3: Calculations

•Test statistic From the computer output, the test statistic is z = 1.82.

•P-value The computer output shows that the P-value is 0.0341.

Step 4: Interpretation The P-value, 0.0341, tells us that it is unlikely that we would obtain a difference in sample proportions as large as we did if the null hypothesis is true. Judges have generally adopted a 5% significance level as their standard for convincing evidence. More than likely, the judge in this case would conclude that the companies who contaminated the well water were responsible for the higher proportion of birth defects in East Woburn.