Math 146: Exercises on hypothesis tests about proportions
(1) State the null and alternative hypotheses. For tests about proportions, the possible
(2) Verify necessary data conditions, and if they are met, summarize the data into an
appropriate test statistic. For tests about proportions:
– Check that the sample is random, or can be considered random in context. This
means the individual responses can be considered independent of each other
– Check that np0 ≥ 10 and n(1 − p) ≥ 10. These conditions give us some assurance
that the normal approximation holds.
standard deviations above or below p0 our sample proportion ˆ
(3) Find the p-value of the test by using software or the normal table. In R, the prop.test
command will automatically compute the p-values. To compute p-values directly:
– for Ha : p < p0, calculate P (Z ≤ z). Look in the normal table or use the R
– for Ha : p > p0, calculate P (Z ≥ z) = 1 − P (Z ≤ z). The R command would be
– for Ha : p = p0, calculate 2P (Z ≥ |z|). The R command is 2*(1-pnorm(abs(z))).
(4) Decide whether the result is statistically significant based on the p-value, and then de-
termine what it means in the context of the problem. Whether the result is statisticallysignificant depends on the chosen significance level. The default is 0.05.
1. Anyone who plays or watches sports has heard of the “home field advantage.” Teams
tend to win more often when they play at home. Or do they? If there were no homefield advantage, the home teams would win about half of all games played. In the2009 Major League Baseball season, there were 2430 regular-season games. It turnsout that the home team won 1333 of the 2430 games, or 54.81% of the time. Couldthis deviation from 50% be explained just from natural sampling variability, or is itevidence to suggest that there really is a home field advantage, at least in professionalbaseball?
2. Advances in medical care such as prenatal ultrasound examination now make it possible
to determine a child’s sex early in pregnancy. There is a fear that in some culturessome parents may use this technology to select the sex of their children. A studyfrom Punjab, India (E.E. Booth, M. Verma, and R.S. Beri, “Fetal Sex Determinationin Infants in Punjab, India: Correlations and Implications,” BMJ 309: 1259-1261),reports that, in 1993, in one hospital, 56.9% of the 550 live births that year were boys. It’s a medical fact that male babies are slightly more common than female babies. The study’s authors report a baseline for this region of 51.7% male live births. Is thereevidence that the proportion of male births has changed? After testing the appropriatehypotheses, calculate a confidence interval to give a sense of the effect size.
3. The question of whether the diabetes drug Avandia increased the risk of heart attach
was raised in a study, which estimated the seven-year risk of heart attack to be 29.9%and reported a P -value of 0.27 for a test of whether this risk was higher than thebaseline seven-year risk of 20.2%.
How should the P -value be interpreted in this

Int. J. Cancer: 110, 284 –290 (2004) Publication of the International Union Against CancerDIETARY PHYTOESTROGEN INTAKE AND PREMENOPAUSAL BREASTCANCER RISK IN A GERMAN CASE-CONTROL STUDYINSEISEN ,* Regina PILLER , Silke HERMANN and Jenny CHANG-CLAUDE1 Unit of Human Nutrition and Cancer Prevention, Technical University of Munich, Freising-Weihenstephan, Germany 2 Department of Clinical E