12.2 Notes

12.2 Notes
12.2 - Comparing Two Proportions
Don't write: In a twosample problem we want to compare two populations or responses to two treatments based on two independent samples. We've done it for means, now we'll look at proportions. We'll have a ton of p's all over the place. p1, p2, p(hat)1, p(hat)2, & a Pvalue.
Confidence Interval for comparing two proportions:
Significance Test for comparing two proportions:
The Changes: we have to make sure both samples are random, we have to have them be independent OF EACH OTHER, and we need both to be approx. normal. We also will have a "new" way of caluclating the standard deviation of the sampling distribution of p1 p2. 1. In the 2001 regular baseball season, the World Series Champion Arizona Diamondbacks played 81 games at home and 81 games away. They won 48 of their home games and 44 of the games played away. We can consider these games as samples from potentially large populations of games played at home and away. where the p(hat)C in the standard error is the pooled proportion of successes in both samples combined. (it's the best estimate for p that we have.)
c. Most people think that it is easier to win at home than away. Perform a significance test to determine whether this is true for the Arizona Diamondbacks.
a. Find the standard error (an estimate of the standard deviation) needed to compute a confidence interval for the difference in the proportions of home games and away games that would be won by the Diamondbacks.
b. Construct and interpret a 90% confidence interval for the difference between the proportion of games that the Diamondbacks win at home and the proportion that they win when on the road.
p1: proportion of games won @ home p2: proportion of games won away
With a Pvalue of 0.26, which is larger than our significance level of 0.05, we FAIL TO REJECT THE null hypothesis and conclude that we don't have enough evidence to say that there's a significant difference between home and away winning percentages.
We are 90% confident that they win as much as 17.7 percentage points more often @ home or lose by as much as 7.8 percentage points @ home.
2. A study of chromosome abnormalities and criminality examined data from 4124 males born in Copenhagen. Each man was classified as having a criminal record or not, using the registers maintained in the local police offices. Each was also classified as having the normal male XY chromosome pair or one of the abnormalities XYY or XXY. Of the 4096 men with normal chromosomes, 381 had criminal records, while 8 of the 28 men with abnormal chromosomes had criminal records. Some experts believe chromosome abnormalities are associated with increased criminality. a. Do these data lend support to this belief? Perform a significance test to answer this question.
2. A study of chromosome abnormalities and criminality examined data from 4124 males born in Copenhagen. Each man was classified as having a criminal record or not, using the registers maintained in the local police offices. Each was also classified as having the normal male XY chromosome pair or one of the abnormalities XYY or XXY. Of the 4096 men with normal chromosomes, 381 had criminal records, while 8 of the 28 men with abnormal chromosomes had criminal records. Some experts believe chromosome abnormalities are associated with increased criminality. b. Construct and interpret a 95% confidence interval for the difference in proportions.
we have already done P, A, so N is a Two Proportion Z Interval.
p1: proportion of males with normal XY chromosomes with a criminal record
p2: proportion of males with XYY or XXY with criminal records.
concerned, but okay
TwoProportion Z Test
We are 95% confident that the interval from 0.36 to 0.0251 contains the true difference in proportion between the proportion of men in Copenhagen with normal chromosomes and a criminal record and men in Copenhagen with abnormal chromosomes and a criminal record. Another way of saying it is that we are 95% confident that the proportion of men with normal chromosomes and a criminal record is between 2.5 percentage points and 36 percentage points less than the proportion of men with abnormal chromosomes and a criminal record.

Download Report