Comparing Two Proportions Chapter 21

In a two-sample problem, we want to compare two populations or the responses to two treatments based on two independent samples. The notation and symbols will be the same as the previous chapter, but there will be 1 and 2 subscripts to make it clear if the data is from Population 1 or Population 2.

We will have a separate SRS from each population or responses from two treatments in a randomized comparative experiment. We compare populations by doing inference (confidence intervals and hypothesis tests) about the difference p 1 – p 2 between the population proportions. We will use the difference between the sample proportions 1 – 2 to estimate that difference.

Conditions 1. When the samples are large, the distribution of 1 – 2 is approximately Normal. To check if the sample size is large enough to say the distribution is close to Normal, check to see the number of successes and failures are each 10 or more in both samples. 2. The mean of the sampling distribution is p 1 – p The standard deviation of the sampling distribution is.

Large-Sample Confidence Intervals for Comparing Proportions Draw an SRS of size n 1 from a large population having proportion p 1 of successes and draw an independent SRS of size n 2 from another large population having proportion p 2 of successes. When n 1 and n 2 are large, an approximate level C confidence interval for p 1 – p 2 is: where the standard error SE of estimate 1 – 2 is the standard deviation for sample values: SE= and z* is the critical value for the standard Normal density curve with area C between –z* and z* we have used in previous chapters.

Example A surprising number of young adults (ages 19 to 25) still live in their parents’ home. A random sample by the National Institutes of Health included 2253 men and 2629 women in this age group. The survey found that 986 of the men and 923 of the women lived with their parents. Find a 95% confidence interval for p 1 – p 2, the difference between the proportions of young men and young women who live with their parents.

List out all the known info: Population 1 _______ Population 2 ______ n 1 =__________x 1 =___________ 1 =____________ n 2 =__________x 2 =___________ 2 =____________ Find 1 – 2 = ___________ Find SE: Find 95% confidence interval: Sentence: We are 95% confident that the percent of young men living with their parents is between _______ and _______ percentage points higher than the percentage of women who live with their parents.

Note In this study, the sample survey selected a single random sample of young adults, not two separate random samples of young men and young women. To get two samples, they divided the single sample by sex. This means they did not know the two sample sizes n 1 and n 2 until after they had all their data. This is a valid way to do these studies as long as your n’s turn out to be large enough.

Lesson 2 Significance Tests for Comparing Two Proportions: Chapter 21

Warm up 1. In 2000, the U.S. Department of Commerce reported the results of a large-scale survey on high school graduation. Researchers contacted more than 25,000 Americans aged 24 years to see if they had finished high school; 9,579 of the 13,460 males and 11,169 of the 12,678 females indicated that they had high school diplomas. a. Create a 90% confidence interval for the difference in graduation rates between males and females.

An observed difference between two sample proportions can reflect an actual difference between the populations, or it may just be due to chance variation in random sampling. Significance tests help us decide if the effect we see in the samples is really there in the populations. The null hypothesis H 0 says there is no difference between the two populations: H 0 : p 1 – p 2 = 0 or H 0 : p 1 = p 2 (this is the one we will write) The alternative hypothesis H a says what kind of difference we expect: H a : p 1 p 2 H a : p 1 ≠ p 2

If,,, then the samples are large enough to use a normal sampling distribution for which means you can use a z-test.  test statistic is  standardized test statistic is z Formula: where and

is called the pooled sample proportion which is used because H 0 states that the observations from both samples are equal and from the same population so we combine them for the test statistic calculation.

Steps for Using a z-Test for a Proportion p 1. Verify samples are large enough (check all four products are ≥5). 2. State H 0 and H a. Mark the claim. 3. Identify α and type of test.

4. Get z 0 from table : α Left tail test Right tail test Two tail test 0.01– ± – ± – ±1.645

5. Find z using formula (above). 6. Draw distribution with point at z, line at z 0, shade direction of test. 7. Make a decision: 1. z in shaded region = reject H 0 2. z outside shaded region = fail to reject H 0 8. Conclusion (evidence) sentence.

CALCULATOR METHOD OR: Use STAT-TESTS-6:2-PropZTest Need to enter x 1, n 1, x 2, n 2, and type of test. Gives z(formula value), P-value, and. Decision: P> α is fail to reject H 0 ; P< α is reject H 0.

Example 1 In a study of 200 randomly selected adult female and 250 randomly selected adult male internet users, 60 of the females and 95 of the males said that they plan to shop online at least once during the next month. At α =0.10, test the claim that there is a difference in the proportion of female Internet users who plan to shop online and the proportion of male Internet users who plan to shop online.

Example 2 A medical research team conducted a study to test the effect of a cholesterol-reducing medication. At the end of the study, the researchers found that of the 4700 subjects who took the medication, 301 died of heart disease. Of the 4300 subjects who took a placebo, 357 died of heart disease. At α =0.01, can you conclude that the death rate is lower for those who took the medication that for those who took the placebo?