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Learn how to compare two populations or groups by determining the conditions for performing inference, constructing and interpreting confidence intervals, and performing significance tests. This section focuses on comparing two proportions.
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2, 4, 6, 8, 12, 16, 18, 24, 36, 38, 46, 54, 58, 60 ,66 Chapter 10: Comparing Two Populations or Groups Section 10.1 Comparing Two Proportions The Practice of Statistics, 4th edition – For AP* STARNES, YATES, MOORE
Chapter 10Comparing Two Populations or Groups • 10.1 Comparing Two Proportions • 10.2 Comparing Two Means
Section 10.1Comparing Two Proportions Learning Objectives After this section, you should be able to… • DETERMINE whether the conditions for performing inference are met. • CONSTRUCT and INTERPRET a confidence interval to compare two proportions. • PERFORM a significance test to compare two proportions. • INTERPRET the results of inference procedures in a randomized experiment.
Comparing Two Proportions • Introduction Suppose we want to compare the proportions of individuals with a certain characteristic in Population 1 and Population 2. Let’s call these parameters of interest p1 and p2. The ideal strategy is to take a separate random sample from each population and to compare the sample proportions with that characteristic. What if we want to compare the effectiveness of Treatment 1 and Treatment 2 in a completely randomized experiment? This time, the parameters p1 and p2 that we want to compare are the true proportions of successful outcomes for each treatment. We use the proportions of successes in the two treatment groups to make the comparison. Here’s a table that summarizes this situation, whether it is a population or treatment.
Comparing Two Proportions • The Sampling Distribution of a Difference Between Two Proportions In Chapter 7, we saw that the sampling distribution of a sample proportion has the following properties: Shape Approximately Normal if np ≥ 10 and n(1 - p) ≥ 10 • To explore the sampling distribution of the difference between two proportions, let’s start with two populations having a known proportion of successes. • At School 1, 70% of students did their homework last night • At School 2, 50% of students did their homework last night. • Suppose the counselor at School 1 takes an SRS of 100 students and records the sample proportion that did their homework. • School 2’s counselor takes an SRS of 200 students and records the sample proportion that did their homework.
Comparing Two Proportions • The Sampling Distribution of a Difference Between Two Proportions Using Fathom software, we generated an SRS of 100 students from School 1 and a separate SRS of 200 students from School 2. The difference in sample proportions was then calculated and plotted. We repeated this process 1000 times. The results are below:
Comparing Two Proportions • The Sampling Distribution of a Difference Between Two Proportions The Sampling Distribution of the Difference Between Sample Proportions Choose an SRS of size n1from Population 1 with proportion of successes p1and an independent SRS of size n2from Population 2 with proportion of successes p2.
Comparing Two Proportions • Example: Who Does More Homework? Suppose that there are two large high schools, each with more than 2000 students, in a certain town. At School 1, 70% of students did their homework last night. Only 50% of the students at School 2 did their homework last night. The counselor at School 1 takes an SRS of 100 students and records the proportion that did homework. School 2’s counselor takes an SRS of 200 students and records the proportion that did homework. School 1’s counselor and School 2’s counselor meet to discuss the results of their homework surveys. After the meeting, they both report to their principals that
Comparing Two Proportions • Example: Who Does More Homework?
Comparing Two Proportions • Confidence Intervals for p1 – p2 If the Normal condition is met, we find the critical value z* for the given confidence level from the standard Normal curve. Our confidence interval for p1 – p2is:
Comparing Two Proportions • Two-Sample z Interval for p1 – p2 Two-Sample zInterval for a Difference Between Proportions
Comparing Two Proportions • Example: Teens and Adults on Social Networks As part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a random sample of 800 U.S. teens about their use of social media and the Internet. A second survey posed similar questions to a random sample of 2253 U.S. adults. In these two studies, 73% of teens and 47% of adults said that they use social-networking sites. Use these results to construct and interpret a 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social-networking sites. State: Our parameters of interest are p1= the proportion of all U.S. teens who use social networking sites and p2= the proportion of all U.S. adults who use social-networking sites. We want to estimate the difference p1 – p2at a 95% confidence level. • Plan: We should use a two-sample z interval for p1 – p2if the conditions are satisfied. • Random The data come from a random sample of 800 U.S. teens and a separate random sample of 2253 U.S. adults. • Normal We check the counts of “successes” and “failures” and note the Normal condition is met since they are all at least 10: • Independent We clearly have two independent samples—one of teens and one of adults. Individual responses in the two samples also have to be independent. The researchers are sampling without replacement, so we check the 10% condition: there are at least 10(800) = 8000 U.S. teens and at least 10(2253) = 22,530 U.S. adults.
Do: Since the conditions are satisfied, we can construct a two-sample z interval for the difference p1 – p2. Comparing Two Proportions • Example: Teens and Adults on Social Networks Conclude: We are 95% confident that the interval from 0.223 to 0.297 captures the true difference in the proportion of all U.S. teens and adults who use social-networking sites. This interval suggests that more teens than adults in the United States engage in social networking by between 22.3 and 29.7 percentage points.
A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used to improve athletic performance. Define the parameters, then construct and interpret a 95% confidence interval for the difference between the population proportions. Explain why we could say there is no difference between these two sample proportions? Interval (-0.007 to 0.0124) We are 95% confident that the interval from -0.007 to 0.0124 captures the true difference in proportion of freshmen and seniors in Illinois who use anabolic steroids. Since 0 is a plausible value, we have evidence that there is no difference between the two proportions.
Many news organizations conduct polls asking adults in the United States if they approve of the job the president is doing. How did President Obama’s approval rating change from August 2009 to September 2010? According to a CNN poll of 1024 randomly selected U.S. adults on September 1-2, 2010, 50% approved of President Obama’s job performance. A CNN poll of 1010 randomly selected U.S. adults on August 28-30, 2009 showed that 53% approved of President Obama’s job performance. • Problem: • (a) Use the results of these polls to construct and interpret a 90% confidence interval for the change in President Obama’s approval rating among all US adults. • (b) Based on your interval, is there convincing evidence that President Obama’s job approval rating has changed?
State: We want to estimate at the 90% confidence level where = the true proportion of all U.S. adults who approved of President Obama’s job performance in September 2010 and = the true proportion of all U.S. adults who approved of President Obama’s job performance in August 2009. Plan: We should use a two-sample z interval for if the conditions are satisfied. Random: The data came from separate random samples. Normal: = 512, = 512, = 535, = 475 are all at least 10. Independent: The samples were taken independently and there are more than 10(1024) = 10,240 U.S. adults in 2010 and 10(1010) = 10,100 U.S. adults in 2009. = –0.03 0.036 = (–0.066, 0.006) We are 95% confident that the interval from –0.066 to 0.006 captures the true change in the proportion of U.S. adults who approve of President Obama’s job performance from August 2009 to September 2010. That is, it is plausible that his job approval has fallen by up to 6.6 percentage points or increased by up to 0.6 percentage points. (b) Since 0 is included in the interval, it is plausible that there has been no change in President Obama’s approval rating. Thus, we do not have convincing evidence that his approval rating has changed.
An observed difference between two sample proportions can reflect an actual difference in the parameters, or it may just be due to chance variation in random sampling or random assignment. Significance tests help us decide which explanation makes more sense. The null hypothesis has the general form H0: p1 - p2= hypothesized value We’ll restrict ourselves to situations in which the hypothesized difference is 0. Then the null hypothesis says that there is no difference between the two parameters: H0: p1 - p2 = 0 or, alternatively, H0: p1 = p2 The alternative hypothesis says what kind of difference we expect. Ha: p1 - p2 > 0, Ha: p1 - p2 < 0, or Ha: p1 - p2 ≠ 0 Comparing Two Proportions • Significance Tests for p1 – p2 If the Random, Normal, and Independent conditions are met, we can proceed with calculations.
Comparing Two Proportions • Significance Tests for p1 – p2 If H0: p1 = p2is true, the two parameters are the same. We call their common value p. But now we need a way to estimate p, so it makes sense to combine the data from the two samples. This pooled (or combined) sample proportion is:
Comparing Two Proportions • Two-Sample z Test for The Difference Between Two Proportions If the following conditions are met, we can proceed with a two-sample z test for the difference between two proportions: Two-Sample z Test for the Difference Between Proportions
Comparing Two Proportions • Example: Hungry Children • Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools. An SRS of 80 students from School 1 found that 19 had not eaten breakfast. At School 2, an SRS of 150 students included 26 who had not had breakfast. More than 1500 students attend each school. Do these data give convincing evidence of a difference in the population proportions? Carry out a significance test at the α= 0.05 level to support your answer. State: Our hypotheses are H0: p1 - p2 = 0 Ha: p1 - p2 ≠ 0 where p1= the true proportion of students at School 1 who did not eat breakfast, and p2= the true proportion of students at School 2 who did not eat breakfast. • Plan: We should perform a two-sample z test for p1 – p2if the conditions are satisfied. • Random The data were produced using two simple random samples—of 80 students from School 1 and 150 students from School 2. • Normal We check the counts of “successes” and “failures” and note the Normal condition is met since they are all at least 10: • Independent We clearly have two independent samples—one from each school. Individual responses in the two samples also have to be independent. The researchers are sampling without replacement, so we check the 10% condition: there are at least 10(80) = 800 students at School 1 and at least 10(150) = 1500 students at School 2.
Do: Since the conditions are satisfied, we can perform a two-sample z test for the difference p1 – p2. Comparing Two Proportions • Example: Hungry Children P-value Using Table A or normalcdf, the desired P-value is 2P(z ≥ 1.17) = 2(1 - 0.8790) = 0.2420. Conclude: Since our P-value, 0.2420, is greater than the chosen significance level of α = 0.05,we fail to reject H0. There is not sufficient evidence to conclude that the proportions of students at the two schools who didn’t eat breakfast are different.
Comparing Two Proportions • Example: Significance Test in an Experiment • High 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 recruited middle-aged men with high cholesterol but no history of other serious medical problems to investigate this question. The volunteer subjects 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. Is the apparent benefit of gemfibrozil statistically significant? Perform an appropriate test to find out. State: Our hypotheses are H0: p1 - p2 = 0 OR H0: p1 = p2 Ha: p1 - p2 < 0 Ha: p1 < p2 where p1is the actual heart attack rate for middle-aged men like the ones in this study who take gemfibrozil, and p2is the actual heart attack rate for middle-aged men like the ones in this study who take only a placebo. No significance level was specified, so we’ll use α = 0.01 to reduce the risk of making a Type I error (concluding that gemfibrozil reduces heart attack risk when it actually doesn’t).
Plan: We should perform a two-sample z test for p1 – p2if the conditions are satisfied. • Random The data come from two groups in a randomized experiment • Normal The number of successes (heart attacks!) and failures in the two groups are 56, 1995, 84, and 1946. These are all at least 10, so the Normal condition is met. • Independent Due to the random assignment, these two groups of men can be viewed as independent. Individual observations in each group should also be independent: knowing whether one subject has a heart attack gives no information about whether another subject does. Comparing Two Proportions • Example: Cholesterol and Heart Attacks Do: Since the conditions are satisfied, we can perform a two-sample z test for the difference p1 – p2. Conclude: Since the P-value, 0.0068, is less than 0.01, the results are statistically significant at the α = 0.01 level. We can reject H0and conclude that there is convincing evidence of a lower heart attack rate for middle-aged men like these who take gemfibrozil than for those who take only a placebo. P-value Using Table A or normalcdf, the desired P-value is 0.0068
Section 10.1Comparing Two Proportions Summary In this section, we learned that… • Choose an SRS of size n1from Population 1 with proportion of successes p1and an independent SRS of size n2from Population 2 with proportion of successes p2. • Confidence intervals and tests to compare the proportions p1and p2of successes for two populations or treatments are based on the difference between the sample proportions. • When the Random, Normal, and Independent conditions are met, we can use two-sample z procedures to estimate and test claims about p1 - p2.
Section 10.1Comparing Two Proportions Summary In this section, we learned that… • The conditions for two-sample z procedures are: • An approximate level C confidence interval for p1 - p2is where z* is the standard Normal critical value. This is called a two-sample z interval for p1 - p2.
Section 10.1Comparing Two Proportions Summary In this section, we learned that… • Significance tests of H0: p1 - p2= 0 use the pooled (combined) sample proportion • The two-sample z test for p1- p2uses the test statistic with P-values calculated from the standard Normal distribution. • Inference about the difference p1 - p2in the effectiveness of two treatments in a completely randomized experiment is based on the randomization distribution of the difference of sample proportions. When the Random, Normal, and Independent conditions are met, our usual inference procedures based on the sampling distribution will be approximately correct.
Looking Ahead… In the next Section… • We’ll learn how to compare two population means. • We’ll learn about • The sampling distribution for the difference of means • The two-sample t procedures • Comparing two means from raw data and randomized experiments • Interpreting computer output for two-sample t procedures
Section 10.2Comparing Two Means Learning Objectives After this section, you should be able to… • DESCRIBE the characteristics of the sampling distribution of the difference between two sample means • CALCULATE probabilities using the sampling distribution of the difference between two sample means • DETERMINE whether the conditions for performing inference are met • USE two-sample t procedures to compare two means based on summary statistics or raw data • INTERPRET computer output for two-sample t procedures • PERFORM a significance test to compare two means • INTERPRET the results of inference procedures
Intervalsσ known, use z* σ unknown, use t*(x1-x2) + z* σ12 + σ22 (x1-x2) + t* Sx12 + Sx22 n1 n2 n1 n21. Decide on whether to use z* or t*2. State parameters and Define the order you are subtracting at [C] % confidence level. 3. State the type of interval. “We are performing a 2 sample z/t interval for the difference of two means”4. Random/Normal/Independent. Check both samples, use CLT for Normality.5. State values for x1, x2, Sx1, Sx2, and df if t interval (use more conservative df value)6. Find the z or t value 7. State correct formula and plug all values into formula and simplify to find interval8. Conclude on interval. “We are [C] % confident that the interval from [here] to [there] captures the true difference in means of [context of the question].
Comparing Two Means • Introduction In the previous section, we developed methods for comparing two proportions. What if we want to compare the mean of some quantitative variable for the individuals in Population 1 and Population 2? Our parameters of interest are the population means µ1and µ2. Once again, the best approach is to take separate random samples from each population and to compare the sample means. Suppose we want to compare the average effectiveness of two treatments in a completely randomized experiment. In this case, the parameters µ1and µ2are the true mean responses for Treatment 1 and Treatment 2, respectively. We use the mean response in the two groups to make the comparison. Here’s a table that summarizes these two situations:
Comparing Two Means • The Sampling Distribution of a Difference Between Two Means In Chapter 7, we saw that the sampling distribution of a sample mean has the following properties: Shape Approximately Normal if the population distribution is Normal or n ≥ 30 (by the central limit theorem). To explore the sampling distribution of the difference between two means, let’s start with two Normally distributed populations having known means and standard deviations. Based on information from the U.S. National Health and Nutrition Examination Survey (NHANES), the heights (in inches) of ten-year-old girls follow a Normal distribution N(56.4, 2.7). The heights (in inches) of ten-year-old boys follow a Normal distribution N(55.7, 3.8). Suppose we take independent SRSs of 12 girls and 8 boys of this age and measure their heights.
Comparing Two Means • The Sampling Distribution of a Difference Between Two Means Using Fathom software, we generated an SRS of 12 girls and a separate SRS of 8 boys and calculated the sample mean heights. The difference in sample means was then calculated and plotted. We repeated this process 1000 times. The results are below:
Comparing Two Means • The Sampling Distribution of a Difference Between Two Means The Sampling Distribution of the Difference Between Sample Means Choose an SRS of size n1from Population 1 with mean µ1and standard deviation σ1 and an independent SRS of size n2from Population 2 with mean µ2and standard deviation σ2.
Comparing Two Means • Example: Who’s Taller at Ten, Boys or Girls? • Based on information from the U.S. National Health and Nutrition Examination Survey (NHANES), the heights (in inches) of ten-year-old girls follow a Normal distribution N(56.4, 2.7). The heights (in inches) of ten-year-old boys follow a Normal distribution N(55.7, 3.8). A researcher takes independent SRSs of 12 girls and 8 boys of this age and measures their heights. After analyzing the data, the researcher reports that the sample mean height of the boys is larger than the sample mean height of the girls.
Comparing Two Means • Example: Who’s Taller at Ten, Boys or Girls? (c) Does the result in part (b) give us reason to doubt the researchers’ stated results?
Comparing Two Means • The Two-Sample t Statistic If the Normal condition is met, we standardize the observed difference to obtain a t statistic that tells us how far the observed difference is from its mean in standard deviation units: The two-sample t statistic has approximately a t distribution. We can use technology to determine degrees of freedom OR we can use a conservative approach, using the smaller of n1– 1 and n2– 1 for the degrees of freedom.
Comparing Two Means • Confidence Intervals for µ1 – µ2 Two-Sample tInterval for a Difference Between Means
Comparing Two Means • Big Trees, Small Trees, Short Trees, Tall Trees The Wade Tract Preserve in Georgia is an old-growth forest of longleaf pines that has survived in a relatively undisturbed state for hundreds of years. One question of interest to foresters who study the area is “How do the sizes of longleaf pine trees in the northern and southern halves of the forest compare?” To find out, researchers took random samples of 30 trees from each half and measured the diameter at breast height (DBH) in centimeters. Comparative boxplots of the data and summary statistics from Minitab are shown below. Construct and interpret a 90% confidence interval for the difference in the mean DBH for longleaf pines in the northern and southern halves of the Wade Tract Preserve. State: Our parameters of interest are µ1 = the true mean DBH of all trees in the southern half of the forest and µ2= the true mean DBH of all trees in the northern half of the forest. We want to estimate the difference µ1 - µ2 at a 90% confidence level.
Plan: We should use a two-sample t interval for µ1 – µ2if the conditions are satisfied. • Random The data come from a random samples of 30 trees each from the northern and southern halves of the forest. • Normal The boxplots give us reason to believe that the population distributions of DBH measurements may not be Normal. However, since both sample sizes are at least 30, we are safe using t procedures. • Independent Researchers took independent samples from the northern and southern halves of the forest. Because sampling without replacement was used, there have to be at least 10(30) = 300 trees in each half of the forest. This is pretty safe to assume. Comparing Two Means • Big Trees, Small Trees, Short Trees, Tall Trees Do: Since the conditions are satisfied, we can construct a two-sample t interval for the difference µ1 – µ2. We’ll use the conservative df = 30-1 = 29. Conclude: We are 90% confident that the interval from 3.83 to 17.83 centimeters captures the difference in the actual mean DBH of the southern trees and the actual mean DBH of the northern trees. This interval suggests that the mean diameter of the southern trees is between 3.83 and 17.83 cm larger than the mean diameter of the northern trees.
An observed difference between two sample means can reflect an actual difference in the parameters, or it may just be due to chance variation in random sampling or random assignment. Significance tests help us decide which explanation makes more sense. The null hypothesis has the general form H0: µ1 - µ2= hypothesized value We’re often interested in situations in which the hypothesized difference is 0. Then the null hypothesis says that there is no difference between the two parameters: H0: µ1 - µ2 = 0 or, alternatively, H0: µ1 = µ2 The alternative hypothesis says what kind of difference we expect. Ha: µ1 - µ2 > 0, Ha: µ1 - µ2 < 0, or Ha: µ1 - µ2 ≠ 0 Comparing Two Means • Significance Tests for µ1 – µ2 If the Random, Normal, and Independent conditions are met, we can proceed with calculations.
Comparing Two Means • Two-Sample t Test for The Difference Between Two Means • If the following conditions are met, we can proceed with a two-sample t test for the difference between two means: Two-Sample tTest for the Difference Between Two Means
Comparing Two Means • Example: Calcium and Blood Pressure Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. The relationship was strongest for black men. Such observational studies do not establish causation. Researchers therefore designed a randomized comparative experiment. The subjects were 21 healthy black men who volunteered to take part in the experiment. They were randomly assigned to two groups: 10 of the men received a calcium supplement for 12 weeks, while the control group of 11 men received a placebo pill that looked identical. The experiment was double-blind. The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative response Here are the data: State: We want to perform a test of H0: µ1 - µ2 = 0 Ha: µ1 - µ2 > 0 where µ1= the true mean decrease in systolic blood pressure for healthy black men like the ones in this study who take a calcium supplement, and µ2= the true mean decrease in systolic blood pressure for healthy black men like the ones in this study who take a placebo. We will use α = 0.05.
Plan: If conditions are met, we will carry out a two-sample t test for µ1- µ2. • Random The 21 subjects were randomly assigned to the two treatments. • Normal With such small sample sizes, we need to examine the data to see if it’s reasonable to believe that the actual distributions of differences in blood pressure when taking calcium or placebo are Normal. Hand sketches of calculator boxplots and Normal probability plots for these data are below: The boxplots show no clear evidence of skewness and no outliers. The Normal probability plot of the placebo group’s responses looks very linear, while the Normal probability plot of the calcium group’s responses shows some slight curvature. With no outliers or clear skewness, the t procedures should be pretty accurate. • Independent Due to the random assignment, these two groups of men can be viewed as independent. Individual observations in each group should also be independent: knowing one subject’s change in blood pressure gives no information about another subject’s response. Comparing Two Means • Example: Calcium and Blood Pressure
Comparing Two Means • Example: Calcium and Blood Pressure Do: Since the conditions are satisfied, we can perform a two-sample t test for the difference µ1 – µ2. P-value Using the conservative df = 10 – 1 = 9, we can use Table B to show that the P-value is between 0.05 and 0.10. Conclude: Because the P-value is greater than α = 0.05, we fail to reject H0. The experiment provides some evidence that calcium reduces blood pressure, but the evidence is not convincing enough to conclude that calcium reduces blood pressure more than a placebo.
We can estimate the difference in the true mean decrease in blood pressure for the calcium and placebo treatments using a two-sample t interval for µ1 - µ2. To get results that are consistent with the one-tailed test at α = 0.05from the example, we’ll use a 90% confidence level. The conditions for constructing a confidence interval are the same as the ones that we checked in the example before performing the two-sample t test. With df = 9, the critical value for a 90% confidence interval is t* = 1.833. The interval is: Comparing Two Means • Example: Calcium and Blood Pressure We are 90% confident that the interval from -0.754 to 11.300 captures the difference in true mean blood pressure reduction on calcium over a placebo. Because the 90% confidence interval includes 0 as a plausible value for the difference, we cannot reject H0: µ1 - µ2 = 0 against the two-sided alternative at the α = 0.10 significance level or against the one-sided alternative at the α = 0.05 significance level.
The two-sample t procedures are more robust against non-Normality than the one-sample t methods. When the sizes of the two samples are equal and the two populations being compared have distributions with similar shapes, probability values from the t table are quite accurate for a broad range of distributions when the sample sizes are as small as n1= n2= 5. Comparing Two Means • Using Two-Sample t Procedures Wisely Using the Two-Sample tProcedures: The Normal Condition • Sample size less than 15: Use two-sample t procedures if the data in both • samples/groups appear close to Normal (roughly symmetric, single peak, • no outliers). If the data are clearly skewed or if outliers are present, do not • use t. • • Sample size at least 15: Two-sample t procedures can be used except in the presence of outliers or strong skewness. • • Large samples: The two-sample t procedures can be used even for clearly • skewed distributions when both samples/groups are large, roughly n ≥ 30.