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C HAPTER 11 Section 11.2 – Comparing Two Means. C OMPARING T WO M EANS Comparing two populations or two treatments is one of the most common situations.

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Presentation on theme: "C HAPTER 11 Section 11.2 – Comparing Two Means. C OMPARING T WO M EANS Comparing two populations or two treatments is one of the most common situations."— Presentation transcript:

1 C HAPTER 11 Section 11.2 – Comparing Two Means

2 C OMPARING T WO M EANS Comparing two populations or two treatments is one of the most common situations encountered in statistical practice. We call such situations two-sample problems. A two sample problem can arise from a randomized comparative experiment that randomly divides subjects into two groups and exposes each group to a different treatment. Comparing random samples separately selected from two populations is also a two sample problem. Unlike the matched pairs designs studied earlier, there is no matching of the units in the two samples and the two samples can be of different sizes.

3 T WO – S AMPLE P ROBLEMS The goal of inference is to compare the responses to two treatments or to compare the characteristics of two populations. We have a separate sample from each treatment or each population.

4 E XAMPLE 11.9 - T WO -S AMPLE P ROBLEMS 1. A medical researcher is interested in the effect on blood pressure of added calcium in our diet. She conducts a randomized comparative experiment in which one group of subjects receives a calcium supplement and a control group receives a placebo. 2. A psychologist develops a test that measures social insight. He compares the social insight of male college students with that of female college students by giving the test to a sample of students of each gender. 3. A bank wants to know which of two incentive plans will most increase the use of its credit cards. It offers each incentive to a random sample of credit card customers and compares the amount charged during the following six months.

5 C ONDITIONS FOR C OMPARING T WO M EANS We have two SRSs, from two distinct populations. Both populations are normally distributed. The means and standard deviations of the populations are unknown. The samples are independent (That is, one sample has no influence on the other.) Matching violates independence, for example. We measure the same variable for both samples.

6 O RGANIZING THE D ATA Call the variable we measure x 1 in the first population and x 2 in the second. We know parameters in this situation. Populatio n VariableMeanStandard deviation 1x1x1 μ1μ1 σ1σ1 2x2x2 μ2μ2 σ2σ2

7 O RGANIZING D ATA ( PART 2) There are four unknown parameters, the two means and the two standard deviations. PopulationSample Size Sample Mean Sample Standard deviation 1 2

8

9

10 S TANDARD E RROR

11 D EGREES OF F REEDOM FOR T WO -S AMPLE P ROBLEMS The two-sample t statistic does not have a t distribution since we replaced two standard deviations by the corresponding standard errors. Therefore, we use two methods for calculating degrees of freedom: Option 1: Use procedures based on the statistic t with critical values from a t distribution (used by calculator). Option 2: Use procedures based on the statistic t with critical from the smaller n – 1.

12 T WO -S AMPLE T D ISTRIBUTIONS

13 C ONFIDENCE I NTERVAL FOR A T WO -S AMPLE T

14 Example: Calcium and Blood Pressure Does increasing the amount of calcium in our diet reduce blood pressure? Examination of alarge sample of people revealed a relationship between calcium intake and blood pressure.The relationship was strongest for black men. Such observational studies do not establishcausation. Researchers therefore designed a randomized comparative experiment. Thesubjects were 21 healthy black men who volunteered to take part in the experiment. Theywere randomly assigned to two groups: 10 of the men received a calcium supplement for 12weeks, while the control group of 11 men received a placebo pill that looked identical. Theexperiment 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 appearsas a negative response Here are the data: We want to perform a test of H 0 : µ 1 - µ 2 = 0 H a : µ 1 - µ 2 > 0 where µ 1 = the true mean decrease in systolic blood pressure for healthy black men like the ones in this study who take a calcium supplement, and µ 2 = 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.

15 Example: Calcium and Blood Pressure If conditions are met, we will carry out a two-sample t test for µ 1 - µ 2. 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.

16 Example: Calcium and Blood Pressure Since the conditions are satisfied, we can perform a two-sample t test for the difference µ 1 – µ 2. P-value Using the conservative df = 10 – 1 = 9, we can use Table C to show that the P-value is between 0.05 and 0.10. Because the P-value is greater than α = 0.05, we fail to reject H 0. 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.

17 Example: Calcium and Blood Pressure 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 µ 1 - µ 2. To get results that are consistent with the one-tailed test at α = 0.05 from 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: 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 H 0 : µ 1 - µ 2 = 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.

18 T HE P OOLED T WO - SAMPLE P ROCEDURES Procedures that average use the statistical term “pooled.” Pooled two-sample t procedures is a situation where the variances of both the samples are assumed to be the same and the sample sizes are the same. This rarely happens and the same results will occur with regular t procedures. On the print out from a computer (p.666), use the unequal line for variances, degrees of freedom, and probability.

19 Homework: Technology toolbox on p.660-661 P.649 #’s 37, 38, 40, 43, & 50 Chapter 11 review: p.675 #’s 64, 68, & 72 Ch.10/11 take home test due Thursday 3/27 DO NOT WORK ON THE TEST TOGETHER!!!


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