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9. Inferences Based on Two Samples Now that we’ve learned to make inferences about a single population, we’ll learn how to compare two populations. For example, we may wish to compare the mean gas mileages for two models of automobiles, or the mean reaction times of men and women to a visual stimulus. In this chapter we’ll see how to decide whether differences exist and how to estimate the differences between population means and proportions. 9.1 Comparing two population means: Independent Sampling One of the most commonly used significance tests is the comparison of two population means 1 and 2 . Two-sample Problems The goal of inference is to compare the responses in two groups. Each group is considered to be a sample from a distinct population. The responses in each group are independent of those in the other group. A two sample problem can arise from a randomized comparative experiment that randomly divides the subjects into two groups and exposes each group to a different treatment. The two samples may be of different sizes. Two-Sample z Statistic Suppose that x1 is the mean of an SRS of size n1 drawn from N( 1 , 1 ) population and that x2 is the mean of an SRS of size n2 drawn from N( 2 , 2 ) population. Then the two-sample z statistic z x1 x2 1 2 12 n1 22 n2 has the standard normal N(0,1) sampling distribution. Large Sample Confidence Interval for 1 2 x1 x2 z 2 12 22 n1 n2 Assumptions: The two samples are randomly selected in an independent manner from the two populations. The sample sizes n1 and n2 are large enough. Example for C.I. of 1 2 Let’s look at Example 9.1 in our textbook (page 431). Example for Test of Significance Let’s look at Examples 9.2 and 9.3 in our textbook (page 434 and 435). In the unlikely event that both population standard deviations are known, the two-sample z statistic is the basis for inference about 1 2 . Exact z procedures are seldom used because 1 and 2 are rarely known. The two-sample t procedures Suppose that the population standard deviations 1 and 2 are not known. We estimate them by the sample standard deviations s1 and s 2 from our two samples. The Pooled two-sample t procedures The pooled two-sample t procedures are used when we can safely assume that the two populations have equal variances. The modifications in the procedure are the use of the pooled estimator of the common unknown variance 2 2 ( n 1 ) s ( n 1 ) s 1 2 2 . s 2p 1 n1 n2 2 This is called the pooled estimator of 2 . When both populations have variance 2 , the addition rule for variances says that x1 x2 has variance equal to the sum of the individual variances, which is 2 n1 2 1 1 2 n2 n1 n2 The standardized difference of means in this equal-variance case is z ( x1 x2 ) ( 1 2 ) 1 1 n1 n2 This is a special two-sample z statistic for the case in which the populations have the same . Replacing the unknown by the estimates s p gives a t statistic. The degrees of freedom are n1 n2 2 . The Pooled Two-Sample t Procedures Suppose that an SRS of size n1 is drawn from a normal population with unknown mean 1 and that an independent SRS of size n2 is drawn from another normal population with unknown mean 2 . Suppose also that the two populations have the same standard deviation. A level C confidence interval 1 2 given by ( x1 x2 ) t * s p 1 1 n1 n2 Here t* is the value for t (n1 n2 2) density curve with area C between –t* and t*. To test the hypothesis H 0 : 1 2 ,compute the pooled two-sample t statistic t x1 x2 1 1 sp n1 n2 In terms of a random variable T having the t( n1 n2 2) distribution, the P-value for a test of H 0 against H a : 0 is P(T t ) H a : 0 is P(T t ) H a : 0 is 2P(T | t | ) Example Take Group 1 to be the calcium group and Group 2 to be the placebo group. The evidence that calcium lowers blood pressure more than a placebo is assessed by testing H 0 : 1 2 H a : 1 2 Here are the summary statistics for the decrease in blood pressure: x s Group Treatment n 1 Calcium 10 5.000 8.743 2 Placebo 11 -0.273 5.901 The calcium group shows a drop in blood pressure, and the placebo group has a small increase. The sample standard deviations do not rule out equal population standard deviations. A difference this large will often arise by chance in samples this small. We are willing to assume equal population standard deviations. The pooled sample variance is s 2p (n1 1) s12 (n2 1) s22 n1 n2 2 (10 1)(8.743) 2 (11 1)(5.901) 2 10 11 2 54.536 . So that s p 54.536 7.385 The pooled two-sample t statistic is x1 x2 t 1 1 sp n1 n2 5.000 (0.273) 5.273 1.634 1 1 3.227 7.385 10 11 The P-value is P(T 1.634 ) , where T has t(19) distribution. From Table, we can see that P lies between 0.05 and 0.10. The experiment found no evidence that calcium reduces blood pressure (t=1.634, df=19, 0.05<P<0.10). Example We estimate that the effect of calcium supplementation is the difference between the sample means of the calcium and the placebo groups, x1 x2 5.273 mm. A 90% confidence interval for 1 2 uses the critical value t*=1.729 from the t(19) distribution. The interval is ( x1 x2 ) t * s p 1 1 n1 n2 = (5.000 (0.273)) (1.729 )(7.385) 1 1 10 11 = 5.273 5.579 We are 90% confident that the difference in means is in the interval (-0.306, 10.852). The calcium treatment reduced blood pressure by about 5.3mm more than a placebo on the average, but the margin of error for this estimate is 5.6mm. Approximate Small-Sample Procedures when both populations have different variance 2 2 ( 1 2 ) Suppose that the population standard deviations 1 and 2 are not known. We estimate them by the sample standard deviations s1 and s 2 from our two samples. Equal Sample Sizes ( n1 n2 n ) The confidence interval for 1 2 is given by ( x1 x2 ) t 2 s12 s22 n To test the hypothesis H 0 : 1 2 , compute the two-sample t statistic t x1 x2 s12 s22 n where t is based on df v n1 n2 2 2(n 1) . Unequal Sample Sizes ( n1 n2 ) The confidence interval for 1 2 is given by ( x1 x2 ) t s12 s22 n1 n2 2 To test the hypothesis H 0 : 1 2 , compute the two-sample t statistic t x1 x2 s12 s22 n1 n2 where t is based on degree of freedom 2 s s n1 n2 . v 2 2 s12 s22 n1 n2 n1 1 n2 2 2 1 2 2 Note: The value of v will generally not be an integer. Round v down to the nearest integer to use the t table. The Two-Sample t Significance test Suppose that an SRS of size n1 is drawn from a normal population with unknown mean 1 and that an independent SRS of size n2 is drawn from another normal population with unknown mean 2 . To test the hypothesis H 0 : 1 2 , compute the two-sample t statistic t x1 x2 s12 s22 n1 n2 and use P-values or critical values for the t(k) distribution, where the degrees of freedom k are the smaller n1 1 and n2 1. Example An educator believes that new directed reading activities in the classroom will help elementary school pupils improve some aspects of their reading ability. She arranges for a third-grade class of 21 students to take part in these activities for an eight-week period. A control classroom of 23 third-graders follows the same curriculum without the activities. At the end of the eight weeks, all students are given a Degree of Reading Power (DRP) test, which measures the aspects of reading ability that the treatment is designed to improve. The summary statistics using Excel are Treatment Group Control Group Mean Standard Error Median Mode Standard Deviation Sample Variance Kurtosis Skewness Range Minimum Maximum Sum Count 51.47619048 2.402002188 53 43 41.52173913 3.575758061 42 42 11.00735685 121.1619048 0.803583546 -0.626692173 47 24 71 1081 21 17.14873323 294.0790514 0.614269919 0.309280608 75 10 85 955 23 Because we hope to show that the treatment (Group 1) is better than the control (Group 2), the hypotheses are H 0 : 1 2 vs. H a : 1 2 The two-sample t statistic is t x1 x2 s12 n1 s22 n2 51.48 41.52 2 11.01 17.15 21 23 2 2.31 The P-value for the one-sided test is P(T 2.31) . The degree of freedom k is equal to the smaller of n1 1 21 1 20 and n2 1 23 1 22 . Comparing 2.31 with entries in Table for 20 degrees of freedom, we see that P lies between 0.02 and 0.01. The data strongly suggest that directed reading activity improves the DRP score (t=2.31, df=20, 0.01<P<0.02). Example We will find a 95% confidence interval for the mean improvement in the entire population of third-graders. The interval is s12 s22 ( x1 x2 ) t * n1 n2 11.012 17.15 2 (51.48 41.52) t * 21 23 9.96 4.31 t * From Example, we have the t(20) distribution. Table D gives t* t0.025, 20 2.086 . With this approximation we have 9.96 4.31 t* 9.96 4.31 2.086 9.96 8.99 (1.0 , 18.9) We can see that zero is outside of the interval (1.0, 18.9). We can say that “ 1 2 is not equal to zero”. 9.2 Comparing two population means: Paired Difference Experiments Matched Pairs t procedures One application of the one-sample t procedure is to the analysis of data from matched pairs studies. We compute the differences between the two values of a matched pair (often before and after measurements on the same unit) to produce a single sample value. The sample mean and standard deviation of these differences are computed. Paired Difference Confidence Interval for D 1 2 Large Sample D xD z 2 nD xD z 2 sD nD Assumption: The sample differences are randomly selected from the population of differences. Small Sample sD nD xD t 2 where t 2 is based on (nD 1) degrees of freedom. Assumptions: 1. The relative frequency distribution of the population of differences is normal. 2. The sample differences are randomly selected from the population of differences. Paired Difference Test of Hypothesis for D 1 2 One-Tailed Test H 0 : D D0 H a : D D0 or H 0 : D D0 H a : D D0 Two-Tailed Test H 0 : D D0 H a : D D0 Large Sample Test statistics z xD D0 D nD xD D0 sD nD Assumption: The sample differences are randomly selected from the population of differences. Small Sample Test statistics xD D0 t sD nD where t 2 is based on (nD 1) degrees of freedom. Assumptions: 1.The relative frequency distribution of the population of differences is normal. 2.The sample differences are randomly selected from the population of differences. Example To analyze these data, we first substract the pretest score from the posttest score to obtain the improvement for each student. These 20 differences form a single sample. They appear in the “Gain” columns in Table 7.1. The first teacher, for example, improved from 32 to 34, so the gain is 34-32=2. To assess whether the institute significantly improved the teachers’ comprehension of spoken French, we test H 0 : D 0 H a : D 0 Here is the mean improvement that would be achieved if the entire population of French teachers attended a summer institute. The null hypothesis says that no improvement occurs, and H a says that posttest scores are higher on the average. The 20 differences have xD 2.5 and sD 2.893 The one-sample t statistic is t xD D0 2.5 0 3.86 sD nD 2.893 20 The P-value is found from the t(19) distribution (n-1=20-1=19). Table shows that 3.86 lies between the upper 0.001 and 0.0005 critical values of the t(19) distribution. The P-value lies between 0.0005 and 0.001. “The improvement in score was significant (t=3.86, df=19, p=0.00053).” Example A 90% confidence interval for the mean improvement in the entire population requires the critical value t 2 1.729 from Table. The confidence interval is xD t 2 sD 2.893 2.5 1.729 nD 20 2.5 1.12 (1.38, 3.62) The estimated average improvement is 2.5 points, with margin of error 1.12 for 90% confidence. Though statistically significant, the effect of the institute was rather small. 9.3 Comparing two population proportions: Independent Sampling Suppose a presidential candidate wants to compare the preference of registered voters in the northeastern United States (NE) to those in the southeastern United States (SE). Such a comparison would help determine where to concentrate campaign efforts. Properties of the Sampling Distribution of ( pˆ1 pˆ 2 ) 1. The mean of the sampling distribution of ( pˆ1 pˆ 2 ) is ( p1 p2 ) that is, E( p ˆ1 pˆ 2 ) = ( p1 p2 ) which means that ( p ˆ1 pˆ 2 ) is an unbiased estimator of ( p1 p2 ) . 2. The standard deviation of the sampling distribution of ( p ˆ1 pˆ 2 ) is p1 (1 p1 ) p1 (1 p1 ) ( pˆ1 pˆ 2 ) n1 n2 3. If the sample sizes n1 and n2 are large, the sampling distribution of ( p ˆ1 pˆ 2 ) is approximately normal. Large-Sample Confidence Interval for ( pˆ1 pˆ 2 ) z 2 ( pˆ1 pˆ 2 ) z 2 ( p1 p2 ) p1 (1 p1 ) p1 (1 p1 ) n1 n2 pˆ1 (1 pˆ1 ) pˆ1 (1 pˆ1 ) n1 n2 Assumption: The two samples are independent random samples. Both samples should be large enough that the normal distribution provides an adequate approximation to the sampling distribution of p̂1 and p̂2 . Large-Sample Test of Hypothesis about ( p1 p2 ) One-Tailed Test H 0 : ( p1 p2 ) 0 H a : ( p1 p2 ) 0 or H 0 : ( p1 p2 ) 0 H a : ( p1 p2 ) 0 Two-Tailed Test H 0 : ( p1 p2 ) 0 H a : ( p1 p2 ) 0 Large Sample Test statistics z ( pˆ 1 pˆ 2 ) ( pˆ pˆ 1 Note: 2) ( pˆ pˆ ) 1 2 p1 (1 p1 ) p1 (1 p1 ) n1 n2 1 1 pˆ (1 pˆ ) n1 n2 x1 x2 where p ˆ n1 n2 Assumption: Same as for large-sample confidence interval for ( p1 p2 ) . Let’s look at Example 9.6 in our textbook (page 471). 9.4 Determining The Sample Size Determination of Sample Size for Estimating (1 2 ) To estimate ( 1 2 ) to within a given bound B with probability (1 ) , use the following formula to solve for equal sample sizes that will achieve the desired reliability: n1 n2 ( z 2 ) 2 ( 12 22 ) B2 You will need to substitute estimates for the 2 2 values of 1 and 2 before solving for the sample size. These estimates might be sample 2 2 variances s1 and s2 from prior sampling, or from an educated guess based on the rangethat is, s R 4. Let’s look at Example 9.8 in our textbook (page 479). Determination of Sample Size for Estimating ( p1 p2 ) To estimate ( p1 p2 ) to within a given bound B with probability (1 ) , use the following formula to solve for equal sample sizes that will achieve the desired reliability: n1 n2 ( z 2 ) 2 ( p1q1 p2 q2 ) B2 You will need to substitute estimates for the values of p1 and p2 before solving for the sample size. These estimates might be based on prior samples, obtained from educated guesses or, most conservatively, specified as p1 p2 0.5. Let’s look at Example 9.9 in our textbook (page 480).