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Independent Samples: Comparing Means Lecture 40 Section 11.4 Mon, Apr 10, 2006 Independent Samples     In a paired study, two observations are made on each subject, producing one sample of bivariate data. Or we could think of it as two samples of paired data. Often these are “before” and “after” observations. By comparing the “before” mean to the “after” mean, we can determine whether the intervening treatment had an effect. Independent Samples     On the other hand, with independent samples, there is no logical way to “pair” the data. One sample might be from a population of males and the other from a population of females. Or one might be the treatment group and the other the control group. The samples could be of different sizes. Independent Samples     We wish to compare population means 1 and 2. We do so by comparing sample meansx1 andx2. More specifically, we will usex1 –x2 as an estimator of 1 – 2. If we want to know whether 1 = 2, we test to see whether 1 – 2 = 0 by computingx1 –x2. The Distributions ofx1 andx2    Let n1 and n2 be the sample sizes. If the samples are large, thenx1 andx2 have (approx.) normal distributions. However, if either sample is small, then we will need an additional assumption.  The populations are normal. Further Assumption  We will also assume that the two populations have the same standard deviation.    Call it . If this assumption is not supported by the evidence, then it should not be made. If this assumption is not made, then the formulas become much more complicated. See p. 658. The Distribution ofx1 –x2   Suppose thatx1 andx2 have normal distributions with means 1 and 2 and standard deviations 1/n1 and 2/n2 (according to the Central Limit Theorem for Means). Thenx1 –x2 is a normal random variable with the following properties: The mean is 1 – 2. 2 2  The variance is    x x 2   x 2   x 2  1  2  1 2 1 2 n1 n2 The Distribution ofx1 –x2  If we assume that 1 = 2, (call it ), then the standard deviation may be simplified to 2 2 1 1    n1 n2 n1 n2  That is,  1 1   x1  x2 is N  1   2 ,   . n1 n2   The Distribution ofx1     x1 is N  1 ,  n1   0 1 The Distribution ofx2    x2 is N   2 ,  n2   0 2 1 The Distribution ofx1 –x2  1 1   x1  x2 is N  1   2 ,    n1 n2   0 1 – 2 2 1 The Distribution ofx1 –x2  If   1 1 x1  x2 is N  1   2 ,   , n1 n2   then it follows that  x1  x2   1   2  Z 1 1   n1 n2 The t Distribution   Let s1 and s2 be the sample standard deviations. Whenever we use s1 and s2 instead of , then we will have to use the t distribution instead of the standard normal distribution, unless the sample sizes are large. Estimating     Individually, s1 and s2 estimate . However, we can get a better estimate than either one if we “pool” them together. The pooled estimate is n1  1s1  n2  1s2 2 sp  n1  n2  2 2 . x1 –x2 and the t Distribution    If we use sp instead of , and the sample sizes are small, then we should use t instead of Z. The number of degrees of freedom is df = df1 + df2 = n1 + n2 – 2. That is x  x   1   2  t (n  n  2)  1 2 1 2 sp 1 1  n1 n2 Hypothesis Testing   See Example 11.4, p. 699 – Comparing Two Headache Treatments. State the hypotheses. H0: 1 = 2  H1: 1 > 2   State the level of significance.   = 0.05. The t Statistic  Compute the value of the test statistic.  The test statistic is x1  x2 t 1 1 sp  n1 n2 with df = n1 + n2 – 2. Computations 9s1  9s2 sp   5.052. 18 22.6  19.4 t  1.416. 1 1 5.052  10 10 2 2 Hypothesis Testing  Calculate the p-value. The number of degrees of freedom is df = df1 + df2 = 18.  p-value = P(t > 1.416) = tcdf(1.416, E99, 18) = 0.0869.  Hypothesis Testing  State the decision.   Accept H0. State the conclusion.  At the 5% level of significance, the data do not support the claim that Treatment 1 is more effective than Treatment 2. The TI-83 and Means of Independent Samples       Enter the data from the first sample into L1. Enter the data from the second sample into L2. Press STAT > TESTS. Choose either 2-SampZTest or 2-SampTTest. Choose Data or Stats. Provide the information that is called for.  2-SampTTest will ask whether to use a pooled estimate of . Answer “yes.” The TI-83 and Means of Independent Samples   Select Calculate and press ENTER. The display shows, among other things, the value of the test statistic and the p-value. Paired vs. Independent Samples  The following data represent students’ calculus test scores before and after taking an algebra refresher course. Student 1 2 3 4 5 6 7 8 Before 85 63 94 78 75 82 45 58 After 92 68 98 83 80 88 53 62 Paired vs. Independent Samples  Perform a test of the hypotheses H0: 2 – 1 = 0  H1: 2 – 1 > 0  treating the samples as independent. Paired vs. Independent Samples  Had we performd a test of the “same” hypotheses H0: D = 0  H1: D > 0   treating the samples as paired, then the p-value would have been 0.000005688. Why so small? Paired vs. Independent Samples  Why is there a difference? 100 90 80 70 60 50 40 1 2 3 4 5 6 7 8 Confidence Intervals    Confidence intervals for 1 – 2 use the same theory. The point estimate isx1 –x2. The standard deviation ofx1 –x2 is approximately sp 1 1     n1 n2  Confidence Intervals  The confidence interval is or 1 1 x1  x2  z       n1 n2  ( known, large samples) 1 1 x1  x2  z  s p     n1 n2  ( unknown, large samples) or x1  x2  t  s p 1 1  n1 n2 ( unknown, normal pops., small samples) Confidence Intervals  The choice depends on Whether  is known.  Whether the populations are normal.  Whether the sample sizes are large.  Example      Find a 95% confidence interval for 1 – 2 in Example 11.4, p. 699. x1 –x2 = 3.2. sp = 5.052. Use t = 2.101. The confidence interval is 3.2  (2.101)(2.259) = 3.2  4.75. The TI-83 and Means of Independent Samples  To find a confidence interval for the difference between means on the TI-83, Press STAT > TESTS.  Choose either 2-SampZInt or 2-SampTInt.  Choose Data or Stats.  Provide the information that is called for.   2-SampTTest will ask whether to use a pooled estimate of . Answer “yes.”