Transcript Treatment 1
Tests of Hypotheses Involving Two Populations 12.1 Tests for the Differences of Means Comparison of two means: 1 and 2 The method of comparison depends on the design of the experiment The samples will either be Independent or Dependent Independent Samples – implies data values obtained from one sample are unrelated to values from the other sample Dependent Samples – implies subjects are paired so that they are as much alike as possible The purpose of pairing is to explain subject to subject variability. (In some studies we apply both treatments to the same subject) If subject to subject variability is large relative to the expected treatment differences then a dependent sample design should be considered 12.3 Tests for Differences of Means of Paired Samples Observe n pairs of observations x ij observation for treatment i in pair j d j x1 j x 2 j difference between treatments 1 and 2 in pair j Pair Treatment 2 Difference 2 x11 x12 x 21 x 22 d1 x11 x21 d 2 x12 x22 3 x13 x 23 d 3 x13 x23 . . . . . . . . . . . . n x1n x2n d n x1n x2 n Sample Mean x1 x2 d x1 x 2 Sample Variance s1 Sample Standard Deviation s1 1 Treatment 1 2 s2 s2 2 sd sd 2 For dependent sample design all analysis is based on differences: d1 , d 2 , d 3 ,..., d n The differences are a sample from a distribution with 2 mean d 1 2 and an unknown variance d We can calculate Confidence Intervals and perform Hypothesis Tests in the same way as with one sample Point Estimate for d 1 2 is d x1 x 2 d d The Test Statistic is T s d n which has a t-distribution with n-1 degrees of freedom Confidence Interval for d 1 2 s E t d d E n Example A marketing expert wishes to prove that a new product display will increase sales over a traditional display Treatment 1: New Method Treatment 2:Traditional Method 12 stores are available for the study There is considerable variability from store to store We will divide the 12 stores into six pairs such that within each pair the stores are as alike as possible Measurement will be cases sold in a one month period Pair New Method Treatment 1 Traditional Method Treatment 2 Difference 1 13 11 2 2 31 29 2 3 20 21 -1 4 19 17 2 5 42 39 3 6 26 22 4 Sample Mean d 2 Sample Variance s d 2.8 Sample Standard Deviation s d 1.6733 2 n6 d 2 s d 1.6733 Perform a Hypothesis test using .05 Conclude with 95% confidence that the new method produces larger sales. Perform and interpret a 95% confidence interval We are 95% confident that the mean increase in sales is between 0.244 cases and 3.756 cases using the new product display. 12.2 Tests of Differences of Means Using Small Samples from Normal Populations When the Population Variances Are Equal but Unknown Notation for Two Independent Samples Sample 1 (Treatment 1) Sample 2 (Treatment 2) n1 n2 x11 , x12 , x13 ,..., x1n1 x 21 , x 22 , x 23 ,..., x 2 n2 Sample Mean x1 x2 Sample Variance s1 2 s2 Sample Standard Deviation s1 s2 Sample Size Data 2 The point Estimate for 1 2 (Independent Samples) is x1 x2 This is the same as what we did with dependent samples The sampling distribution for x1 x2 x x 1 2 1 2 2 x1 x2 2 1 n1 2 2 n2 If we sample from populations with means of 1 and 2 , and standard deviations of 1 and 2 respectively, then x1 x 2 1 2 Z 2 1 n1 2 2 n2 is approximately standard normal for large n. The form of a confidence interval and a hypothesis test for two independent samples depends on what we assume If we sample from populations 1 and 2, the samples are independent, and 1 2 2 then 2 x1 x 2 1 2 T S 2 p n1 S 2 p n2 has a t-distribution with d . f . n1 n2 2 Where s p 2 n1 1s1 2 n2 1s 2 2 n1 n2 2 This means s p s p 2 2 2 is pooled estimate of 1 2 Thus for a Hypothesis Test the Test Statistic is x1 x 2 1 2 T S 2 p n1 S 2 p n2 and a Confidence Interval is calculated using 2 S2 S E t p p n1 n2 x1 x2 E Example A study is designed to compare gas mileage with a fuel additive to gas mileage without the additive. A group of 10 Ford Mustangs are randomly divided into two groups and the gas mileage is recorded for one tank of gas. Treatment 1 (with additive) Treatment 2 (without additive) n1 5 n2 5 26.3, 27.4, 25.1, 26.8, 27.1 24.5, 25.4, 23.7, 25.9, 25.7 x1 26.54 x2 25.04 s1 0.813 s 2 0.848 Sample Size Data Sample Mean Sample Variance 2 2 Do we have enough evidence at .05 to prove that the 2 2 Additive increases gas mileage. Assume 1 2 Conclude with 95% confidence that the additive improves gas mileage Compute and interpret a 95% Confidence Interval for 1 2 We are 95% confident that the additive will increase gas mileage by an amount between 0.177 and 2.823 miles. What if we do not assume 1 2 ? 2 It then makes no sense to compute s p 2 2 A reasonable variable is T x1 x 2 ( 1 2 ) s1 2 n1 s 2 2 n2 Unfortunately the distribution of T is unknown. For large n1 and n 2 , T is approximately standard normal. For small n1 and n 2 , the distribution of T can be approximated by a t-distribution using a complicated formula for the degrees of freedom.