11- Chapter Eleven McGraw- © 2005 The McGraw-Hill Companies, Inc., All Chapter Eleven 11- Two-Sample Tests of GOALSHypothesis When you have completed this chapter, you will be able to: ONE Conduct a test of hypothesis about the difference between two independent population means TWO Conduct a test of hypothesis regarding the difference in two population proportions THREE Conduct a test of hypothesis about the mean difference between paired or dependent observations Chapter Eleven continued 11- Two Sample Tests of Hypothesis GOALS When you have completed this chapter, you will be able to: FOUR Understand the difference between dependent and independent samples 11- Comparing two populations Does the distribution of the differences in sample means have a mean of 0? If both samples contain at least 30 observations we use the z distribution as the test statistic The samples are from independent populations No assumptions about the shape of the populations are required X1 − X The formula for z= computing the s12 s 22 + value of z is: n1 n2 Comparing two populations 11- Two cities, Bradford and Kane are separated only by the Conewango River There is competition between the two cities The local paper recently reported that with a standard deviation the mean household income of $7,000 for a sample of in Bradford is $38,000 with 35 households At the 01 a standard deviation of significance level can we $6,000 for a sample of 40 conclude the mean income households The same in Bradford is more? article reported the mean income in Kane is $35,000 EXAMPLE 11- Step State the decision rule The null hypothesis is rejected if z is greater than 2.33 or p < 01 Step State the null and alternate hypotheses H0: µB < µK H1: µB > µK Step Find the appropriate test statistic Because both samples are more than 30, we can use z as the test statistic Step State the level of significance The 01 significance level is stated in the problem Example continued 11- Step 5: Compute the value of z and make a decision z= The p(z > 1.98) is 0239 for a one-tailed test of significance $38,000 − $35,000 ($6,000) ($7,000) + 40 35 = 1.98 Because the computed Z of 1.98 < critical Z of 2.33, the p-value of 0239 > α of 01, the decision is to not reject the null hypothesis We cannot conclude that the mean household income in Bradford is larger Example continued 11- Two Sample Tests of Proportions investigate whether two samples came from populations with an equal proportion of successes The two samples are pooled using the following X + X formula pc = n1 + n2 where X1 and X2 refer to the number of successes in the respective samples of n1 and n2 The value of the test statistic is computed from the following formula z= p1 − p pc (1 − pc ) pc (1 − pc ) + n1 n2 where X1 and X2 refer to the number of successes in the respective samples of n1 and n2 11- Are unmarried workers more likely to be absent from work than married workers? A sample of 250 married workers showed 22 missed more than days last year, while a sample of 300 unmarried workers showed 35 missed more than five days Use a 05 significance level Example 11- 10 The null and the alternate hypotheses H0: π U < π M The null hypothesis is rejected if the computed value of z is greater than 1.65 or the p-value < 05 H1: π U > π M The pooled proportion 35 + 22 pc = 300 + 250 = 1036 Example continued 11- 12 Small Sample Tests of Means The t distribution is used as the test statistic if one or more of the samples have less than 30 observations The required assumptions Both populations must follow the normal distribution The populations must have equal standard deviations The samples are from independent populations Small Sample Tests of Means 11- 13 Finding the value of the test statistic requires two steps Step One: Pool the sample standard deviations 2 ( n − ) s + ( n − ) s 2 sp = n1 + n2 − Step Two: Determine the value of t from the following formula t= X1 − X 2 s p 1 + n1 n2 Small sample test of means continued A recent EPA study compared the highway fuel economy of domestic and imported passenger cars A sample of 15 domestic cars revealed a mean of 33.7 mpg with a standard deviation of 2.4 mpg 11- 14 A sample of 12 imported cars revealed a mean of 35.7 mpg with a standard deviation of 3.9 At the 05 significance level can the EPA conclude that the mpg is higher on the imported cars? Example 11- 15 Step Find the appropriate test statistic Both samples are less than 30, so we use the t distribution Step State the null and alternate hypotheses H0: µD > µI H1: µD < µI Step State the level of significance The 05 significance level is stated in the problem Example continued 11- 16 Step The decision rule is to reject H0 if t critical z of –1.71, the pvalue of 0567 > α of 05, H0 is not rejected There is insufficient sample evidence to claim a higher mpg on the imported cars Example continued Independent samples are samples that are not related in any way 11- 19 Dependent samples are samples that are paired or related in some fashion If you wished to buy a car you would look at the same car at two (or more) different dealerships and compare the prices If you wished to measure the effectiveness of a new diet you would weigh the dieters at the start and at the finish of the program Hypothesis Testing Involving Paired Observations 11- 20 Use the following test when the samples are dependent: d t= sd / n where d is the mean of the differences sd is the standard deviation of the differences n is the number of pairs (differences) Hypothesis Testing Involving Paired Observations An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis A random sample of eight cities revealed the following information At the 05 significance level can the testing agency conclude that there is a difference in the rental charged? 11- 21 City Hertz ($) Avis ($) Atlanta 42 40 Chicago 56 52 Cleveland 45 43 Denver 48 48 Honolulu 37 32 Kansas City 45 48 Miami 41 39 Seattle 46 50 EXAMPLE 11- 22 Step Ho: µ d = H1: µ d = Step The stated significance level is 05 Step The appropriate test statistic is the paired t-test Step H0 is rejected if t < -2.365 or t > 2.365; or if p-value < 05 We use the t distribution with n-1 or degrees of freedom Step Perform the calculations and make a decision Example continued 11- 23 City Hertz Avis d d2 Atlanta 42 40 Chicago 56 52 16 Cleveland 45 43 Denver 48 48 0 Honolulu 37 32 25 Kansas City 45 48 -3 Miami 41 39 Seattle 46 50 -4 16 Example continued 11- 24 Σd 8.0 d= = = 1.00 n sd = Σd t= ( Σd ) − n n −1 d sd n = 78 − = 3.1623 = −1 1.00 3.1623 = 0.894 Example continued 11- 25 P(t>.894) = 20 for a one-tailed t-test at degrees of freedom Because 0.894 is less than the critical value, the p-value of 20 > a of 05, not reject the null hypothesis There is no difference in the mean amount charged by Hertz and Avis Example continued 11- 26 Two types of dependent samples Disadvantage of dependent samples: Degrees of freedom are halved The same subjects measured at two different points in time Matched or paired observations Advantage of dependent samples: Reduction in variation in the sampling distribution Comparing dependent and independent samples ... number of pairs (differences) Hypothesis Testing Involving Paired Observations An independent testing agency is comparing the daily rental cost for renting a compact car from Hertz and Avis A random... at two different points in time Matched or paired observations Advantage of dependent samples: Reduction in variation in the sampling distribution Comparing dependent and independent samples... you would weigh the dieters at the start and at the finish of the program Hypothesis Testing Involving Paired Observations 11- 20 Use the following test when the samples are dependent: d t= sd /