SECTION 8.3 Testing the Difference Between Means (Dependent Samples) 441 Using the table at the left, you can calculate d and sd as shown below. d = Σd n = 86 16 = 5.375 sd = HΣd2 - c 1Σd22 n d n - 1 = H1874 - 862 16 16 - 1 ≈ 9.7014 The standardized test statistic is t = d - md sd 1n Use the t@test. ≈ 5.375 - 0 9.7014 116 Assume md = 0. ≈ 2.216. You can check this result using technology, as shown below using StatCrunch. STATCRUNCH Paired T hypothesis test: μD = μ1 - μ2 : Mean of the difference between Last year and This year H0 : μD = 0 HA : μD ≠0 Hypothesis test results: Difference Mean Std. Err. DF T-Stat P-value Last year–This year 5.375 2.4253436 15 2.2161808 0.0426 The figure at the left shows the location of the rejection region and the standardized test statistic t. Because t is in the rejection region, you reject the null hypothesis. Interpretation There is enough evidence at the 1% level of significance to conclude that the legislator’s performance rating has changed. TRY IT YOURSELF 2 A medical researcher wants to determine whether a drug changes the body’s temperature. Seven test subjects are randomly selected, and the body temperature (in degrees Fahrenheit) of each is measured. The subjects are then given the drug and, after 20 minutes, the body temperature of each is measured again. The results are listed below. At a = 0.05, is there enough evidence to conclude that the drug changes the body’s temperature? Assume the body temperatures are normally distributed. Subject 1 2 3 4 5 6 7 Initial temperature 101.8 98.5 98.1 99.4 98.9 100.2 97.9 Second temperature 99.298.498.299.098.6 99.797.8 Answer: Page A41 Before After d d2 60 54 6 36 54 46 8 64 78 68 10 100 84 58 26 676 91 83 8 64 25 38 -13 169 50 38 12 144 65 53 12 144 68 78 -10 100 81 72 9 81 75 76 -1 1 45 48 -3 9 62 48 14 196 79 83 -4 16 58 51 7 49 63 58 5 25 Σ = 86 Σ = 1874 −3 −2 −4 −1 0 1 2 3 4 t t0 = 2.131 −t0 = −2.131 t ≈ 2.216 α= 0.025 1 2 α= 0.025 1 2 1 − = 0.95 α
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