双语统计学.doc

Example 13.1 .that high-fiber cereals reduce the likelihood of various forms of cancer. .another advantage of eating their productpotential weight reduction for dieters. Can the scientist conclude at the 5% significance level that his belief is correct? (p396) Solution : The data are obviously interval. This problem objective/data type combination tells us that the parameter to be tested is the difference between two means, 1 2. Then, the test is known as the unequal-variance t-test. I. Formulate the HypothesesH0: 1 2; H1: 1 2. II. Select the Test Statistic III. Identify the Rejection Region Select the significance level = 0.05, referring to the t-value table, and a critical value t() = t0.05(123) = 1.658. Hence, the rejection region is: t-1.658 . IV. Make the Statistical Decision On the sample data, the value of the test statistic t is: and it does fall into the rejection region, then we reject H0 . As a result, we conclude that there is sufficient evidence to infer that consumers of high-fiber cereal do eat fewer calories at lunch than do nonconsumers.Example 13.4 . Matched pairs experiment depending on GPA.(p416) Solution : The experimental design tells us that the parameter of interest is the mean of the population of differences, which we label D. Note that 1 2 = D but that we test D because of the way the experiment was performed. I. Formulate the Hypotheses H0: D= 0; H1: D 0 II. Select the Test StatisticIII. Identify the Rejection Region . VI. Make the Statistical Decision .Example 15.1 . The marketing manager has to decide how to market the new product. She can create advertising that emphasizes convenience, quality, or price. To facilitate a decision, she conducts an experiment in three different small cities. . Solution : According to the conditions in the problem, we use ANOVA test. It is known that the dependent variable (response variable) is weekly sales (packages), and the factor is the advertising strategy and there are three levels. I. Formulate the Hypotheses H0: 1=2=3 ; H1: At least two means differ. II. Select the Test StatisticMST = SST / dfT , MSE= SSE/ dfE , dfT = k 1, dfE = n - k III. Identify the Rejection Region Select the significance level =0.05, referring to the F-value table, and a critical value F(dfT, dfE)= F0.05(2, 57)= 3.15. Hence, the rejection region is: F 3.15. IV. Make the Statistical Decision At first, we calculate the sum of squares, SS:According to the sample data, xij = 36,784 x2ij = 23,115,540; T.1 = 11,551, T.2 = 13,060, T.3 = 12,173, T2.j = 452,171,130. Then, SST = 57,513, SSE = 506,983, SS(total)= 564,496.Thus, MST= SST / dfT = 28,756, MSE = 8,894, and F = MST / MSE = 3.23.When = 0.05, we can reject H0 . So, there is enough evidence to infer that the mean weekly sales differ between the three cities. V. Tabulate ANOVASource of Variation Sum of Squares df MS F-StatisticTreatments 57,513 2 28,756 3.23 error 506,983 57 8,894total 564,495 5915.2 . cholesterol, which can lead to heart attacks. . Four such drugs. . Solution: According to the conditions in the problem, we use ANOVA test. It is known that we identify the experimental design as randomized block, and response variable is the cholesterol reduction, the treatments are the drugs, and the blocks are the 25 similar groups of men. I. Formulate the Hypotheses H0: 1=2=3=4 ; H1: At least two means differ. II. Select the Test StatisticMST = SST / dfT , MSE= SSE/ dfE , dfT = k 1, dfE = n k b + 1 III. Identify the Rejection Region Select the significance level =0.05, referring to the F-value table, and a critical value F(dfT, dfE)= F0.05(3, 72)= 2.76. Hence, the rejection region is: F 2.76. IV. Make the Statistical Decision At first, we calculate the sum of squares, SS: According to the sample data,SS(Total) = 5,187.2, SST = 196.0 SSB = 3,848.7 SSE = 1,142.5 MST = 65.3 MSE = 15.9 So, F = MST / MSE = 65.3 / 15.9 = 4.1 . When =0.05, we do reject H0 .V. Tabulate ANOVAVariation Source SS df MS F Treatment T 196.0 3 65.3 4.1 2.76 Block B 3,848.7 24 160.3 10.1 1.67 Error 1,142.5 72 15.9 Total 5,187.2 99A Type I error occurs when you conclude that difference exist when, in fact, they do not . A T