exam2 - STAT 350 – Spring 2009 Exam 2 SOLUTION Last Name:

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Unformatted text preview: STAT 350 – Spring 2009 Exam 2 SOLUTION Last Name: ________________________ First Name: ____________________ Your Seat: __________ Section Time (circle): 10:30 11:30 4:30 Note: • You are responsible for upholding the Honor Code of Purdue University. This includes protecting your work from other students. • Show your work on all questions. Unsupported work will not receive full credit. Credit will not be given for dumb luck. Showing work includes defining any random variable or event you use in the solution. Showing work also includes identifying any named distribution you are using the solution and the values of all relevant parameters. You do not need to show work for multiple choice questions. • Decimal answers should be exact or to at least four significant digits. • Unless otherwise stated, assume the significance level for any hypothesis test is 0.05. • Standard Normal (Z) and values/probabilities must be taken from the tables provided. Probabilities, p-values, critical values, etc., for χ2, t, and F distributions must also be taken from the tables provided, unless this information is available on SAS output provided, in which case the values from SAS must be used. • You are allowed the following aids: a one-page (8.5×11 inch) cheat sheet, a scientific calculator, and pencils. • Turn off and put away your cell phone before the exam begins! Question Points Possible Points Missed 1 31 2 3 20 4 18 5 16 6 5 Total STAT 350 Exam 2 10 100 Page 1 of 1 1. Apples and oranges were compared for the quantities of fiber. The results of PROC TTEST are given below. The TTEST Procedure Statistics Variable fruit N fiber fiber fiber apple orange Diff (1-2) 15 18 Lower CL Mean Mean Upper CL Mean Lower CL Std Dev Std Dev Upper CL Std Dev Std Err 45.849 53.703 -27.14 52.716 65.751 -13.03 59.583 77.798 1.0702 9.0781 18.18 15.859 12.4 24.227 19.782 19.555 36.32 26.299 3.2016 5.7104 6.9158 T-Tests Variable Method Variances fiber fiber Pooled Satterthwaite Equal Unequal DF t Value Pr > |t| 31 26.2 -1.88 -1.99 0.0689 0.0570 Equality of Variances Variable Method fiber Folded F Num DF Den DF F Value Pr > F 17 14 3.82 0.0150 a. (6 pts; 3 pts each part) Is it appropriate to assume equal variances (i.e., σapple = σorange)? (i) Just answer "yes" or "no". No (ii) Give the p-value you used to determine this. 0.0150 b. (15 pts; 5 pts each part) For the following questions, assume that the null hypothesis is µapple - µorange = 0. Be sure your answers below are consistent with your answer to part (a). (i) Assuming the alternative hypothesis was µapple - µorange < 0, what is the appropriate p-value? 0.0570/2 = 0.0285 (ii) Assuming the alternative hypothesis was µapple - µorange > 0, what is the appropriate p-value? 1 – 0.0570/2 = 0.9715 (iii) Assuming the alternative hypothesis was µapple - µorange ≠ 0, what is the appropriate p-value? 0.0570 STAT 350 Exam 2 Page 2 of 2 1. (continued) c. (10 pts) You have just picked an apple. Give a 95% prediction interval for the quantity of fiber in this apple. 52.716 ± 2.145(12.4) 1 + 1 15 → (25.2467, 80.1863) 2. (10 pts; 5 pts each part) Basics of hypothesis testing a. You are testing the null hypothesis μ = 50 against the alternative hypothesis μ ≠ 50 at the α = 0.10 significance level. You have 10 observations (x1, x2, . . . , x10). Give the critical value for your test and sketch the rejection region. df = 10 – 1 = 9 tcrit = ±1.833 (Table IV) The rejection region is the area ≤-1.833 and the area ≥+1.833 b. You are testing the null hypothesis μ = 50 against the alternative hypothesis μ ≠ 50. You have 16 observations (x1, x2, . . . , x16). The value of your test statistic is 1.8. What is the p-value for this test? df = 16 – 1 = 15 p-value = (0.046)×2 = 0.092 (Table VI) STAT 350 Exam 2 Page 3 of 3 3. A clinical trial was conducted to test the effectiveness of DeChol, a new medication intended to reduce cholesterol. 400 subjects, aged 50 to 65, volunteered to participate. Subjects were randomly assigned to take either DeChol or a placebo (though they did not know which they are on). The trial lasted 5 years. In addition to measuring cholesterol levels of the subjects, many other variables were recorded (e.g., blood pressure, weight) including any medical complaints (e.g., dry mouth, headaches) or medical conditions or complications (e.g., cancers, heart attacks, death) for the purpose of finding side-effects. Below is a table showing the number of subjects in each treatment group that either had or did not have a heart attack during the clinical trial. Heart Attack No Heart Attack 12 188 DeChol Placebo 4 196 a. (8 pts) Fill in the table below with the expected counts. Heart Attack No Heart Attack 8 192 DeChol placebo 8 192 b. (6 pts) What is the value of the appropriate test statistic? (12 − 8 ) χ= 2 8 2 ( 4 − 8) + 8 2 (188 − 192 ) + 192 2 (196 − 192 ) + 192 2 = 4.16667 c. (2 pts) What is the critical value for this test (α = 0.05)? df = (2-1)(2-1) = 1 Critical Value = 3.84 d. (4 pts) Based on your analysis above, what is your conclusion? Circle the single most appropriate choice from the options below. The test statistic (4.1667) is greater than the critical value (3.84), therefore we REJECT the null hypothesis of independence. Because this data was based on an EXPERIMENT (random assignment of the drug/placebo), we can conclude the difference is CAUSED by the treatment (DeChol vs. placebo) There is no statistically significant association between the treatment group and the whether or not a person has a heart attack. In other words, there is no reason to believe taking DeChol alters ones risk of having a heart attack. There is a statistically significant association between the treatment group and the whether or not a person has a heart attack, but we cannot conclude that the difference in heart attack rates between the two groups is caused by DeChol. We can conclude that taking DeChol causes a statistically significant change in the likelihood of having a heart attack. STAT 350 Exam 2 Page 4 of 4 4. (18 pts; 3 pts each part) Multiple Choice. For each part, circle the single most appropriate answer from the options provided. a. Assume that you are testing H0: μ = 5 versus Ha: μ ≠ 5 at α = 0.05 and you rejected the null hypothesis. If you had used the same data to instead test H0: μ = 5 versus Ha: μ > 5 at α = 0.05, would you have also rejected the null hypothesis? Note that you do not know whether x was <5 or >5, so we do not know if the test statistic is positive or negative. definitely yes definitely no maybe b. You want to test the null hypothesis that μ = 25 against the alternative hypothesis that μ ≠ 25. The 95% CI for μ is (23.2, 47.6). Therefore, the p-value of your test must be greater than 0.05. See Lecture K, page # 14 definitely yes definitely no maybe c. In hypothesis testing, the p-value is the probability that the null hypothesis is true. See Lecture K, page # 4 True False d. When conducting a t-test, a paired t-test can be used as long as there are the same number of observations in both groups. The data must be inherently paired True False f. You are testing the null hypothesis μ1 - μ2 = 0 against the alternative hypothesis μ1 - μ2 ≠ 0 at the α = 0.01 significance level. Your p-value is 0.0345. the p-value is greater than the significance level(alpha), therefore we fail to reject the null You conclude that the two group means are not significantly different You conclude that the two group means are significantly different g. If you conduct a 1-way ANOVA with only 2 treatment levels, this is equivalent to See Lecture L, page # 16-17 conducting an independent sample Satterthwaite t-test conducting an independent sample pooled t-test conducting a paired t-test STAT 350 Exam 2 Page 5 of 5 5. Kristen is a food scientist studying methods of reducing lactose content in ice cream. In her experiment, she used lactase (an enzyme that breaks down lactose) was taken from 4 different microbial sources (Kluyveromyces fragilis, K. lactis, Aspergillus niger, and A. oryzae). She also used ice cream with 2 different levels of butterfat (10% and 15%). She made 3 batches of ice cream for each combination of lactase source and butterfat content. One of the measurements she was interested in was the texture of the product. Volunteers each tasted each batch of ice cream and rated the texture on a scale of 1 to 20 (20 being the best). Kristen took the average score for each batch as the measure of texture. The SAS results from the ANOVA are given below. Some information has been omitted and replaced with asterisks (*). The omitted values have been replaced for the solution (in blue) The ANOVA Procedure Class Level Information Class Levels Values lactase 4 an ao kf kl butterfat 2 10 15 Number of Observations Read Number of Observations Used 24 24 The ANOVA Procedure Dependent Variable: texture DF Sum of Squares Mean Square F Value Pr > F Model 7 588.5330292 84.0761470 21.14 <.0001 Error 16 63.6276667 3.9767292 Corrected Total 23 652.1606958 Source R-Square Coeff Var Root MSE texture Mean 0.902436 19.26816 1.994174 10.34958 Source lactase butterfat lactase*butterfat STAT 350 Exam 2 DF Anova SS Mean Square F Value Pr > F 3 1 3 25.8171458 543.8776042 18.8382792 8.6057153 543.8776042 6.2794264 2.16 136.77 1.58 0.1322 <.0001 0.2335 Page 6 of 6 5 (continued) Was there a significant lactase effect? a. (7 pts) Give the value of the appropriate test statistic. = MS (lactase) SS (lactase) (4 − 1) 25.817 3 8.6057 = = = = 2.164 MS ( Error ) SS (error ) 16 63.628 16 3.9767 b. (3 pts) Give the degrees of freedom for this test statistic. numerator df = 3, denominator df = 16 c. (2 pts) Give the critical value for this test (α = 0.05). 3.24 d. (4 pts) Based on your analysis above, what is your conclusion? Circle one of the options below. The test statistic (2.164) is less than the critical value (3.24), therefore fail to reject null There is NO statistically significant difference in the mean texture among the 4 sources of lactase studied The mean texture for each lactase source is significantly different from each of the other lactase sources. At least one of the lactase sources has a mean texture that is significantly different from the other lactase sources. 6. (5 pts) Assume you have conducted an analysis of variance (ANOVA) and a multiple comparison procedure is warranted. When is it appropriate to use Dunnett’s procedure instead of Tukey’s? You must only be concerned with comparing the other treatment groups to a control (or baseline) treatment – not comparing the other treatments to each other. Note – it is not enough to say that there is a control (or that you wish to compare the other groups to the control) – you can still use Tukey’s test when there is a control and Tukey’s will tell you if the other treatment are different than the control. STAT 350 Exam 2 Page 7 of 7 ...
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