April 8, 2004Statistics 512 Midterm 2 Spring 2004Statistics 512Midterm 2April 8, 2004The following rules apply.1. You may bring with 3 sheets of paper, double-sided with any information you need.2. You may use a calculator.3. You may not collaborate or copy.4. Failure to comply with item 3 could lead to reduction in your grade, ordisciplinary action.5. The data used for this exam was made up by Prof. Altman and was notbased on scientific results.I have read the rules above and agree to comply with them.Signature ________________________________________________Name (printed) ___________________________________________problem your score total points128222Total 501Statistics 512 Midterm 2 Spring 20041. A Chicago health-food executive began serving a 15-month prison sentence for repackaging 25-cent bakery doughnuts and reselling them for $1 each as low-fat. (from The Washington Times, Jan. 7, 2004).However, since there is a large market for diet snacks, the search for a low-fat, low calorie donut (that tastes good) continues. Donuts manufacturing consists of 3 main steps. The ingredients are mixed into dough. The dough is put into a machine which extrudes it into the familiar donut shape, and the resulting ring is fried. The amount of oil absorbed depends on the dough recipe, shape and the type of oil used.A food company is conducting an experiment to produce a lower fat donut. They intend to use 2 dough recipes, 2 extrusion shapes and 3 types of oil. The response variable is theamount of oil absorbed.a) Write a factor effects model for this experiment assuming that all effects are fixed and that the amount of oil absorbed is normally distributed. (Include all constraints and distribution assumptions.)2Statistics 512 Midterm 2 Spring 2004b) Explain how the experiment should be randomized for a balanced design, assuming that a total of 72 absorption measurements will be taken.c) Assume that the experiment was properly designed. Some of the SAS output is available in the handout "SAS Output for Midterm 2". Fill in the missing details in the ANOVA table. ABCDEF3Statistics 512 Midterm 2 Spring 2004d) Fill in the missing details in the Table of Type 3 SS. Note that there are some blank entries you do not need to fill.GHThe investigator plotted the residuals versus the fitted values and did a normal probabilityplot. She concluded that the data fit the ANOVA assumptions.e) Is there evidence of a 3-way interaction? Do a statistical test to support your claim.Null and Alternative Hypothesestest statisticp-valueConclusion (in words)4Statistics 512 Midterm 2 Spring 2004f) Is there evidence of a 2-way interaction between type of oil and recipe? Do a statisticaltest to support your claim.Null and Alternative Hypothesestest statisticp-valueConclusion (in words)g) Is there evidence of any effect of shape? Explain your answer5Statistics 512 Midterm 2 Spring 2004h) The investigator looked at the interaction plot below. (The x-axis is the levels of "oil" and the y-axis is predicted values.)Why are there 2 "+" for oil=1 and oil=2, but only 1 for oil=3?r e c i p e 1 2p1 31 41 51 61 71 81 92 02 12 22 3o i l1 2 36Statistics 512 Midterm 2 Spring 2004 i) Based on the interaction plot, the investigator decided that the effect of recipe=1, oil=1is the same as the effect of recipe=1, oil=2.Do a formal test of this (without a multiple comparisons procedure - which should of course be done).Null and Alternative Hypothesestest statisticp-valueConclusion (in words)7Statistics 512 Midterm 2 Spring 2004j) Based on the interaction plot, the investigator decided that recipe=1, shape=0, oil=2 was the lowest fat combination. The company can advertise a "lower fat" product if the mean fat content is less than 16 gm. Use a test (at a=0.05) to determine if donuts made with recipe=1, shape=0 and fried in oil=2 can be advertised as "lower fat".8Statistics 512 Midterm 2 Spring 20042. Students living in residence at college face many challenges - among them is the need to study under noisy conditions. A psychologist hypothesized that students without siblings might have more problems studying under these conditions than those who comefrom larger families.From the new freshman class, he obtained a sample of 10 students with no siblings, 10 with only older siblings and 10 with at least 1 younger sibling. He tested all of the students on high school math under quiet conditions in a classroom setting (pre-test). He then retested the students in a residence lounge. He put the TV on a music video station at medium volume, and had other students wandering in and out, holding conversations, and adjusting the TV during the test.He decided that the most appropriate analysis of the data was Analysis of Covariate, with the pre-test score as the covariate, and the test score as the response. The "treatments" were (1)="no sibs" ,(2)="older sibs only" , (3)="at least 1 younger sib" and the factor is called SIBGROUP.a) Clearly the experimental units (students) cannot be assigned at random to the treatments. What assumption needs to be made in order to do statistical inference with this sample. ("Normally distributed" should NOT be part of this discussion.)9Statistics 512 Midterm 2 Spring 2004b) The investigator did a one-way ANOVA on the pre-test scores. Suppose that he found a statistically significant difference among the 3 treatment groups. How would this affectconclusions from this study?10Statistics 512 Midterm 2 Spring 2004c) A plot of the data is below. What can you conclude about the relationship between response, pretest and treatment from this plot?s i b g r o u p 1 2 3r e s p o n s e6 07 08 09 01 0 0p r e t e s t7 0 8 0 9 0 1 0 011Statistics 512 Midterm 2 Spring 2004d) The investigator decided to test for an interaction between the treatment and covariate. The computer output for this is below.Is there a statistically significant interaction effect?p-value for testConclusionClass Levels Valuessibgroup 3 1 2 3Number of observations 30Dependent Variable: response Sum ofSource DF Squares Mean Square F Value Pr > FModel 5 1341.876125 268.375225 7.27 0.0003Error 24 885.990541 36.916273Corrected Total 29 2227.866667Source DF Type I SS Mean Square F Value Pr > Fsibgroup 2 130.466667 65.233333 1.77