Questions 8 - 16 concern the following idea: A doughnut company is trying to market its doughnuts to health conscious consumers. It tried four different oils for the deep frying process to determine which one had the least fat absorption. Twenty doughnuts were fried in each of the oils, and the amount of fat absorbed was recorded. The fat is measured as a % of the final weight of the doughnut. Let H1, M2, M3 and u4 be the population mean amount of fat after frying in oil #1, #2, #3 and #4 respectively, and o?, o3, o? and o? be their respective population variances. An ANOVA was performed on the data and some of the output is given below, with some entries obscured with ?? mean sd n 1 36.988 2.271 20 UN 32.734 3.185 20 33.882 3.175 20 39.031 3.907 20 summary (AnovaModel.7) Df Sum Sq Mean Sq F value Pr(>F) type 3 497.0 165.66 16.3 2.83e-8 Residuals ?? 772.3 10.16 Fit: aov(formula = oil ~ type, data = donuts) Linear Hypotheses: Estimate Std. Error tvalue Pr(>|t[) 2 -1=0 -4.254 1.008 -4.220 F) group 3 1.5769 0.2019 76To analyse whether or not the mean fat absorption is the same for each oil, what are the correct set of hypotheses? a. Ho: M1 = M2 = M3 = 14 H1: not all means are equal Ob. Ho: H1 = M2 = 13 = 14 H1: all means are not equal OC. Ho: H1 = 12 = M3 = 14 H1: all means are different Od. Ho: H1 = M2 = 13 = M4 H1: M1 * M2 # 13#14What is the missing number in the DF for Residuals? a. 79 Ob. 76 Oc. 80 Od. 77What is the global ANOVA p-value associated with this analysis? (Note that when reporting p-values, R sometimes uses a non-standard format to represent a number, in answering this question you need to express that notation as a decimal number) Oa. 2.83e-8 Ob. 0.015 c. 0.0000000283 Od. 0.2019What is the estimate of the variance in the amount of fat in the doughnuts over all doughnuts? Oa. 10.16 Ob. 165.66 Oc. 1.5769 Od. 16.3If we examined a qq plot of the residuals, which of the following best describes what we would be looking for? A '~ - " a. We would want to see the points roughly fall in lines of the same height, which would help support the assumption of equal variance A '~ , " b' We would want to see a roughly bell shaped graph, which would help support the assumption of normality A '~ , " C- We would want to see a patternless spread of points across the graph, which would help support the assumption of equal variance A '~ , " d- We would want to see the points roughly fall along the straight line, which would help support the assumption of normality Read the p-value from Levene's test . Which of the following is the correct decision and conclusion based on that p-value (use a = 0.05)? a. p 0.05 so we can reject H1, meaning there is significant evidence to support the idea of equal variance. c. p 0.05 so we can not reject Ho, meaning there is no significant evidence against the idea of equal variance.Part of the ANOVA output above is the Tukey multiple comparison of means. Using this Tukey output, which of the following would be the correct conclusion when trying to find the oil(s) that are the best? Oa. Oil #2 is the best oil as it has the lowest sample mean. b. Oils #1 and #4 have the highest sample means and are not significantly different to each other (p = 0.188) but are both significantly different to oils #2 and #3. Thus oils #1 and #4 are the best. C. Oils #2 and #3 have the lowest sample means and are not significantly different to each other (p = 0.667) but are both significantly different to oils #1 and #4. Thus oils #2 and #3 are the best. Od. Oil #4 is the best oil as it has the highest sample mean.If I was to calculate a 95% CI for the mean fat for oil 3, what would be the standard deviation that I used in the calculation of that confidence interval? a. 3.187 Ob. 10.16 Oc. 10.081 d. 3.175If the 95% CI for the mean amount of fat with oil #1 was calculated as (35.57, 38.41), which ofthe following would be the correct interpretation? F'x 5.31 Every doughnut fried with oil #1 has a 95% chance of having the mean fat content being between 35.57% and 38.41%. A '. . 9 b' If we take one doughnut and fry it with oil #1 there is a 95% chance that its fat content will be between 35.57% and 38.41%. ' C- We are 95% sure that the mean fat content of doughnuts fried with oil #1 is between 35.57% and 38.41%. F'x '~ . 1' d- If we fry 100 batches of 10 doughnuts in oil #1, and for each batch calculate the sample mean fat content, then 95% of those sample means would be between 35.57% and 38.41%