Assignment 3 – Due Wednesday, February 15Assignment 3 – Due Wednesday, February 15Two-way ANOVA and Follow-UpA group of researchers are interested in determining the effect of coffee drinking and smoking on kidney function (see, for example, http://www.bnl.gov/bnlweb/pubaf/pr/2003/bnlpr090803.htm). The researchers classified a simple random sample of 30 adults as non-coffee drinkers, moderate coffee drinkers, or as heavy coffee drinkers. They also classified these 30 subjects smokers or non-smokers. The level of the enzyme monoamine oxidase B (MAO B) was measured in each of the subject’s kidneys. This crucial enzyme breaks down neurotransmitters and dietary amines, and too much or too little MAO B can adversely affect health and even personality. The data can be downloaded from the class web page at www.math.montana.edu/~parker/courses/STAT217/assn3_data.xlsThe first column in this Excel spreadsheet is the MAO B level. The second column specifies the three levels ofcoffee drinking (non-coffee drinkers are 1’s, moderate coffee drinkers are 2’s, and heavy coffee drinkers are 3’s). And the third column specifies the level of smoking (non-smokers are 0’s, smokers are 1’s). 1. Identify the following (1 pt):- Population of interest: human adults- Response variable: level of the enzyme MAO B- Factor A: coffee- Factor B: smoking- The values of the following:I = 3 J = 2 N = 30- x215 = 37.8072. Perform a two way ANOVA in MINITAB, and turn in all of the MINITAB output which you use to answer the following questions. 3. (1/2 pt) What is the two-way ANOVA model? Be sure to specify the distribution of the error term.),0(~Nwherexijkijkijijk4. (1/2 pt) There are two implicit assumptions in the two-way ANOVA model. What are these two assumptions?1. Constant Variance Assumption: The enzyme levels have the same standard deviation  for each population.2. The data within a group is normally distributed, ),(~ijijkNx.5. (1 pt) Use the MINITAB output to check the two assumptions from #4. What do you conclude? Do the assumptions appear to be satisfied? How did you check the assumptions?The ‘Residuals vs Fitted Values plot shows that there may be a problem with the constant standard deviation assumption across all IJ = 6 groups. In particular, there is only one non-coffee drinking smoker.The ‘Normal Probability Plot’ shows that the residuals follow a linear pattern, and so we will assume that the data within a group is normally distributed.6. (1 pt) From the MINITAB output, identify:- An unbiased estimate of 12 is 39.05. In terms of the problem, 12 represents the mean enzyme level of non-coffee drinking smokers.- An unbiased estimate of  is 7.54. In terms of the problem,  represents the standard deviationwithin each group.- An unbiased estimate of 3 is 31.59. In terms of the problem, 3 represents the mean enzyme level of the heavy coffee drinkers (over all smokers and non-smokers).The MINITAB output used to answer this question was this table of the sample means:cofee Mean SE Mean1 48.18 4.3542 41.76 1.8323 31.59 2.666smoking0 50.72 2.2171 30.30 2.857cofee*smoking1 0 57.31 4.3541 1 39.05 7.5412 0 49.57 2.5142 1 33.95 2.6663 0 45.28 4.3543 1 17.89 3.0797. (1/2 pt) Based on the interaction plots, does there appear to be a significant interaction effect of coffee drinking and smoking on the mean enzyme levels? What about the interaction plots helped you make your decision? There is evidence of non-parallelism in the interaction plots. Hence, the interaction may be significant.8. (1 pt) Conduct the interaction test.- State the appropriate Ho and Ha in terms of this problem. H0: There is no interaction effect on the mean enzyme levelHa: There is an interaction effect on the mean enzyme level- What is the value of the test statistic? FAB = 1.66 - What is the distribution of the test statistic assuming that Ho is true? FAB ~ F(2,24)- What is the p-value for this test? 0.211- State your decision at a significance level of .05. Since p-value=0.211 > .05, then we Fail To Reject H0- State your practical conclusion in terms of this problem. The evidence fails to suggest that there is a significant coffee-smoking interaction effect on the meanlevel of MAO B in the kidneys.9. (1/2 pt) Based on your conclusion of the interaction test, can you now legitimately test for main effects of factors A and B? Why or why not? Since we failed to reject the null hypothesis in our test for an interaction effect, we can now proceed with a test of the main effects of coffee and smoking on the mean levels of MAO B in the kidneys.10. (1 pt) Conduct the main effect test for coffee drinking.MINITAB gives the two-way ANOVA table:Analysis of Variance for enzyme, using Adjusted SS for TestsSource DF Seq SS Adj SS Adj MS F Pcofee 2 2222.5 813.3 406.7 7.15 0.004smoking 1 2593.8 1813.3 1813.3 31.88 0.000cofee*smoking 2 189.0 189.0 94.5 1.66 0.211Error 24 1365.0 1365.0 56.9Total 29 6370.2S = 7.54145 R-Sq = 78.57% R-Sq(adj) = 74.11%- State the appropriate Ho and Ha in terms of this problem. H0: There is NOT an effect due to coffee on the mean enzyme levelHa: There is an effect due to coffee on the mean enzyme level- What is the value of the test statistic? FA = 7.15- What is the distribution of the test statistic assuming that Ho is true? FA ~ F(2,24)- What is the p-value for this test? The p-value is 0.004.- State your decision at a significance level of .05. Since p-value=0.004 < .05, then we REJECT H0- State your practical conclusion in terms of this problem. The evidence suggests that coffee affects the mean level of the enzyme MAO B in the kidneys.- (1/2 pt) Does smoking have a significant effect on the mean levels of the enzyme MAO B in the kidneys? WHY OR WHY NOT?For the main effect test of smoking on the mean level of the enzyme MAO B in the kidneys, the teststatistic is FB = 31.88. The p-value is tiny, 0.000, which suggests that smoking affects the mean level of the enzyme MAO B in the kidneys.11. (2 pts) Follow Up Test- Is it appropriate to conduct a follow-up test to determine the effect of coffee drinking on the mean level of the enzyme MAO B