Test 1 BNAL306 Take Home Portion Due Monday October 3 2016 10 points This problem is from 11.13 in the text. A pet company has a business objective of expanding its product line beyond its current kidney- and shrimp-based cat foods. The company developed two new products, one based on chicken livers and the other based on salmon. The company conducted an experiment to compare the two new products with its two existing ones, as well as the generic beef-based product sold in a supermarket chain. For the experiment, a sample of 50 cats from the population at a local animal shelter was selected. Ten cats were randomly assigned to each of the five products being tested. Each of the cats was then presented with 3 ounces of the selected food in a dish at feeding time. The researchers defined the variable to be measured as the number of ounces of food that the cat consumed within a 10-minute interval that began when the filled dish was presented. The results for this experiment are stored in CatFood. (If you want to see the table with the data you can look at 11.13 in the text) 1) At the 0.05 significance level, is there evidence in a difference between the mean amount consumed between the kidney and generic beef cat food? 2) At the 0.01 significance level, is there evidence in a difference between the mean amount consumed between the shrimp and salmon cat food? 3) At the 0.05 significance level, is there evidence in a difference between the mean amount consumed among the 5 products? If there is a difference, run the Tukey test to see where the differences are. Discuss the differences. For each test follow the 5 step p-value method. include the minitab output. Please clearly label each step. Also \f\f\f1) At the 0.05 significance level, is there evidence in a difference between the mean amount consumed between the kidney and generic beef cat food? Step 1: Null and alternative hypothesis In this case we want to test the difference between the mean amount consumed between the kidney and generic beef cat food. So the null hypothesis would be that there is no difference in the mean amount consumed between the kidney and generic beef cat food and alternative would be there is a difference. Symbolically this can be written as, H0: kidney - beef = 0 and H1: kidney - beef 0 Step 2: Significance level The test is performed with a 0.05 or 5% significance level. The sample sizes for both the samples are 10. Step 3: Appropriate test We can see that this is a two tailed test for a two independent sample mean. The sample sizes are small(less than 30) and the population standard deviations are unknown therefore a t-test for two independent samples would be the best test to use. Step 4: Test Statistic Using Minitab, the obtained output is given below, the test statistic value is 6.58 with associated p-value 0.000 (i.e. less than 0.001). Step 5: Conclusion As the p-value is smaller than the significance level of 0.05 so the null hypothesis is rejected in conclusion that there is a significant difference in the mean amount consumed between the kidney and generic beef cat food at 0.05 significance level. MINITAB output: 2) At the 0.01 significance level, is there evidence in a difference between the mean amount consumed between the shrimp and salmon cat food? Step 1: Null and alternative hypothesis In this case we want to test the difference between the mean amount consumed between the shrimp and salmon cat food. So the null hypothesis would be that there is no difference in the mean amount consumed between the shrimp and salmon cat food and alternative would be that there is a difference. Symbolically this can be written as, H0: Shrimp - Salmon = 0 and H1: Shrimp - Salmon 0 Step 2: Significance level The test is performed with a 0.01 or 1% significance level. The sample sizes for both the samples are 10. Step 3: Appropriate test We can see that this is a two tailed test for a two independent sample mean. The sample sizes are small(less than 30) and the population standard deviations are unknown therefore a t-test for two independent samples would be the best test to use. Step 4: Test Statistic Using Minitab, the obtained output is given below, the test statistic value is 4.27 with associated p-value 0.001. Step 5: Conclusion As the p-value is smaller than the significance level of 0.01 so the null hypothesis is rejected in conclusion that there is a significant difference in the mean amount consumed between the shrimp and salmon cat food at 0.01 significance level. MINITAB output: 3) At the 0.05 significance level, is there evidence in a difference between the mean amount consumed among the 5 products? If there is a difference, run the Tukey test to see where the differences are. Discuss the differences. Step 1: Null and alternative hypothesis In this case we want to compare between all the 5 group means. The null hypothesis would be that the group means are equal and alternative hypothesis would be that there is a difference in the mean amount of the groups i.e. H0: kidney = Shrimp = Chicken Liver = Salmon = Beef and H1: At least one mean is different Step 2: Significance level The test is performed with a 0.05 or 5% significance level. The sample sizes for each of the 5 samples are 10. Step 3: Appropriate test We can see that this is a test for more than 2 group means thus assuming normality the one way ANOVA would be the most appropriate test. Step 4: Test Statistic Using Minitab output, the test statistic value is 20.81 with associated p-value less than 0.001. Step 5: Conclusion As the p-value is smaller than the significance level of 0.05 so the null hypothesis is rejected in conclusion that at least one group mean is significantly different. Tukey's post hoc analysis is performed which indicated that the means for the groups Salmon and Beef are significantly different from the other group means. MINITAB output: One-way ANOVA: Kidney, Shrimp, Chicken Liver, Salmon, Beef Method Null hypothesis Alternative hypothesis Significance level All means are equal At least one mean is different = 0.05 Equal variances were assumed for the analysis. Factor Information Factor Factor Levels 5 Values Kidney, Shrimp, Chicken Liver, Salmon, Beef Analysis of Variance Source Factor Error Total DF 4 45 49 Adj SS 3.659 1.978 5.637 Adj MS 0.91474 0.04397 F-Value 20.81 P-Value 0.000 Model Summary S 0.209682 R-sq 64.90% R-sq(adj) 61.78% R-sq(pred) 56.67% Means Factor Kidney Shrimp Chicken Liver Salmon Beef N 10 10 10 10 10 Mean 2.4560 2.4090 2.3680 2.0280 1.7540 StDev 0.1215 0.1592 0.1621 0.2333 0.3147 95% (2.3225, (2.2755, (2.2345, (1.8945, (1.6205, CI 2.5895) 2.5425) 2.5015) 2.1615) 1.8875) Pooled StDev = 0.209682 Tukey Pairwise Comparisons Grouping Information Using the Tukey Method and 95% Confidence Factor Kidney Shrimp Chicken Liver Salmon Beef N 10 10 10 10 10 Mean 2.4560 2.4090 2.3680 2.0280 1.7540 Grouping A A A B C Means that do not share a letter are significantly different. Tukey Simultaneous 95% CIs Tukey Simultaneous 95% CIs Difference of Means for Kidney, Shrimp, ... Shrimp - Kidney Chicken Live - Kidney Salmon - Kidney Beef - Kidney Chicken Live - Shrimp Salmon - Shrimp Beef - Shrimp Salmon - Chicken Live Beef - Chicken Live Beef - Salmon -1.0 -0.8 -0.6 -0.4 -0.2 0.0 0.2 If an interval does not contain zero, the corresponding means are significantly different. Interval Plot of Kidney, Shrimp, ... Interval Plot of Kidney, Shrimp, ... 95% CI for the Mean 2.50 Data 2.25 2.00 1.75 1.50 Kidney Shrimp Chicken Liver The pooled standard deviation was used to calculate the intervals. Salmon Beef