10.26 A partially completed ANOVA table for a completely ran- NW domized design is shown here: Source df SS MS F Treatments 6 18.4 Error Total 41 45.2 a. Complete the ANOVA table. b. How many treatments are involved in the experiment? c. Do the data provide sufficient evidence to indicate a difference among the population means? Test, using a = .10. d. Find the approximate observed significance level for the test in part c, and interpret it.
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- Complete the following ANOVA table for an experiment that involved five treatments with sample sizes of 12, 12, 14, 15, 14:A test was conducted of two overnight mail delivery services. Two samples of identical deliveries were set up so that both delivery services were notified of the need for a delivery at the same time. The hours required to make each delivery follow. Do the data shown suggest a difference in the delivery times for the two services? Use a .05 level of significance for the test. Use Table 1 of Appendix B. Click on the datafile logo to reference the data. DATA file Service Delivery 1 2 24.5 26.0 28.0 21.0 18.0 36.0 25.0 21.0 24.0 26.0 31.0 28.0 25.5 32.0 20.0 19.5 28.0 29.0 22.0 23.5 29.5 30.0 What is the z-statistic? If required, round your answer to two decimal places. Enter negative values as negative number, if necessary. -.31 What is the p-value? If required, round your answer to two decimal places. .76 Conclude. 12345 7890:=Weight loss: In a study to determine whether counseling could help people lose weight, a sample of people experienced a group-based behavioral intervention, which involved weekly meetings with a trained interventionist for a period of six months. The following data are the numbers of pounds lost for 14 people. Assume the population is approximately normal. Perform a hypothesis test to determine whether the mean weight loss is greater than 13 pounds. Use the =α0.05 level of significance and the P -value method with the TI-84 Plus calculator. 21.2 27.5 6.5 22.8 20 11.5 16.2 19.7 36.3 32.5 11.1 34.8 22.4 17.9 Send data to Excel Part: 0 / 5 0 of 5 Parts Complete Part 1 of 5 (a) State the appropriate null and alternate hypotheses. H0 : =μ13 H1 : >μ13 This hypothesis test is a ▼right-tailed test.…
- Answer this question using the One-Way ANOVA with Equal Sample Sizes: Weights (kg) of poplar trees were obtained from trees planted in a sandy and dry region. The trees were given different treatments identified in the table below. Use a 0.05 significance level to test the claim that the four treatment categories yield poplar trees with the same mean weight. Is there a treatment that appears to be most effective in the sandy and dry region? (Show the solution) No Treatment Fertilizer Irrigation Fertilizer and Irrigation 0.60 1.16 0.65 0.22 1.11 0.93 0.08 2.13 0.07 0.30 0.62 2.33 0.07 0.59 0.01 1.74 0.44 0.17 0.03 0.12Take Test: 4.5.Case Study x * Question Completion Status: Here is a link to a data set concerning three treatment groups: A, B, & C. In order to perform an ANOVA on the data to assess if A, B, and C are different, what assumptions must be met? Enter the number of all correct answer choices in numerical order separated by commas in the blank below. 1. The variance between the three groups must be equal. 2. The data in each group must be normally distributed. 3. The members of the sample populations were randomly chosen 4. The variance between each group must be unequal. 5. The variances of each group must overlap with the variance of at least one other group. Using the data analysis add in for Excel (installation deduce if A, B, and C are different. To do this, click the data analysis tool icon, then select "single factor ANOVA", enter the range of cells containing the data, and click okay. Excel will generate a results table you can use to answer the next two questions. structions can…Use a repeated-measures ANOVA with a = .05 to determine whether there are significant mean differences between the two treatments. Calculate n? as a measure of effect size. State the results as they would appear in a research report.
- A survey conducted in a small business yielded the results shown in the table. Test the claim that health care coverage is independent of gender. Use a 0.05 significance level. What is the value of the test statistic? Men Women Health insurance 50 20 No health insurance 30 10 O A. x2 = 0.1637 O B. x2 = 3.841 O C. x2 = 5.821 O D. x2 = 8.263 %3DWhat source of variation is found in an ANOVA summary table for a within subjects design that is not in an ANOVA summary table for a between subjects design. What happens to this source of variation in a between-subjects design?To help assess the health risks of second-hand smoke, the levels of cotinine (a metabolite of nicotine) were measured in mmol/l in the urine of seven subjects prior to exposure to second-hand smoke and shortly after a two-hour exposure to secondary cigarette smoke. Did the exposure significantly increase the cotinine level? What will be the appropriate statistical analysis to use in this problem? a. z-Test for Comparing Two Means from Independent Populations b. t-Test for Comparing Two Means when the Samples are Dependent c. z-test for comparing the difference between two proportions d. t-Test for Comparing Two Means from Independent Samples e. F-test in Comparing the Difference Between Two Variances