In a test of weight loss programs, 111 subjects were divided such that 37 subjects followed each of 3 diets. Each was weighed a year after starting the diet and theresults are in the ANOVA table below. Use a 0.05 significance level to test the claim that the mean weight loss is the same for the different diets.Source of VariationBetween GroupsWithin GroupsTotalP-value25.38918 0.8127 0.446353F crit3.080387dfMSF50.7783373.9773424.7552108 31.24053110Should the null hypothesis that all the diets have the same mean weight loss be rejected?O A. Yes, because the P-value is greater than the significance level.O B. No, because the P-value is greater than the significance level.OC. No, because the P-value is less than the significance level.O D. Yes, because the P-value is less than the significance level.

Solution: From the given information, the claim is that the mean weight loss is the same for the…

In a test of weight loss programs, 111 subjects were divided such that 37 subjects followed each of 3 diets. Each was weighed a year after starting the diet and the results are in the ANOVA table below. Use a 0.05 significance level to test the claim that the mean weight loss is the same for the different diets. Source of Variation Between Groups Within Groups Total df F F crit P-value 2 25.38918 0.8127 0.446353 3.080387 MS 50.778 3373.977 108 31.24053 3424.755 110 Should the null hypothesis that all the diets have the same mean weight loss be rejected? O A. Yes, because the P-value is greater than the significance level. O B. No, because the P-value is greater than the significance level. OC. No, because the P-value is less than the significance level. O D. Yes, because the P-value is less than the significance level.

In a test of weight loss programs, 111 subjects were divided such that 37 subjects followed each of 3 diets. Each was weighed a year after starting the diet and the results are in the ANOVA table below. Use a 0.05 significance level to test the claim that the mean weight loss is the same for the different diets. Source of Variation Between Groups Within Groups Total df F F crit P-value 2 25.38918 0.8127 0.446353 3.080387 MS 50.778 3373.977 108 31.24053 3424.755 110 Should the null hypothesis that all the diets have the same mean weight loss be rejected? O A. Yes, because the P-value is greater than the significance level. O B. No, because the P-value is greater than the significance level. OC. No, because the P-value is less than the significance level. O D. Yes, because the P-value is less than the significance level.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Q: Using the data (provided below this question) apply ANOVA to determine if the mean response for value for the dollar is the same for different types of cell phones. Use a 5% significance level if the problem doesn’t give an alpha. Note: Please only provide the t stat, p-value, and critical value, no need to explain the whole process. Gender Carrier Type Usage Signal strength Value for the Dollar Customer Service M AT&T Smart High 5 4 4 M AT&T Smart High 5 4 2 M AT&T Smart Average 4 4 4 M AT&T Smart Very high 2 3 3 M AT&T Smart Very high 5 5 2 M AT&T Smart Very high 4 3 5 M AT&T Smart Very high 3 4 4 F AT&T Smart Very high 3 2 3 F AT&T Smart Very high 4 3 4 M AT&T Smart Very high 3 3 1 M Other Smart Average 1 2 4 M Sprint Smart Very high 3 5 4 M Sprint Smart Very high 3 5 3 F Sprint Smart Average 2 5 4 F Sprint Smart Average 3 5 4 M Verizon Smart Average 4 3 3 F Verizon Smart Very high 4 3 2 M Verizon…
Please answer and show all work. Thank you!!!
A research assigns 21 adolescents to rate the likability of a person described in a vignette in hqvinf one of three personality traits. An equal number of participents are assigned to each group. If Fobt = 4.08 for this study, then what was the decision at a .05 level of significance for a one-way between subjects ANOVA? df bg= df E= F crit=
Is there a treatment that appears to be most effective in the sandy and dry region?
Is narcissism a more common personality trait today than it was a few decades ago? It is known that the mean population score on the Narcissistic Personality Inventory (NPI) for students attending University of South Alabama around 20 years ago was μ= 15 (Twenge, 2010). Interested in the narcissism levels of students in the year 2020, a researcher administers the NPI to a random sample of 25 University of Alabama sophomores this Spring term. The mean NPI score from the researcher’s sample of sophomores is M = 16.5, with s = 3.4. 1. Write the null and alternative hypotheses in symbols. Possible symbols for your answer: H0, H1, μ, M, σ. 2. Calculate the standard error. 3. Find the critical value for the test statistic, assuming alpha = .05 (Use largest [i.e., most conservative] value if exact value not given in the chart) a) 2.064 b) 1.96 c) 1.98 d) 2.000
Bone mineral density (BMD) is a measure of bone strength. Studies show that BMD declines after age 45. The impact of exercise may increase BMD. A random sample of 59 women between the ages of 41 and 45 with no major health problems were studied. The women were classified into one of two groups based upon their level of exercise activity: walking women and sedentary women. The 39 women who walked regularly had a mean BMD of 5.96 with a standard deviation of 1.22. The 20 women who are sedentary had a mean BMD of 4.41 with a standard deviation of 1.02. Which of the following inference procedures could be used to estimate the difference in the mean BMD for these two types of women
A data set includes the counts of chocolate chips from three different types of Chips Ahoy cookies. The accompanying StatCrunch display shows results from analysis of variance used with those three types of cookies. Use a 0.05 significance level to test the claim that the three different types of cookies have the same mean number of chocolate chips. ANOVA table Source ns on Viewer Error Total DF SS MS F-Stat P-value 2 1235.5500 617.77500 62.2790 <0.0001 77 763.80000 9.9194805 79 1999.3500 Ho: H₁ H₂ H3 Determine the alternative hypothesis. H₁: At least one of the means is different from the others Determine the test statistic. The test statistic is 62.28. (Round to two decimal places as needed.) Determine the P-value. The P-value is (Round to three decimal places as needed.)
Is there a relationship between one’s gender and whether one owns a dog, cat, or reptile? Use the data provided in the table below to answer the following question. -Show all the relevant statistical output How do I enter this data in SPSS and which test do I run? I input one column for each level but I am so confused and my output is weird looking. Do I run the One Way ANOVA in order to be able to calculate the effect size? Dog Cat Reptile Row Totals Male 20 17 11 48 Female 25 23 5 53 Column totals 45 40 16 101
ARE THE FOLLOWING DATA EQUALLY DISTRIBUTED ACROSS CATEGORIES? TEST AT.05 SIGNIFICANCE LEVEL. ABICID 35 15 10 45 WHAT STATISTICAL TEST WAS USED? OBSERVED VALUE ? CRITICAL VALUE? ACCEPT OR REJECT NULL? DO THE GROUPS DIFFER SIGNIFICANTLY?