F.6,005%24S'nttps://ng.cengage.com/static/nb/ui/evo/index.html?deploymentld%3D5993922324514086591623647962&elSBN=9780357035542&id%3D1231809269&snapshotld%=2490480&Maps M GmailNewsATranslate* CENGAGE MINDTAPE Re>>Q Search this courseProblem Set: Chapter 12 Introduction to Analysis of VarianceSuppose researchers at a sleep clinic are interested in the effect of surgical treatment for pediatric sleep apnea on hyperactive behavior. They study 21children without sleep apnea, 21 children with untreated sleep apnea, and 21 children who have been surgically treated for sleep apnea. Hyperactivitytudy Toolsis measured using the Conners Rating Scales.nsThe sample means and sums of squares of the scores for each of the three groups are presented in the following table.ess TipsGroupSample MeanSum of Squaresess TipsNo Sleep Apnea0.410.5120Untreated Sleep Apnea0.720.8820Treated Sleep Apnea0.480.7220ackA-The researchers perform an analysis of variance (ANOVA) at a = 0.05 to test the hypothesis that the treatment means are equal. The results arepresented in the following ANOVA table.ANOVA TableSource of VariationSum of SquaresDegrees of FreedomMean Square0.555115.73Between Treatments1.11022.0.0353Within Treatments2.116009Total3.2262624:10 AM11/30/2021pe here to searchdyhomeendort scdelete114unubackspacelock4.P.enterH.asnedW N Σ00I3.30FLml!deploymentld%3D5993922324514086591623647962&elSBN=9780357035542&id%3D1231809269&snapshotld%3D2490480&AMaps M GmailNewsA Translate* CENGAGE MINDTAP>>国Q Search this coursProblem Set: Chapter 12 Introduction to Analysis of VarianceThe ANOVA yielded a significant F statistic, so the null hypothesis is rejected. Since there are more than two groups, the researchers are interested indetermining which groups are different. The Tukey's Honestly Significant Difference (HSD) test will be used to evaluate the pairs. First, use the tabley Toolsgiven below to determine the appropriate value of q at a = 0.05. The q value for this problem is 3.40The Studentized Range Statistic (q)df for Error Term2.2.953.58sdļ4.024.64242.923.53964.553.492.89684.452.863.443.824.372.833.40093.764.28The top value is a = .05; the bottonm (bold) value is a = .01. The number of treatments is listed across. The df for the error term is in%3D%3Dthe left column, where the "error term" is another name for the within-treatments variance.4:11 AM各ツ <11/30/2021e here to searchE.a.dyendSunprt scdelete11404backspaceunumlock米$44.7.PUenterasned

There are 3 groups Group 1 - No sleep Apnea Group 2 - Untreated sleep Apnea Group 3 - Treated sleep Apnea

Answered: The ANOVA yielded a significant F…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Professor Nord stated that the mean score on the final exam from all the years he has been teaching is a 79%. Colby was in his most recent class, and his class’s mean score on the final exam was 82%. Colby decided to run a hypothesis test to determine if the mean score of his class was significantly greater than the mean score of the population. α = .01. What is the mean score of the population? What is the mean score of the sample? Is this test one-tailed or two-tailed? Why?
For the following question- It is thought that having a relatively consistent length of message in any given conversation indicates a pattern of helpful, well-balanced communication. In fact, some new research suggests that to ensure a productive conversation the maximum message length should be no more than 8 words greater than the minimum message length in any conversation. Test whether the conversations in the sample meet the criteria suggested by this research study. which type of statistical test would be best used to conduct this test= 1. χ 2 goodness of fit 2. χ 2 test for independence (2-way table) 3. Correlation 4. Anova 5. T test
Use the following R output to answer the question. Use the following R output to answer the question. >chisq.test(data) Pearsons Chi-square test data:data X-squared=4.7194, df=2, p-value=0.09445 If the significance level is 5%, what conclusion can you make? Group of answer choices A. fail to reject the alternative hypothesis B. cannot determine C. fail to reject the null hypothesis D. reject the null hypothesis and accept the alternative hypothesis
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Time to Complete the Course Right 45 47 50 49 50 50 48 44 Left | 45 | 43 48 47 50 52 | 44 42 Assume a Normal distribution. What can be concluded at the the a = 0.01 level of significance level of significance? For this study, we should use t-test for the difference between two dependent population means a. The null and alternative hypotheses would be: p1 p2 Ні: p1 p2 b. The test statistic t v v = 1.199 (please show your answer to 3 decimal places.) c. The p-value = [1338 your answer to 4 decimal places.) d. The p-value is > va e. Based on this, we should fail to reject f. Thus, the final conclusion is that ... * (Please show v the null hypothesis. O The results are statistically insignificant at a = 0.01, so there is insufficient evidence to conclude that the population mean time to complete the obstacle course with a patch over
my answer for part b) of 6.24 was: (4.3, 9.26) so the variability in Hours is equal to 4.3 hours to 9.26 hours and in minutes its 258 to 555.6 minutes. I also posted 6.24 the previous exercise so that you have all the information but I need help with 6.25.
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