UTMQUESTION 2UTMpaint. Variouswere cured, and each sample was measured for hardness. The results from a one-wayIn a study, an engineer wants to compare the hardness of five blends ofUTMsamples of paint blends were each applied to a piece of metal. The pieces of metalUTMANOVA analysis are partially displayed in the following table:2UTM UTM UTMTM &UTMAverageGroupsPlain Blend 1CountSumPlain Blend 2VarianceUTMUT84.6UTM UT16.9Plain Blend 364.31.58UTM 4Plain Blend 416.14UTM67.40.80Plain Blend 516.9697.90.50ANOVA16.3UTMSource of Variation70.31.63UTM&UT14.1UTM0.37UTMMSBetween GroupsdfWithin Groups26.6FTotal46.4i. State two assumptions used when performing a one-way ANOVA.UTMii. Calculate the respective values for x1, x1, x3, and a4 in the ANOVA tableUTMabove. Show your working.UTM UTUTMUTMUTMhave any significant effect on the hardness of metal at 5% level of significance?iii. Using your answer obtained in part (ii), can you conclude that paint blendsUTMUTMJustify your answer by carrying out the appropriate hypothesis test.UTMUTM &UTiv. Suppose more samples are added to the study and the p-value is found to be0.35. Will you change the conclusion you have obtained in partand state your new conclusion if there is any.

The one – way ANOVA, can be conducted to check the mean difference between three or four groups. The…

Answered: UTM Anova: Single Factor Groups UTM…

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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Prices and mileage of 44 Toyota Prius cars. Prices ranged from $17,043 to $48,935. Mileage ranged from 403 to 70,838. While I did not share the cars were 2017 to 2020 models. Please show all work. Does this model meet the Normality assumption? Provide evidence and a discussion of support for meeting the assumption. Does this model meet the Equal Variance (Homoscedasticity) assumption? Provide evidence and a discussion of support for meeting the assumption. Price Mileage 17167 67380 24500 8137 20700 14907 20700 13906 18995 39268 21985 3292 19995 32630 19983 36074 19477 35484 20980 23954 24940 7001 24940 7991 25480 3476 17960 41051 25980 403 24500 8137 18588 66109 18788 51022 19200 32753 21985 38838 24512 2307 22997 33671 18480 27233 18480 27758 17043 70838 17761 21967 18488 11686 17793 26080 17900 36655 18100 46264 20500 27112 22410 17384 22817 2860 21660 17060 17575 54848 20250 21084 20259 21183…
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The graphs show a violation of the assumption of (check all that apply) normality homogeneity of variance
Table Physical Health Complaints and Burnout among Social Workers: One-Way Analysis of Variance Results Physical Health Complaints Low Burnout (n = 92) Middle Burnout (n = 93) High Burnout (n = 91) F(dfs) Respiratory infections M SD 5.7 2.3 7.0 3.0 8.8 3.7 23.8(2,273)* Overall Physical Health M SD 31.7 9.4 39.7 10.6 47.8 14.1 44.1(2,272)* * p < .05 Written Assignment Based on the information provided in the article excerpts and table, write a paragraph or two that describes and interprets the findings related to the one-way ANOVA presented in the table. Be sure to include sample size, relevant descriptive statistics, patterns in the data and an interpretation of the findings presented with supportive statistical information included. You do not need to talk about every number in the table, but you do need to include relevant statistical information to support your statements.
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Microsoft Edge 3 https://elearn.squ.edu.om/mod/quiz/attempt.php?attempt3D1894461&cmid3890604&page=12 QU E-LEARNING SYSTEM (ACADEMIC) Business Statistics 1 || Spring21 Time left 1:46:05 Quiz Question 13 The total cost for a production process is $600 minus two times the number of units produced. The mean and variance of the number of produced are 200 and 20, respectively. So, the mean and variance of the total cost are ot yet swered arked out of a. 600, 200 Nu 1 Flag estion ОБ. 400, 180 O c. 1000, 560 O d. 200, 80 NEXT PAGE PREVIOUS PAGE
In an analysis of variance where the total sample size for the experiment is nT and the number of populations is k, the mean square due to error is a. SSE/(nT-k) b. SSTR/(nT-k) c. SSE/(k-1) d. SSTR/k
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The graphs show a violation of the assumption of (check all that apply) normality homogeneity of variance

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