O(d) Compute the mean square due to error. (Round your answer to two decimal places.)(e) Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to four decimal pSourceof VariationDegreesof FreedomMeanSquareTreatmentsErrorTotalSumof Squares25741O HO HA HB HcH: HA HB HCX216Hoi HAHB "cH: Not all the population means are equal.18O Ho: Not all the population means are equal.H₂HAHHCO Ho: MAHB HcH₂HA HB HCOHO: HA HB HCHai HA HB #C(f) At the a 0.05 level of significance, test whether the means for the three treatments are equal.State the null and alternative hypotheses.✔• HO HA" HB - исH: Not all the population means are equal.O Ho: Not all the population means are equal.Ha: MAHB HcxShow Transcribed Textx(f) At the a= 0.05 level of significance, test whether the means for the three treatments are equarState the null and alternative hypotheses.FO Ho: At least two of the population means are equal.H: At least two of the population means are different.p-valueFind the value of the test statistic. (Round your answer to two decimal places.)Find the p-value. (Round your answer to four decimal places.)p-value=State your conclusion.O Do not reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal.Reject Ho. There is sufficient evidence to conclude that the means for the three treatments are not equal.O Do not reject Ho. There is not sufficient evidence to conclude that the means for the three treatments are notequal.O Reject Ho. There is not sufficient evidence to conclude that the means for the three treatments are not equal. You may need to use the appropriate technology to answer this question.The following data are from a completely randomized design.SamplemeanSamplevarianceA161141166144147171Treatment155B142156142137124 138145C141127121140150122133158.8 109.6 132.8(a) Compute the sum of squares between treatments.2574X(b) Compute the mean square between treatments.143X(c) Compute the sum of squares due to error.

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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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The accompanying data table contains chest deceleration measurements (in g, whereg is the force of gravity) from samples of small, midsize, and large cars. Shown are the technology results for analysis of variance of this data table. Assume that a researcher plans to use a 0.05 significance level to test the claim that the different size categories have the same mean chest deceleration in the standard crash test. Complete parts (a) and (b) below. Click the icon to view the table of chest deceleration measurements A Click the icon to view the table of analysis of variance results. ..... a. What characteristic of the data specifically indicates that one-way analysis of variance should be used? O A. There are three samples of measurements. O B. The population means are approximately normal. O C. The measurements are categorized according to the one characteristic of size. O D. Nothing specifically indicates that one-way analysis of variance should be used. b. If the objective is to test…
8. Hypothesis Test of Equal Variances. Test the given claim. Use a significance level of a = 0.05 and assume that all populations are normally distributed. Claim: IQ scores of statistics students vary less than IQ scores of other students. Statistics students: n = 28, x = 118, s = 10.0 Other students: n = 25, x = 112, s = 12.0
A researcher takes sample temperatures in Fahrenheit of 17 days from New York City and 18 days from Phoenix. Test the claim that the mean temperature in New York City is different from the mean temperature in Phoenix. Use a significance level of α=0.05. Assume the populations are approximately normally distributed with unequal variances. You obtain the following two samples of data. New York City Phoenix 99 94.2 95.5 72 93.2 86.8 102 122.1 85.4 114.4 80 94.7 86.4 89.7 75.4 104.7 79.5 77.6 83.4 106.8 64.3 98.6 65.5 91.5 87.7 82 104 97.7 74.3 64.9 59.5 82 82.8 72 116.2 The Hypotheses for this problem are: H0: μ1 = μ2 H1: μ1μ2 Find the p-value. Round answer to 4 decimal places. Make sure you put the 0 in front of the decimal. p-value =
You work for a soft-drink company in the quality control division. You are interested in the variance of one of your production lines as a measure of consistency. The product is intended to have a mean of 12 ounces, and your team would like the variance to be as low as possible. You gather a random sample of 16 containers. Estimate the population variance at a 80% level of confidence. 11.82 11.83 11.85 11.89 11.9 11.97 12 12 12 12.01 12.02 12.04 12.05 12.11 12.16 12.18 Variance of Data: 0.012 Note: Round all values to 3 decimals and use those rounded values in subsequent calculations. a) Find the lower and upper x critical values at 80% confidence: Lower: Upper: b) Report your confidence interval for o?: Lower Bound: Upper Bound:
(a) Perform the analysis of variance (ANOVA) at the 5% significance level to identify is there any difference in the mean between the three populations. 1 2 3 94 71 84 96 82 87 87 87 93 89
Ten randomly chosen participants in a Math Training Program took a pretest and a posttest. Test at 0.05 level of significance whether the variance of the posttest scores is significantly different from the variance of the pretest scores. Use the data the following data. [pretest: 5, 6,7,7,7,8,8,9,9,10] [posttest: 6,7,9,8,7,913,10,11,6]
Heights The heights (cm) in the following table are from Data Set 1 “Body Data” in Appendix B. Results from two-way analysis of variance are also shown. Use the displayed results and use a 0.05 significance level. What do you conclude? XLSTAT
please provide accurate answer
Speed Dating Listed below are attribute ratings of males by females who participated in speed dating events (from Data Set 18 “Speed Dating” in Appendix B). Use a 0.05 significance level to apply the methods of two-way analysis of variance. What do you conclude?
The article cited in Exercise 19 provides measures of electrical conductivity (in uS/cm). The results are presented in the following table. Sample Size 49 Location Mean SD Upstream 463 208 Midstream 363 98 31 Downstream 647 878 30 Can a one-way analysis of variance be used to determine whether conductivity varies with location? Or is one of the necessary assumptions violated? Explain.
The number of traffic accidents in a city was recorded for a year is as reported below. What is the sample variance? Jan Feb Mar Apr May Jun | Jul Oct Aug Sep Nov 18 18 20 19 37 28 48 36 38 41 35 O 131.61 ○ 117.00 ○ 116.00 O 120.63

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