Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 36 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was usedby 12 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 12,120; SSTR = 4,540.(a) 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 places.)SourceSumMeanDegreesof FreedomFp-valueof Variationof SquaresSquareTreatments454022709.88Error758033229.70Total1212035(b) Use a = 0.05 to test for any significant difference in the means for the three assembly methods.State the null and alternative hypotheses.Ho: H1 = "2 = H3Ha: H1 * 42 + H3Ho: H1 # H2 # 43Ha: H1 = #2 = #3Ho: Not all the population means are equal.Ha: H1 = 42 = #3Ho: At least two of the population means are equal.Ha: At least two of the population means are different.Ho: H1 = "2 = H3Ha: Not all the population means are equal.Find 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 =

Null and alternative hypothesis: Test statistic:

Answered: Three different methods for assembling…

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 leader of two postpartum women’s support groups is interested in the depression levels of the women in her groups. She administers the Center for Epidemiologic Studies Depression Scale (CES-D) screening test to the members of her groups. The CES-D is a 20-question self-test that measures depressive feelings and behaviors during the previous week. The mean depression level from the screening test for the 10 women in the first group is μ₁ = 16; the mean depression level for the 14 women in the second group is μ₂ = 10. Without calculating the weighted mean for the combined group, you know that the weighted mean is:
State the two-way ANOVA identity for degrees of freedom and interpret its meaning with regard to partitioning the total degrees of freedom in the data.
The data is presented in parallel format. Before running the ANOVA in Excel, you will need to put it into serial (column) format. You will need to create a categorical variable for the first factor (days) and for the second factor (time of day). This is a critical step.The data presented shows exam scores for students randomly assigned to the sections meeting on these days at these times of the day: Thursday Friday Morning Afternoon Morning Afternoon 70.878.15978.246.646.476.559.376.370 65.678.359.359.778.680.892.780.865.392.3 90.890.569.177.36977.558.677.877.558.6 63.372.854.254.472.166.26372.56672.2 Conduct a 2-way ANOVA. First, find the group means for each level of the factors. (Report answers accurate to 2 decimal places.)Days: MThursday=MThursday= MFriday=MFriday= Time of day: Mmorning=Mmorning= Mafternoon=Mafternoon= Report the results of the ANOVA for the main & interaction effects. (Report P-values accurate to 4 decimal places and…
An analysis of variances produces dftotal = 16 and dfwithin = 14. For this analysis, how many treatment conditions are being compared?
A paint company has large outlet stores in two cities. Ten samples of the weekly sales (in tons) for the two cities are as follows. City A: n = 10, X10 = 148.5, S1 = = 4.3. City B: n = 10, Y 10 = 144.2, S2 = 3.8. At a = 0.05 level of significance, test if the weekly sales are different. The following ¨answers" have been proposed. (a) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using an F test, we reject the null hypothesis. (b) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using an F test, we do not reject the null hypothesis. (c) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using a i test, we do not reject the null hypothesis. (d) The null hypothesis is that the weekly sales are the same and the alternative is that the they are not the same. By using a i test, we reject the…
On a particular proposal of national importance, males and females cast votes as included in the following table In Favor Opposed Undecided Male 90 65 43 Female 115 74 32 Suggest an appropriate test for the data and analyze it at 1% level of significance.
In a random sample of males, it was found that 29 write with their left hands and 212 do not. In a random sample of females, it was found that 70 write with their left hands and 444 do not. Use a 0.01 significance level to test the claim that the rate of left-handedness among males is less than that among females. Complete parts (a) through (c) below.
The proprietor of a greenhouse claims that its pots of lilies average 6.0 buds, with s = 0.4 buds. If n = 29 pots of lilies from a large no of pots of lilies has a mean of 5.8 buds, does this deny the proprietor’s claim of the mean number of buds at 0.015 level of significance?
A researcher intends to estimate the effect of a drug on the scores of human subjects performing a task of psychomotor coordination. The members of a random sample of 9 subjects were given the drug prior to testing. The mean score in this group was 9.07, and the sample variance was 17.68. An independent random sample of 10 subjects was used as a control group and given a placebo prior to testing. The mean score in this control group was 15.21, and the sample variance was 27.48. Assuming that the population distributions are normal with equal variances, find a 90% confidence interval for the difference between the population mean scores The confidence interval is<, -,O (Round to two decimal places as needed.)
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