SAMPLE (show all steps) ONLY ANSWER PROBLEM 1 device 1 2 3 4 5 6 a -.307 -.294 .
ID: 3362584 • Letter: S
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SAMPLE (show all steps) ONLY ANSWER PROBLEM 1
device 1 2 3 4 5 6 a -.307 -.294 .079 .019 -.136 -.324 b -.176 .125 -.013 .082 .091 .459 c .137 -.063 .240 -.050 .318 .154 d -.042 .690 .201 .166 .219 .407 Problem 1: For the data from problem number 8.6 in the textbook calculate the Kruskal-Wallis test statistic H by hand. You may use the rank sums given below (which were obtained using JMP's "oneway analysis nonparametric - Wilcoxon test"). Also estimate the p-value and state your conclusion ( = 0.05) Wilcoxon / Kruskal- Wallis Tests (Rank Sums) (Mean-Me Level Count Score Sum Score Mean an0)/Std0 2.833 0.000 0.633 2.167 32.000 75.000 85.000 108.000 5.3333 12.5000 14.1667 18.0000 4. Problem 2: 8.32 (b) from the textbook. Are there significant differences between the varieties? If yes, identify them. Use JMP.Explanation / Answer
Problem 2)
Here we have to test the hypothesis that,
H0 : All means are equal.
H1 : Atleast one of the mean is differ than 0.
Assuma alpha = level of significance = 0.05
JMP steps for One way ANOVA :
From an open JMP data table, select Analyze > Fit Y by X.
Click on a continuous variable from Select Columns, and Click Y, Response (continuous variables have blue triangles).
Click on a categorical variable and click X, Factor (categorical variables have red or green bars).
Click OK. The Oneway Analysis output window will display.
Click on the red triangle, and select Means/Anova.
JMP will plot means diamonds (95% confidence intervals for each mean), and will generate:
• The Summary of Fit.
• The Analysis of Variance (Anova) table.
• Means for Oneway Anova, containing summary statistics and confidence intervals for each mean (based on the pooled estimate of the standard error).
Output :
test statistic = 0..55
P-value = 0.7357
P-value > alpha
Accept H0 at 5% level of significance.
COnclusion : There is not sufficient evidence to say that the population means are differ.
Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 4 -0.388 -0.097 0.036041 Column 2 4 0.458 0.1145 0.176563 Column 3 4 0.507 0.12675 0.013383 Column 4 4 0.217 0.05425 0.008456 Column 5 4 0.492 0.123 0.038449 Column 6 4 0.696 0.174 0.127973 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 0.184051 5 0.03681 0.550964 0.735701 2.772853 Within Groups 1.202593 18 0.066811 Total 1.386644 23Related Questions
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