Table Analyzed HVC VolumeTwo-way ANOVA OrdinaryAlpha0.05Source of Variation.AgerGender.Interaction 3.893.0.1020 0.8341 ns69.29 0.0001ANOVA tableSSDFInteraction 0.01103 110.00028900.1962 10.07566 12AgerGenderResidualof total variation P value P value summary Significant?0.2107 nsDifference between column meansMean of MaleMean of FemaleDifference betweenSE of difference.95% CI of difference.Interaction CIDifference between row means.Mean of 25 d0.1725Mean of 55 d0.1810Difference betweenSE of difference95% CI of differenceData summarymeans0.28750.06600means 0.22150.039700.03970Mean diff, A1 B1 0.16900.2740Mean diff, A2 B2(A1-B1) (A2-B2)95% CI of difference.(B1A1) (B2A2)95% CI of difference.MSNumber of columns (Gender)Number of rows (Ager)Number of values. 16F (DFn, DFd)0.01103 F (1, 12) 1.7490.00028900.1962 F (1, 12)- 31.130.0063051NoNoYes0.1350 to 0.3080-0.008500-0.09500 to 0.07800-0.1050-0.2780 to 0.068010.1050-0.06801 to 0.278022P valueP-0.2107F (1, 12) 0.04584P-0.0001P=0.8341

A two way analysis of variance (ANOVA) was conducted to investigate the effect of two categorical…

Answered: Table Analyzed HVC Volume Two-way ANOVA…

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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An educational psychologist is examining response times to an on-screen stimulus. The researcher believes there might be a weak effect from age, but expects a more pronounced effect for different color contrasts. She decides to examine a black on white (B/W) combination compared to 2 alternatives: red on white (R/W) and yellow on blue (Y/B). Here is the data for response times (in milliseconds): Color Scheme B/W R/W Y/B 10 17 23 26 44 44 22 21 25 22 21 4 16-17 11 32 34 31 24 20 16 20 21 36 8 8 24 37 23 11 39 36 23 17 28 32 36 31 Age 18-19 19 15 45 15 35 36 13 47 23 16 19 31 24 30 25 16 40 16 27 25 42 23 28 35 20-21 21 18 42 27 19 39 39 28 27 18 26 38 Using MS Excel, conduct a 2-way ANOVA with a = 0.05. Fill in the summary table. The last column in the table below is read as, "Partial Eta-squared " and is a concept that was NOT covered in the lectures. Partial n2 is used when there is a chance the data is not independent (click here to read more if interested 2) Computing the values for…
P-values should be accurate to 4 decimal places and all other values accurate to 3 decimal places. Source S df MS F-ratio P-value Partial n? Age (A) 2 Color (B) Interaction (A × B) 4 Error 63 Provide the conclusions for this 2-way ANOVA. What is the conclusion regarding the main effect for age (A or rows): O There appears to be an effect due to age. O There is not enough evidence to reject the assumption that the marginal means across age groups are equal. What is the conclusion regarding the main effect for color (B or columns): There appears to be an effect due to color. There is not enough evidence to reject the assumption that the marginal means across color groups are equal. What is the conclusion regarding an interaction effect between age and color (A x B): There appears to be an interaction effect between age and color. There is not enough evidence to reject the assumption of a lack of an interaction.
/transcript Call centers typically have high turnover. The director of human resources for a large bank has compiled data on about 70 former employees at one of the bank's call centers. The accompanying table shows this data. In writing an article about call center working conditions, a reporter has claimed that the average tenure is no more than two years. Formulate and test a hypothesis using these data to determine if this claim can be disputed. Use a level of significance of 0.05. Click the icon to view the call center data. Is there sufficient evidence at the 0.05 level of significance that the average tenure is no more than two years? Determine the null hypothesis, Ho, and the alternative hypothesis, H₁. Ho ... H₁: (Type whole numbers.) le Le
Analysis of Variance results: Where: Age > 5Responses: AgeFactors: Sex Response statistics by factor Sex n Mean Std. Dev. Std. Error female 121 10.490083 4.2222704 0.38384277 male 155 9.0806452 3.0091656 0.241702 ANOVA table Source DF SS MS F-Stat P-value Sex 1 134.98939 134.98939 10.466692 0.0014 Error 274 3533.79 12.897044 Total 275 3668.7794 Tukey HSD results (95% level) female subtracted from Difference Lower Upper P-value male -1.4094375 -2.2670907 -0.55178425 0.0014 In one complete sentence, interpret your confidence interval. Make sure in the interpretation it is clear the direction of the difference (i.e. which gender weighs more and by how much. Explain why a confidence interval provides more information to a tiger researcher than the results of a hypothesis test
pls Answer No. 15
here any statistically significant differences in the characteristics shown among atments? Justify your answer Your Answer: the Question 42 Not yet graded /3.5 pts The table below summarizes baseline characteristics of patients participating in a clinical trial. a) Are there any statistically significant differences in baseline characteristics between treatment groups? Justify your answer. Characteristic Placebo (n = 125) Experimental ( n=125) %3D Mean (+ SD) Age 54 + 4.5 53 + 4.9 0.7856 % Female 39% 52% 0.0889 % Less than High School Education 24% 22% 0.0386 % Completing High School 37% 36% % Completing Some College 39% 42% Mean (+ SD) Systolic Blood Pressure 136 + 13.8 134 + 12.4 0.4736 Mean (+ SD) Total Cholesterol 214 + 24.9 210 + 23.1 0.8954 % Current Smokers 17% 15% 0.0441 % with Diabetes 8% 3% 0.0938 Your Answer: Question 43 Not vot graded (34 nts
Pls answer Activity 2 , table 2
Treatment II III 2 N = 24 G = 72 EX2 = 292 3 M = 2 M = 3 T = 16 T= 24 T= 32 SS = 16 SS = 24 SS = 20 Use an ANOVA with a = .05 to determine whether there is a significant mean difference between the two treatments. (Round to two decimal places where needed.) Source df MS Feritical Between treatments Within treatments Total F Distribution Numerator Degrees of Freedom = 6 Denominator Degrees of Freedom = 16 0.0 1.0 3.0 4.0 5.0 6.0 7.0 9.0 11.0 120 Conclusion: Fail to reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are no significant differences among the three treatments. Reject the null hypothesis; there are significant differences among the three treatments. o Fail to reject the null hypothesis; there are significant differences among the three treatments.
What key fact permits you to determine percentages for a normally distributed variable by first converting to z-scores and then determining the corresponding area under the standard normal curve?
The key assumptions of the ANOVA include homogeneity of variance, normality of population distributions and: Independence of observations The DV must be measured on a nominal level The IV must be measured on a continuous level Variability of residuals
Question 15 What proportion of a Standard Normal Model (Z-model) have z-scores between z 0.57 and z = 2.38? O a. 0.7157 b. 0.9913 O c. 0.5700 O d. 0.2756 Moving to another question will save this response. Stat100_Practice..pdf Writing Prompt 3.docx Online Syllabus J...pdf Bio- ips-D056AE4A- Ips-531 E23F-45...p4.movE6-4E6 OCT 22
use the date form the second picture to answer question 3 please help me with question 3

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