Analysis of Variance results:Responses: HDLFactors: Exercise LevelResponse statistics by factorExercise Level .HighLowMean Std. Dev. Std. Error909750.50515513.778095 1.3989536113 54.1946913.866621 1.30446256.4 18.3081 1.9298432MediumANOVA tableSourceDFSMSF-Stat P-valueExercise Level 21673.4324836.716213.5709029 0.0293Error29769591.564234.3150329971264.997TotalIs the result significant? What hypothesis did you test? Did you acceptor reject your hypothesis and why? What is your conclusion here aboutHDL levels and exercise in this study?

There are 3 independent samples which are High, Low and Medium. We have to test whether there is a…

Answered: Analysis of Variance results:…

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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Question: How to describe normal distribution and skewed in the histogram and density plots in general. Also, explain any correlation we observe between the response and predictor variables in any scatter plots in general.
Assume the samples are random and independent, the populations are nomally distributed, and the population variances are equal. The table available below shows the prices (in dollars) for a sample of automobile batteries. The prices are classified according to battery type. At a = 0.10, is there enough evidence conclude that at least one mean battery price is different from the others? Complete parts (a) through (e) below. E Click the icon to view the battery cost data. (a) Let u1. P2. H3 represent the mean prices for the group size 35, 65, and 24/24F respectively. Identify the claim and state Ho and H. H Cost of batteries by type The claim is the V hypothesis. Group size 35 Group size 65 Group size 24/24F 101 111 121 124 D 146 173 182 278 124 140 141 89 (b) Find the critical value, Fo, and identify the rejection region. 90 79 84 The rejection region is F Fo, where Fo = (Round to two decimal places as needed.) (c) Find the test statistic F. Print Done F= (Round to two decimal places as…
DO MALES AND FEMALES DIFFER ON ATTITUDES TOWARD CRIME ? MALES N=12 FEMALES N=13 CRIME ATTITUDE MALES 77, 97, 28, 55, 75, 43, 21, 70, 91, 22, 58, 45 FEMALES 32,31, 89, 21, 79, 67, 41, 24, 43, 87, 64, 28, 21 MEAN SD MALE FEMALE • DV= 1 •T-VALUE • df • SIGNIFICANT (Y/N). • TEST STATISTIC RAN • ACCEPT OR REJECT NULL HYPOTHESIS. • CONCLUSION:
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…
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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
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WALKING RUNNING Mean 7.181818182 7.3636364 Variance 4.563636364 6.4545455 Observations 11 11 Pooled Variance 5.509090909 Hypothesized Mean Difference 0 df 20 t Stat -0.181668105 P(T<=t) one-tail 0.428835886 t Critical one-tail 1.724718243 P(T<=t) two-tail 0.857671772 t Critical two-tail 2.085963447 If the t-Stat value is less than the t Critical two-tail value I can reject the null, right? I am still a little confused on what a few numbers represent.

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