Conduct a Post hoc Tukey test using the following ANOVA tablesSource ofVariationSums ofSquaresdfMean SquaresFFactors orTreatment23.28437.7614.764Error32.573201.629 Total55.85723 Keto DietGymNo-Exercise – No DietIntermittent Fasting 91111.513.25 9.5101211.5 8.2511.259.2512.55 1113.512.511.75 12.2511.513.7512.5 8.51211.2513Overall MeanMeans9.75011.54211.70812.42511.356

Give Information: Consider the given dataset: Keto Diet Gym No-Exercise – No Diet Intermittent…

Answered: Conduct a Post hoc Tukey test using…

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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Suppose we want to test that the average age of a The following table shows the Excel output for this test. Pstudent is not 27 years old. t-Test: Paired Two Sample for Means Score Dummy Mean 25.89474 Variance 5.43275 Observations Pearson Correlation Hypothesized Mean Difference 19 19 #DIV/O! 27 df 18 t Stat P(T<=t) one-tail t Critical one-tail P(T<=t) two-tail t Critical two-tail -2.06696 0.02671 1.7341 0.05343 2.1009 Select the p-value for this test. 2.1009 O 0.05343 0.02671
Want to explain these outputs?
How many computers in a random sample of 155 household the simple mean number of personnel computers was 1.15 assume the population standard deviation is 0.78 construct a 99.5% confidence interval for the mean number of personal computers run the answer to the least two decimal places a 99.5% conference interval for the main number of personal computers is blank <mean<blank
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402562 2600368.qx3zqy7 Start A real estate agent believes that the mean home price in the northern part of a county is higher than the mean price in the southern part of the county and would like to test the claim. A simple random sample of housing prices is taken from each region. The results are shown below. Mean Southern Northern 154.889 167.833 354.458 521.912 Variance Observations Pooled Variance 438.185 0 Hypothesized Mean Difference t Stat 1.885 P= Ex: 1.234 PIT<-t) one-tail 0.036 t Critical one-tail 1.691 P(T<=t) two-tail 0.072 t Critical two-tail 2.032 2 -1 1 2 Confidence Level 95% t = Ex: 1.234 n= Ex: 9 Degrees of freedom: df Point estimate for the southern part of the county: 1 Ex: 1.234 Point estimate for the northern part of the county: ₂ = J. 3 10 @ 05 (CC
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"What i v nighest level of education you've completed?" and "If you're employed full-time, how many hours do you spend working at your job during a typical week?" In a recent year, 1125 respondents answered both questions. The summary statistics are given in the chart below. (The sample data consist of the times, in hours per week, that were given by the respondents.) Sample Sample Sample size Groups mean variance Less than h.s. 257 43.1 110.1 High school 292 44.2 84.9 Bachelor's 294 42.7 105.8 Graduate 282 43.6 100.8 Send data to calculator Send data to Excel To decide if there are any differences in the mean hours per week worked by these different groups, we can perform a one-way, independent-samples ANOVA test. Such a test uses the following statistic. Variation between the samples F = Variation within the samples For the data from the survey, F - 1.22. (a) Give the p-value corresponding to this value of the F statistic. Round your answer to at least three decimal places. (b) From…

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Source ofVariation	Sums ofSquares	df	Mean Squares	F
Factors or Treatment	23.284	3	7.761	4.764
Error	32.573	20	1.629
Total	55.857	23

	Keto Diet	Gym	No-Exercise – No Diet	Intermittent Fasting
	9	11	11.5	13.25
	9.5	10	12	11.5
	8.25	11.25	9.25	12.55
	11	13.5	12.5	11.75
	12.25	11.5	13.75	12.5
	8.5	12	11.25	13	Overall Mean
Means	9.750	11.542	11.708	12.425	11.356