DISCUSSION FILE # 16I'm trying to determine if there actually is a difference in 7-iron distance between 3 golf balls I'm thinking aboutplaying. I conducted a study by 15 shots with each type of ball hitting (at random across the ball types) andnoting the total distance of each shot. The results are shown below.What is the conclusion at the alpha = .05 level and why?Analysis of Variance results:Data stored in separate columns.Column meansColumn nBall 1Ball 2Ball 315 151.46666 0.821535515ANOVA tableError15TotalMean Std. ErrorSource dfTreatments 2152.2 0.7571878154.8 0.9719886SS92.04444MS44 552.5777646.0222242 460.53333 10.965079F-Stat P-value0.0218dal4 Reject the null hypothesis because there's a lot of error in the modelO Fail to reject the null hypothesis because the F stat is so smallReject the null hypothesis because the F stat is sufficiently largeFail to reject the null hypothesis because there's sufficient error in the modelReject the null hypothesis because the p-value is so large

I'm trying to determine if there actually is a difference in the 7-iron distance between the 3 golf…

Answered: DISCUSSION FILE # 16 I'm trying to…

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 a car's city miles per gallon rating can be determined by the car's weight. The means and standard deviations of these variables and the correlation coefficient between them are reported in the following table: What proportion of the variability in cars' miles per gallon ratings is explained by the least squares line with weight?
Heights The heights (cm) in the following table are from Data Set 1 “Body Data” in Appendix B. Results from two-way analysis of variance are also shown. Use the displayed results and use a 0.05 significance level. What do you conclude? XLSTAT
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Actuaries use various parameters when evaluating the cost of a life insurance policy. The variance of the life spans of a population is one of the parameters used for the evaluation. Each year, the actuaries at a particular insurance company randomly sample 30 people who died during the year (with the samples chosen independently from year to year) to see whether the variance of life spans has changed. The life span data from this year and from last year are summarized below. Current Last Year Year *1 = 75.8 x2 = 76.2 s3 = 47.61 s3= 92.16 (The first row gives the sample means and the second row gives the sample variances.) Assume that life spans are approximately normally distributed for each of the populations of people who died this year and people who died last year. Can we conclude, at the 0.05 significance level, that the variance of the life span for the current year, of, differs from the variance of the life span for last year, o,? Perform a two-tailed test. Then complete the…
The F-statistic is a ratio score resulting in a measure of the estimated proportion of variance due to the independent variable. True or false
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HYPERTENSION. The data below was collected from over 200 infants at 1 month of age. In this study a research nurse visited the home of each child and took their blood pressure using a special apparatus for measuring infant bp. She also noted whether the infant was asleep or awake and whether or not the child was agitated when the bp was taken. Assuming equal variances, is there a significant mean difference in SBP between the quiet sleep infants and the awake and quiet infants? 1. What is the appropriate test to conduct in the above situation? 2. What is the degree of freedom? 3. What is the computed test statistic? 4. What is the correct conclusion for this test? 5. What can be implied from the conclusion above?
Hi there, I am having a difficult time synthesizing this information I obtained from excel. I compiled the data utilizing a two-sample t-Test assuming equal variances. 1. Who is most likely to have prevalent coronary heart disease? Test if there are significant differences in the following risk factors between persons with and without prevalent coronary heart disease (CHD). Summarize the statistical results in the table below and then compare risk factors in persons with and without prevalent CHD. Patient Characteristic* History of CHD (n = 194) No History of CHD (n = 4240) p-value* Age, years 57.48 (7.42) 49.58 (8.57) < 0.001 Systolic blood pressure, mmHg 145.00 (27.03) 132.35 (22.03) < 0.001 Diastolic blood pressure, mmHg 87.14 (14.33) 82.90 (11.91) < 0.001 Total serum cholesterol, mg/dL 243.20 (45.61) 236.70 (44.59) 0.049 Body mass index 26.83 (4.45) 25.80 (4.08) < 0.001 * Mean (Standard deviation).…
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