Question

Who are taller, pro football players or pro basketball players? A random sample 45 pro football players resulted in a mean height of x-6.179 feet. A random sample 40 pro basketball players resulted in a mean height of y-6.453 feet. It is recognized that the true standard deviation of pro football players heights is Ох-0.47 feet while it is recognized that the true standard deviation of pro basketball players heights is ay0.55 feet. The true (unknown) mean of football players heights is Hx feet, while the true (unknown) mean of basketball players heights is y feet. 


Who are taller, pro football players or pro basketball players? A random sample 45 pro football players resulted in a mean height of x-6.179 feet. A random sample 40 pro basketball players resulted in a mean height of y-6.453 feet. It is recognized that the true standard deviation of pro football players heights is Ох-0.47 feet while it is recognized that the true standard deviation of pro basketball players heights is ay0.55 feet. The true (unknown) mean of football players heights is Hx feet, while the true (unknown) mean of basketball players heights is y feet. Type Sample Size Sample Mean Standard Deviation Football (x Basketball ml40 6.179 6.453 0.55 a)Calculate the variance of the random variable X which is the mean of the heights of the 45 football players. b)Calculate the variance of the random variable Y, which is the mean of the heights of the 40 basketball players c) Calculate the variance of X -Y? alculat thestandar devato e) If we wish to create a 94% confidence interval for Ax- fly then what is the z critical value used? f)Create a 94% confidence interval for μΧ-ly-( g) what is the length of your 94% confidence interval for Ax . μ h) If we used this data to test Ho: μ μ,-0 against the alternative Ho: μ μ, < 0 then what would the value of the calculated test statistic Z have been? i) If we used this data to test Ho: Ax-Hy-0 against the alternative Ha: μ μ-0 then what would the p value have been? j)If we used this data to test Ho: μ μ,-o against the alternative Ha: μ μ-0 then what would the p value have been? k) Copy your R script for the above into the text box here.


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Answer #1

a) Variance of Xbar = 6.179/sqrt(45)= 0.92111
b) Variance of Ybar = 6.453/sqrt(40) = 1.02031
c) Variance of Xbar - Ybar = 0.92111+ 1.02031 = 1.94142
d) SD of Xbar - Ybar = 1.39335
e) Z critical value = +/- 1.881

Distribution Plot Normal, Mean 0, StDev-1 0.4 0.3 0.2 0.03 0.03 0.0 1.881 1.881

g)

The 94% confidence interval is (6.179 - 6.453) +/- 1.881 * 1.39335 = (-2.89489135,2.34689135)

h)

Test Statistic Z = (6.179-6.453) / 1.39335 = -0.1967

i) P-value = 0.4220

j) P-value = 0.8440

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