Question
. A kinesiology major wanted to predict O2max based on the mile run. To develop the...
. A kinesiology major wanted to predict O2max based on the mile run. To develop the regression equation, she obtained O2max values (ml/kg/min) in the exercise physiology laboratory on 18 students. Two days later, she measured the same 18 students on the mile walk–run with scores reported as total time in seconds. The data are as follows.
a. Using a computer, calculate the means, standard deviation, and correlation coefficient between the two variables.
b. What is the probability that the correlation coefficient happened by chance? Why is the coefficient negative?
c. What is the slope, the Y-intercept, and the standard error of the estimate for the regression line?
d Estimate V.O2max for a person who runs the mile in 540 seconds. If you wanted your estimate to be accurate at the 95% LOC, what is the error factor in your estimate?
as -X Subject Mile walk-run VO,max 1 2 3 4 5 6 7 8 9 10 11 At ht 250 315 420 410 436 511 460 510 530 586 591 600 626 643 650 675 710 720 60.3 57.2 55.4 51.4 52.5 45.6 38.4 41.5 39.6 33.2 37.7 40.1 32.0 35.4 33.7 35.9 27.4 25.3 un 12 te at 13 14 15 16 17 18 nti er —
Answers
a) Mean: Mile-walk run=535.7222, VO2max=41.25556
Standard deviation: Mile-walk run=133.4386, VO2max= 10.31077 and the correlation = -0.94b) If we test the hypothesis,
. The test statistic is given by:
, here the test statistic follows a t-distribution with n-2=16 degrees of freedom. The p-value of the test statistic is approximately 0. Thus there is approximately 0 that the correlation occurred by chance. The coefficient is negative because the two variables have a negative linear relationship.
c) The y-intercept=80.16692, the slope= -0.07263, and the standard error of the estimate is 3.626.
d) The estimated VO2max for a person who runs the mile in 540 seconds is 40.94485. The error factor in the estimate is 1.81282 if we want to estimate to be accurate at the 95% LOC.
We were unable to transcribe this imageWe were unable to transcribe this image> x=C(250,315,420,410,436,511,460,510,530,586,591,600,626,643,650,675,710,720) > y=c(603,572,554,514,525,456,384,415, 396,332, 377,401,320,354,337,359,274,253);y=y/10 > mx=mean(x); my=mean(y); sx=sd(x);sy=sdy) > mx; my;sx;sy [1] 535.7222 [1] 41.25556 [1] 133.4386 [1] 10.31077 > cor.test(x,y) Pearson's product-moment correlation data: x and y t = -11.021, df = 16, p-value = 6.9911-09 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: -0.9777680 -0.8431509 sample estimates: cor -0.9399983 > fit=lm(y~); fit Call: Im(formula = y - x) Coefficients: (Intercept) 80.16692 X -0.07263 > newdat = data.frame (x=540) > predict(fit, newdata =newdat, interval = "confidence") fit lwr upr 1 40.94485 39.13203 42.75767 > err= 40.94485-39.13203; err [1] 1.81282
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