Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

In a study of fitness, researchers were interested in the effects of weight and

ID: 3268880 • Letter: I

Question

In a study of fitness, researchers were interested in the effects of weight and oxygen capacity on the time to run 1.5 miles in working adults. Data were collected on 31 individuals. Weight was measured in kg and Runtime in minutes. Higher values of oxygen capacity (Oxy) are associated with better aerobic fitness. (a) Figure 1 and Table 1 give the results from fitting different Simple Linear Regressions (SLR's). What conclusions are suggested by each SLR? (b) Based on your conclusions from Figure 1 and Table 1, find a 95% confidence interval Root for the mean time for an individual whose oxygen capacity is 51 to run 1.5 miles. (c) Can the correlation between Runtime and 0xy be determined from Table 1? If so, compute it: if not, explain why. (d) If both weight and oxygen capacity are used as explanatory variables, would you expect your conclusions as to the usefulness of weight and oxygen capacity to explain average running time to be the same as those in part(a)? Explain.

Explanation / Answer

a)

The variable Oxy is statisitcally significant as the p value is less than 0.05 , hence we can reject the null hypothesis and conclude that the variable is significant

The F value is 83.91 and the corresponding p value is less than 0.05 , hence the model is statistically signifcant

The R2 value is 0.7433 , hence the model is able to capture 74.33% variation of the data

b)

we calculate the regresion line as

Y = -0.2245*Oxy +21.229

so -0.2245*51 +21.229 , using the standard errors and considering Z = 1.96 for 95% CI we calculate

-0.2245*51 +- 1.96*0.0245 = -11.49 , -11.40

21.229 +- 1.96*1.16775 = 18.94 , 23.51

so the interval is -11.49 +18.94 and  -11.40 +23.51

7.45 , 12.11

c)

yes we know that R2 is 0.7433 , so correlation r can be calculted as sqrt(0.7433) = 0.8621

d) Weight is not a statistically signifcant variable as the p value is 0.4412 , which is not less than 0.05 . hence we can fail to reject the null hypothesis and coclude that weight is not significant for the model

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote