Because cardiac deaths appear to increase after heavy snowfalls, astudy was desi

Question

Because cardiac deaths appear to increase after heavy snowfalls, astudy
was designed to compare cardiac demands of snow shoveling to thoseof using an
electric snow thrower. Ten subjects cleared tracts of snow usingboth methods,
and their maximum heart rates were recorded during both activities.The following results were obtained (based on data from "CardiacDemands of Heavy Snow Shoveling," by Franklin et a., Journal ofthe American Medical Association, Vol. 273, No. 11):

Manual snow shoveling maximum heart rates: n = 10, x =175, s = 15
Electric snow thrower maximum heart rates: n= 10, x = 175,s = 15

a. Find the 95% confidence interval estimate of the population meanfor those people who shovel snow manually.
b. Find the 95% confidence interval estimate of the population meanfor those people who use the electric snow thrower.
c. If you are a physician with concerns about cardiac deathsfostered by manual snow shoveling, what single value in theconfidence interval from part (a) would be of greatest concern?
d. Compare the confidence intervals from parts (a) and (b) andinterpret your findings.

Explanation / Answer

Given that , Manual snow shoveling maximum heart rates: n =10, x = 175, s = 15                   Electricsnow thrower maximum heart rates: n= 10, x = 175, s= 15 a) To find the 95% confidence interval estimate of thepopulation mean for those people who shovel snow manully The formla is given by (Xbar ± t/2 /n) =(175 ± t0.025*15/10)                                                                =(175+2.262*4.7435 , 175-2.262*4.7435)=(175+10.7297 ,175-107297)                                                                =(185.7297 , 164.2703)                                        (b) To find the 95% confidence interval estimate ofthe population mean for those people who use the electric snowthrower. The formla is given by (Xbar ± t/2 /n) =(175 ± t0.025*15/10)                                                                =(175+2.262*4.7435 , 175-2.262*4.7435)=(175+10.7297 ,175-107297)                                                                =(185.7297 , 164.2703) (c) 185.7297 may be the single value in the C-I wouldbe of greatest concern (d) There is not much differnece in the two methods. Theyboth might yield sameresults.                                                                  The formla is given by (Xbar ± t/2 /n) =(175 ± t0.025*15/10)                                                                =(175+2.262*4.7435 , 175-2.262*4.7435)=(175+10.7297 ,175-107297)                                                                =(185.7297 , 164.2703) (c) 185.7297 may be the single value in the C-I wouldbe of greatest concern (d) There is not much differnece in the two methods. Theyboth might yield sameresults.                                                                 

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