How heavy a load (pounds) is needed to pullapart pieces of douglas fir 4 inches

Question

   How heavy a load (pounds) is needed to pullapart pieces of douglas fir 4 inches long and 1.5 inchessquare? Here are data from students doing a laboratoryexcercise.
33,190   31,860   32,590   26,520   33,280 32,320   33,020   32,030   30,460   32,700 23,040   30,930   32,720   33,650   32,340 24,050    30,170   31,300   28,730   31,920 You want to estimate the mean load needed to pull apart thepieces of wood. to within ± 1000 pounds with 95 %confidence. how large a sample is needed?    How heavy a load (pounds) is needed to pullapart pieces of douglas fir 4 inches long and 1.5 inchessquare? Here are data from students doing a laboratoryexcercise.
33,190   31,860   32,590   26,520   33,280 32,320   33,020   32,030   30,460   32,700 23,040   30,930   32,720   33,650   32,340 24,050    30,170   31,300   28,730   31,920 You want to estimate the mean load needed to pull apart thepieces of wood. to within ± 1000 pounds with 95 %confidence. how large a sample is needed?

Explanation / Answer

Given n=20, s=3018.504 (standard deviation based on thedata) E=1000, =0.05, t(0.025,df=n-1=19)=2.09 (check student t table) The sample size is n=(t*s/E)^2 = (2.09*3018.504/1000)^2 = 39.79 Take n=40

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