compute the value of the correlation coefficient (r) - The correlation coefficie

Question

compute the value of the correlation coefficient (r) - The correlation coefficient: r = 0.85

1) test the significance of the correlation coefficient (report your p value) 2) give a brief explanation of the type of relationship (e.g., direction, strength). 3) determine the slope and intercept of the regression line 4) predict the number of push-ups for an individual who can perform 22 pull-ups 5) determine what percent of the variation in # of push-ups can be explained by variation in # of pull-ups (i.e., the coefficient of determination) 6) report the standard error of the estimate.

Subject # Pull-ups # Push-ups 1 0 8 2 1 14 3 5 16 4 6 12 5 8 24 6 10 35

Explanation / Answer

from above

1) here r =0.85

and n=6

for tstat =r((n-2)/(1-r2))1/2 =3.2642

for above p value =0.031

hence above value of r is significant at 0.031 level or below.

2)the relationship seems to +ve with weak strength due to low value of p

3) slope =Sxy/Sxx =2.1447

intercept =Ybar-Xbar*slope =7.443

4) Yhat=7.443+22*2.1447=54.63

5)% of Variation  in # of push-ups can be explained by variation in # of pull-ups =r2=0.7271

6) std error =(SSy/(n-2))1/2 =5.728

x y (X-Xbar)^2 (Y-Ybar)^2 (X-Xbar)(Y-Ybar) Yi' (Y-Yi')^2 1 0 8 25.00 103.36 184.69 7.442982 0.31026854 2 1 14 16.00 17.36 71.53 9.587719 19.468221 3 5 16 0.00 4.69 28.53 18.16667 4.69444444 4 6 12 1.00 38.03 75.03 20.3114 69.0794283 5 8 24 9.00 34.03 -59.31 24.60088 0.3610534 6 10 35 25.00 283.36 -137.47 28.89035 37.3278124 Total 30.00 109.00 76.00 480.83 163.00 131.241228 mean 5.00 18.17 SSx Ssy Sxy

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