7.2-9. Students in a semester-long health-fitness programhave their percentage of body fat measured at the begin-ning of the semester and at the end of the semester. Thefollowing mcasurements give thesc percentages for tenmen and for ten women:MalesFemalesPre-program Post-program Pre-program Post-program%%%%11.109.9722.9022.8919.5015.8031.6033.4714.0013.0227.7025.758.309.2821.7019.8012.4011.5119.3618.007.897.4025.0322.3312.1010.7026.9025.268.3010.4025.7524.9012.3111.4023.6321.8010.0011.9525.0624.28(a) Find an approximate 90% confidence interval for themean of the difference in the percentages for themales.(b) Find an approximate 90% confidence interval for themean of the difference in the percentages for thefemalcs.(c) On the basis of these data, have these percentagesdecreased?(d) If possible, check whether each set of differencescomes from a normal distribution.

Given information: Males Females Pre program Post program Pre program Post program 11 9.97 22.9 22.89 19.5 15.8 31.6 33.47 14 13.02 27.7 25.75 8.3 9.28 21.7 19.8 12.4 11.51 19.36 18 7.89 7.4 25.03 22.33 12.1 10.7 26.9 25.26 8.3 10.4 25.75 24.9 12.31 11.4 23.63 21.8 10 11.95 25.06 24.28a) Pre program Post program Difference(d) d-d¯2 11 9.97 1.03 0.351649 19.5 15.8 3.7 10.64717 14 13.02 0.98 0.294849 8.3 9.28 -0.98 2.007889 12.4 11.51 0.89 0.205209 7.89 7.4 0.49 0.002809 12.1 10.7 1.4 0.927369 8.3 10.4 -2.1 6.436369 12.31 11.4 0.91 0.223729 10 11.95 -1.95 5.697769 Sum 115.8 111.43 4.37 26.79481 Mean 11.58 11.143 0.437 2.679481 d¯=∑dn=4.3710=0.437s=∑(d-d¯)2n-1=26.7948110-1=1.725456783≈1.73S.E=sn=1.725510=0.545637344≈0.5456D.F=n-1=10-1=9 Use the excel formula, "=T.INV.2T(0.1,9)" for determining the critical value:tCritical = 1.833 C.I=x¯d±tα2*S.E=0.437±1.833*0.5456=0.437±1.0002C.I=(-0.563,1.437) 90% confidence interval for the mean of the difference in the percentages for the males is C.I = (-0.563, 1.437).b) Pre program Post program Difference(d) d-d¯2 22.9 22.89 0.01 1.221025 31.6 33.47 -1.87 8.910225 27.7 25.75 1.95 0.697225 21.7 19.8 1.9 0.616225 19.36 18 1.36 0.060025 25.03 22.33 2.7 2.512225 26.9 25.26 1.64 0.275625 25.75 24.9 0.85 0.070225 23.63 21.8 1.83 0.511225 25.06 24.28 0.78 0.112225 Sum 249.63 238.48 11.15 14.98625 Mean 24.963 23.848 1.115 1.498625 d¯=∑dn=11.1510=1.115s=∑(d-d¯)2n-1=14.9862510-1=1.290402607≈1.29S.E=sn=1.2910=0.40806≈0.4081D.F=n-1=10-1=9 Use the excel formula, "=T.INV.2T(0.1,9)" for determining the critical value:tCritical = 1.833 C.I=x¯d±tα2*S.E=1.115±1.833*0.4081=1.115±0.74798C.I=(-0.367,1.863) 90% confidence interval for the mean of the difference in the percentages for the females is C.I = (-0.367, 1.863).c) 90% confidence interval for the mean of difference in the percentages for the males is C.I = (-0.563, 1.437). 90% confidence interval for the mean of difference in the percentages for the females is C.I = (-0.367, 1.863). Since both the 90% confidence interval of the difference in the percentages include 0, hence there is no change in these percentages. On the basis of these data, these percentages has not decreased.