Question 7, 10.2.33 Female college student participation in athletics has increased dramatically over the past few decades. Sports medicine providers are aware of some unique health concerns of athletic women,including disordered eating. A study compared disordered-eating symptoms and their causes for collegiate female athletes (in lean and nonlean sports) and nonathletes. The sample mean of thebody dissatisfaction assessment score was 13.1 (s = 8.1) for 17 lean sports athletes (those sports that place value on leanness, including distance running, swimming, and gymnastics) and 7.4(s = 5.9) for the 69 nonlean athletes. Assume equal population standard deviations. Complete parts a and b below.a. Find the standard error for comparing the means.se = 1.727 (Round to three decimal places as needed.)b. Construct a 95% confidence interval for the difference between the mean body dissatisfaction for lean sport athletes and nonlean sport athletes. Interpret.The 95% confidence interval is (2.2658, 9.1342(Round to two decimal places as needed.)

Given Let sample 1 be denoted be of lean athletes and sample 2 be denoted by non-lean athletes. Sample mean, x¯1=13.1 Sample standard deviation, s1=8.1 Sample size, n1=17 Sample mean, x¯2=7.4 Sample standard deviation, s2=5.9 Sample size, n2=69 It is assumed that population standard deviations are equal.a. Computing pooled standard deviations sp=n1-1s12+n2-1×s22n1+n2-2=17-1×8.12+69-1×5.9217+69-2=6.378 Computing standard error se=sp×1n1+1n2=6.378×117+169=1.727b. df=n1+n2-2=17+69-2=84 α=1-CL=1-0.95=0.05 tc=t1-α2,df The t-critical value is computed using excel. The function used is =T.INV(1-0.05/2,84) and the output value is approximately 1.989. The snip is attached below: CI=x¯1-x¯2-tc×se, x¯1-x¯2+tc×se=13.1-7.4-1.989×1.727, 13.1-7.4+1.989×1.727=2.266, 9.134≈2.27, 9.13 The 95% CI is (2.27, 9.13).