Question 5 ( B ) The IQ scores for a random sample of subjects with low lead levelsin their blood and another random sample of subjects with highlead levels in their blood were collected. The statistics aresummarized in the accompanying table. Assume that the twosamples are independent simple random samples selected fromnormally distributed populations. Do not assume that thepopulation standard deviations are equal. Complete parts (a) to (c)91.26016 15.79292Low Lead Level H1High Lead Level 2952786.11758 8.87896below.scores.b. Construct a confidence interval appropriate for the hypothesis test in part (a).Round to one dacimal place as needed.)

Given Data : Provided that population variances are not equal Assuming significance level = 0.05 Confidence level = 0.95 Degrees of Freedom : df=(s12/n1+s22/n2)2(s12/n1)2n1+1+(s22/n2)2n2+1−2=79.721 df = \frac{ (s_1^2/n_1 +s_2^2/n_2)^2 }{ \frac{(s_1^2/n_1)^2}{n_1+1} + \frac{(s_2^2/n_2)^2}{n_2+1}} - 2 = 79.721 Critical value = tc=t1−α/2;n−1=1.99t_c = t_{1-\alpha/2; n-1} = 1.99 Standard error : se=s12n1+s22n2=15.793295+8.879227=2.355 \begin{array}{ccl} se & = & \displaystyle \sqrt{ \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} } \\\\ \\\\ & = & \displaystyle \sqrt{ \frac{15.793^2}{95} + \frac{8.879^2}{27}} \\\\ \\\\ & = & 2.355 \end{array} Corresponding confidence interval: =(Xˉ1−Xˉ2−tc×se,Xˉ1−Xˉ2+tc×se)=(91.26−86.118−1.99×2.355,91.26−86.118+1.99×2.355)=(0.456,9.829) The 95%95\% confidence interval : μ1−μ2\mu_1 - \mu_2 is 0.456<μ1−μ2<9.829 0.456 < \mu_1 - \mu_2 < 9.829