Question

3) Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed.

A 95% confidence interval for a difference in means μ1-μ2 if the samples have n1=80 with x¯1=269 and s1=48and n2=120 with x¯2=229 and s2=49, and the standard error is SE=6.99.

Round your answers to three decimal places.

The 95% confidence interval is ______________ to ____________

______________________________________________________________________

5) How Much More Effective Is It to Test Yourself in Studying?

We have found that students who study by giving themselves quizzes recall a greater proportion of words than students who study by reading. Also, we see that there is an effect, but often the question of interest is not ‘‘Is there an effect?” but instead ‘‘How big is the effect?” To address this second question, use the information given below to find a 95% confidence interval for the difference in proportions p1-p2, where p1 represents the proportion of items correctly recalled by all students who study using a self-quiz method and p2 represents the proportion of items correctly recalled by all students who study using a reading-only approach. Assume that the standard error for a bootstrap distribution of such differences is also about 0.07.

The proportion of items correctly recalled was 0.15 for the reading-study group and 0.42 for the self-quiz group.

Round your answers to two decimal places.

The 95% confidence interval is ________________ to __________________

Any help with these questions is greatly appreciated! :)

Answer 1

Solution Given that ni =80 Vi=269 Si = 48 ng = 120 Xa - 227 Sq = 49 Standard Card, SE = 6.99 To find the 95% confidence interval for Mitte Here d=1- confidence Interval Q-1-95.4 =195 Q = -0.95 =0.05 20.05 N 212= . = 20.025 - 1.96 Tharga We know that The 954. (Ifor Meade = (Fi- 82) where St (standard ford) - Here given SE = 6.99-

Now the 95% CI for Morda =(269-229) +1.96 X 6.99 : -- (40)+ 13.7004 = (40–13.7004 ,40+ 13.7004) = (26.299, 53.700) Therefore the 95% confidence. Interval is (26.299 to 53.700) 0-5 Solution Given Pa = 0.15 Pa 0.42 Standard Entero? (s.F) = 0.07 Here ar la confidence interval =1954 =1-95 d = i-0-95 = 0.05 24 - 25 = 20.025 F412 = 1.96] Now the 95% confidence Interival for Pi-l2 is . Pîrê + Zviz XS.E.

95%CI -0.42-0.15)+1.96 x0.07 = (0.27) +0.1372 -(0-27-6-1372, 0.27401312) (0-1328, 044072) = (0.13, 0:48) Therefore the 95% confidence Interval is 0.3 to 0.41