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Question 3 --/1 View Policies Current Attempt in Progress 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 90% confidence interval for the difference in proportions P - P2, where p, represents the proportion of items correctly recalled by all students who study using a self-quiz method and pa 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 90% confidence interval is i e Textbook and Media Attempts: 0 of 3 used Save for Later Submit Answer Send to Gradebook < Prev Next > * Previous Next

Answer 1

the proportion of items correctly recalled $\hat{p_{1}}$ = 0.15

the proportion of items reading study group $\hat{p_{2}}$ = 0.42

standard error (SE)=0.07

z at 90 % confidence is 1.645

confidence interval =$\hat{p_{1}}-\hat{p_{2}}\pm z *SE$

= (0.15-0.42)$\pm$ 1.645*0.07

= -0.27$\pm$0.12

= (-0.39,-0.15)

the 90 % confidence interval is -0.39 to-0.15