Chapter 5, Section 2, Exercise 047 How Much More Effective Is It to Test Yoursel

Question

Chapter 5, Section 2, Exercise 047 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 Pi 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. Ass also about 0.07. ume that the standard error for a bootstrap distribution of such differences is 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 to

Explanation / Answer

(a)

Data given is:

p1' = 0.42, p2' = 0.15

Standard error, SE = 0.07

The 90% CI is:

(p1' - p2') - (1.64*SE) < p1 - p2 < (p1' - p2') + (1.64*SE)

(0.42 - 0.15) - (1.64*0.07) < p1 - p2 < (0.42 - 0.15) + (1.64*0.07)

0.1552 < p1 - p2 < 0.3848

(b)

Data given is:

Sample proportion, p' = 53/194 = 0.273

Standard error, SE = 0.032

The 95% CI is:

p' - (1.96*SE) < p < p' + (1.96*SE)

0.273 - (1.96*0.032) < p < 0.273 + (1.96*0.032)

0.210 < p < 0.335

(c)

Data given is:

Sample proportion, p' = 50/411 = 0.122

Standard error, SE = 0.016

The 99% CI is:

p' - (2.576*SE) < p < p' + (2.576*SE)

0.122 - (2.576*0.016) < p < 0.122 + (2.576*0.016)

0.081 < p < 0.163

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