4 - 4/17/2011 www.lecturebook.com/homeworks/sho Assignm ent...

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Assignment Stats 250 W11 HW 4 Graded Point Total: 29 out of 30 a. 1 out of 1 points Let p 1 represent the population proportion for all such boys that regularly go on-line. Similarly, p 2 represents the population of all such girls that do so. Assuming that each sample is a random sample from the populations of interest and that the two samples are independent, verify the remaining conditions necessary for computing a confidence interval for p 1 p 2 . Your Answer: n1 1, n1(1- 1), n2 2, n2(1- 2) should be preferably at least 10. 1 = 156/236 = .6610 n1 1 = .6601 * 236 = 156 n1(1- 1) = 236*(1-.6610) = 80 2 = 152/272 = .5588 n2 2 = 272* .5588 = 152 n2(1- 2) = 272*(1-.5588) = 120 n1 1, n1(1- 1), n2 2, n2(1- 2) are all greater than 10. Solution: The sample sizes are sufficiently large enough. For a confidence interval, we need to have that all of the quantities n 11 =156, n 1 (1 - 1 ) =80, n 22 =152, and n 2 (1 - 2 ) =120 be at least 10. And we do! b. 3 out of 3 points Construct a 95% confidence interval for the difference in these two population proportions, namely p 1 p 2 . Your Answer: Standard Error: s e 1 2) = sqrt ( 1*(1 1)/n1)+ 2*(1 2)/ n2)) Required Assignment Stats 250 W11 HW 4 Question 1 (Chapter 10) Several years ago, Newsweek conducted a survey of 508 teenagers (age 12-17) and asked questions regarding their thoughts on technology and its impact on their lives. Of the 236 boys surveyed, 156 stated they regularly go on-line. Of the 272 girls surveyed, 152 stated they regularly go on-line. 4/17/2011 www.lecturebook.com/homeworks/sho… lecturebook.com/homeworks/show/39 1/12
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Standard Error: s.e.( 1 - 2) = sqrt (( 1*(1- 1)/n1)+( 2*(1- 2)/ n2)) =sqrt((.6610*(1-.6610)/236)+(.5588*(1-.5588)/272)) = .0431 Confidence interval: =( 1 - 2) +/- z*s.e.( 1 - 2) =(.6610 - .5588) +/- 1.96 * .0431 =0.1022 +/- 0.084476 =(0.017724, 0.186676) Solution: So the 95% confidence interval for the difference in population proportions (boys less girls) that go on-line regularly is: (0.0178, 0.1866) or 1.78% to 18.66%. a. 1 out of 1 points Provide an estimate of p 1 , including an appropriate symbol for your answer. Your Answer: 1 = 57/100 = .57 Solution: Estimate of p 1 is 1 = 57/100 = 0.57 b. A 95% confidence interval for the difference in population rates, p 1 p 2 , was calculated to be (-0.05, 0.23). In order for this confidence interval to be valid, certain conditions must apply. The problem states that the samples are independent random samples. Verify the remaining condition(s). Question 2 (Chapter 10) High School Sports The Department of Education wants to estimate the difference in rates at which high school boys and girls participate in school sponsored sports. Let p 1 represent the population proportion of all high school girls that participate in a school sport and let p 2 represent the population proportion of all high school boys that participate in a school sport. Independent random samples were collected and the following results were obtained. Participate in Sports? Gender Yes No Total 1 = Girls 57 43 100 2 = Boys 48 52 100 4/17/2011 www.lecturebook.com/homeworks/sho… lecturebook.com/homeworks/show/39 2/12
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1 out of 1 points Your Answer: The remaining conditions are that n1 1, n1(1- 1), n2 2, n2(1- 2) should be preferably at least 10.
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This note was uploaded on 01/26/2012 for the course STATS 250 taught by Professor Gunderson during the Winter '10 term at University of Michigan.

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4 - 4/17/2011 www.lecturebook.com/homeworks/sho Assignm ent...

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