# 7 - www.lecturebook.com/homeworks/sho Assignm ent Stats 250...

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Assignment Stats 250 W11 HW 7 Graded Point Total: 30 out of 30 a. 1 out of 1 points A study was conducted to compare the grade point averages of sophomores who live in campus dormitories with the grade point averages of sophomores who live off campus. Twenty sophomores were randomly selected from campus dormitories at a college, and 20 other sophomores were randomly selected from students who live off campus. 1. Paired Data 2. Two Independent Samples Solution: 2. Two Independent Samples b. 1 out of 1 points An experiment was performed to compare the reaction time to two types of traffic signs. One sign type is Prohibitive (e.g., No Left Turn) and the other sign type is Permissive (e.g., Left Turn Only). Ten drivers were used in the study. Each driver was presented with 40 traffic signs, 20 of each type, in a random order. For all 40 signs, reaction time was recorded and the mean of each sign-type calculated. 1. Paired Data 2. Two Independent Samples Solution: 1. Paired Data c. A study was conducted to evaluate the average gain in strength due to a weight-training program. Thirteen individuals each took a strength test both before and again after they participated in a six-week weight- training program. Required Assignment Stats 250 W11 HW 7 Question 1 (Chapter 11) Independent or Paired? For each scenario below determine if it is a paired data scenario or based on two independent samples. 4/17/2011 www.lecturebook.com/homeworks/sho… lecturebook.com/homeworks/show/45 1/10

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1 out of 1 points 1. Paired Data 2. Two Independent Samples Solution: 1. Paired Data a. 2 out of 2 points See Textbook for Exercise: 11.a. Your Answer: 1- 2=8.1-4.5 = 3.6 with 1 being the sample mean overall duration of symptoms of the placebo group, and 2 being the sample mean overall duration of symptoms of the zinc lozenge group. unpooled s.e.( 1- 2) = sqrt(s1^2/n+s2^2/n) = sqrt(1.8^2/23+1.6^2/25) =.4932 Solution: Let 1 = Zinc Lozenge group and 2 = Placebo group. Then the estimate is 1 - 2 = 8.1 - 4.5 days = 3.6 days. The unpooled standard error is given by: b. 3 out of 3 points See Textbook for Exercise: 11.b. Your Answer: ( 1- 2) +/- t*(s.e.( 1- 2) ) with df =min(n1-1,n2-1) =3.6 +/- t*(.4932) df = 22 t*=2.07 95 confidence interval is: =3.6 +/- 2.07(.4932) =[2.580,4.621] With 95% confidence we would estimate that, on average, the difference in the population mean overall duration of symptoms between people who take a placebo pill and people who take zinc l t b b t 2 580 d 4 621 d ( l b d ti i l Question 2 (11.44) See Textbook for Exercise: 11.44. 4/17/2011 www.lecturebook.com/homeworks/sho… lecturebook.com/homeworks/show/45 2/10
lozenge to be between 2.580 and 4.621 days. (placebo group duration - zinc lozenge group duration). Solution: Again we will let 1 represent the placebo group and 2 represent the zinc lozenge group. The confidence interval would be computed following the basic structure of: Sample estimate ± Multiplier × Standard error.

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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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7 - www.lecturebook.com/homeworks/sho Assignm ent Stats 250...

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