6 - www.lecturebook.com/homeworks/sho Assignm ent Stats 250 W11 HW 6 Required Assignment Stats 250 W11 HW 6 Graded Point Total 25 out of 30

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Assignment Stats 250 W11 HW 6 Graded Point Total: 25 out of 30 a. 0.5 out of 1 points Feedback: dbar is reserved for sample mean difference-- shouldn't be used here. See Textbook for Exercise: 11.a. Your Answer: population mean difference, μ . This is the population mean difference in height between college women and their mothers. (daughter-mother) Solution: The parameter of interest is μ d = the population mean difference in heights (in inches) between all college women and their mothers (where difference = daughter height - mother height). b. 2.5 out of 3 points Feedback: wrong notation. See Textbook for Exercise: 11.b. Your Answer: (0+2+-1+3+3)/5 = 1.4 = assuming difference to be heights of daughter - mother sqrt(((0-1.4)^2+(2-1.4)^2+(-1-1.4)^2+(3-1.4)^2+(3-1.4)^2) / (5-1)) = 1.81659 = s s/sqrt(n) = 1.81659/sqrt(5) = .8124 = s.e.( ) t* multiplier for a 95% confidence interval with 5-1 degrees of freedom =2.78 Confidence Interval = +/-t*s.e.( ) = 1.4+/-2.78(.8124) =[ 8585 3 658 Required Assignment Stats 250 W11 HW 6 Question 1 (11.37) See Textbook for Exercise: 11.37. 4/17/2011 www.lecturebook.com/homeworks/sho… lecturebook.com/homeworks/show/43 1/8
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
=[-.8585,3.658] Solution: The sample estimate is = 1.4. The standard error is s.e.( ) = s d /sqrt(n) = 1.817/sqrt(5) = 0.813. The multiplier is t* = 2.78; coming fromTable A.2 using df = n - 1 = 5 - 1 = 4 and the .95 column. Thus, the 95% confidence interval for the population mean difference in heights (daughter less mother) is given by: 1.4 ± (2.78 × 0.813), or 1.4 ± 2.26, or -0.86 to 3.66. c. 1 out of 1 points Feedback: Technically, this corresponds to a TWO sided test, you you should say different from, rather than "taller". See Textbook for Exercise: 11.c. Your Answer: Because 0 in included in our interval above, we do not have sufficient evidence to conclude that the all college women are taller on average than their mothers. Solution: With 95% confidence we would estimate that, on average, college women are somewhere between 0.86 inch shorter and 3.66 inches taller than their mothers. These results apply for the population of mother-daughter pairs represented by this sample. d. 2 out of 2 points See Textbook for Exercise: 11.d. Your Answer: The researchers could have gotten a larger sample size, or they could have lowered the confidence level. Solution: They could have used a larger sample size, which would have reduce both the standard error and the multiplier (with the same confidence level). They could also use a lower confidence level; for instance they could find a 90% confidence interval instead of a 95% confidence interval. 1 out of 2 points Feedback: Need to be clear that the normality assumption is regarding the population of differences. Question 2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 8

6 - www.lecturebook.com/homeworks/sho Assignm ent Stats 250 W11 HW 6 Required Assignment Stats 250 W11 HW 6 Graded Point Total 25 out of 30

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online