This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Problem Set Problem 9.38 (p. 385) A volunteer working at an animal shelter conducted a study on the effect of catnip on cats at the shelter. She recorded the number of negative interactions each of the cats made in 15 minute periods before and after being given a teaspoon of catnip. The paired measurements were collected on the same day within 30 minutes of one another; the data are given in the accompanying table: Cat Before ( 1 Y ) After ( 2 Y ) Difference Amelia Bathsheba 3 6-3 Boris 3 4-1 Frank 1-1 Jupiter Lupine 4 5-1 Madonna 1 3-2 Michelangelo 2 1 1 Oregano 3 5-2 Phantom 5 7-2 Posh 1 1 Sawyer 1-1 Scary 3 5-2 Slater 2-2 Tucker 2 2 Mean 1.8 2.8-1 SD 1.66 2.37 1.20 (a) Construct a 95% confidence interval for the difference in mean number of negative interactions. 1 2 1, 1.2, 0.3098 14 1 (2.145)(0.3098): 1.66 0.34 d d s SE df = - = = =-- <- < - (b) Construct a 95% confidence interval the wrong way, using the independent sample method. How does this interval differ from the one in part (a)? 2 2 1 2 1.66 /15 2.37 /15 0.747 14( use the complicated formula, or df=28) 1 (2.145)(0.747): 2.603 0.603 SE df or = + = =-- <- < The interval much wider than in (a). Problem 9.39 (p. 385) Refer to exercise 9.38. Compare the before and after populations using a t-test at 0.05 = . Use a non-directional alternative. 1- mean number of negative interactions in cats 2 -mean number of negative interactions in cats after catnip H0: The means of negative interactions before and after catnip are the same( 1=2) HA: They are different. ( 12) Use paired non-directional t-test with df=14. Reject H0 if ts>2.145. SE=1.2/15=0.3098, ts=-3.23, so reject H0. Catnip changes (increases) the mean number of negative interactions in cats at significance level 0.05. (P<0.01) Problem 9.40 (p. 385) Refer to Exercise 9.38. (a) Compare the before and after populations using a sign test at 0.05 = . Use a non-directional alternative. (b) Calculate the exact P-value for the analysis in part (a) Let p be the probability that the count of negative interactions is higher before catnip. H0: p=0.5 Probability of having more negative interactions is the same before and after catnip. HA: p0.5 The probabilities are different. Use non-directional sign test with n=12. Has folded binomial distribution with n=12, p=0.5 under H0. Will reject H0 if Bs<=critical value of Table 7=10. N+=2, N-=10, Bs=10<=10 so reject H0 at 0.05 significance level. At 0.05 significance level the counts of negative interactions are different (higher) after catnip. P=2x(0.5)^12[66+12+1]=0.0385 (this is the exact P-value in (a)) Problem 10.4 (p. 400) At a Midwestern hospital there were a total of 932 births in 20 consecutive weeks. Of these births, 216 occurred on weekends. Do these data reveal more than chance deviation from random timing of the births? (Test for goodness of fit, with two categories of birth weekday and weekend. Use a nondirectional alternative and let = 0.05.)= 0....
View Full Document
This note was uploaded on 04/23/2011 for the course STAT 511 taught by Professor Bud during the Fall '08 term at Purdue University-West Lafayette.
- Fall '08