DOC PREVIEW
UCLA STATS 13 - Final Exam Review Problem Solutions

This preview shows page 1-2-3-4-5 out of 15 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
View full document
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 15 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

All Problems are from: Myra L. Samuels and Jeffrey A. Witmer,Statistics for the Life Sciences, 3rd edition, Prentice-Hall (2003)12.612.1212.1512.22a: From Exercise 12.6, s_Y|X = sqrt[SS(resid)/df] = sqrt[16.7812/(12-2)] = 1.3 The standard error of the slope is12.27a: r = 82.8977/sqrt[(28465.7)(.363708)] = .8147Stat 13 Final Exam Review Problem Solutionshttp://www.stat.ucla.edu/~dinov/courses_students.htmlAll Problems are from: Myra L. Samuels and Jeffrey A. Witmer, Statistics for the Life Sciences, 3rd edition, Prentice-Hall (2003)Chapter 10 – 2 Tests10.4: H0: Timing of births is random (Pr(weekend) = 2/7)HA: Timing of births is not random (Pr(weekend) not= 2/7). Weekend WeekdayObserved 216 716Expected 266.29 665.71Difference -50.29 +50.29Chi-Square = Sum of (O-E)2/E = (-50.29)2/266.29 + (+50.29)2/665.71 = 13.3With df = 1, Table 8 gives http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables.0001 < P-value < .001. There is sufficient evidence to conclude that the timing of births is not random. 10.5 Let WF and DF denote white and dark feathers; let SC and LC denote small and large comb.H0: The model is correct; that is, Pr(WF,SC) = 9/16, Pr(WF,LC) = 3/16, Pr(DF,SC)=3/16, Pr(DF,LC)=1/16.HA: The model is incorrect; that is, Probabilities are not as specified by H0. OBS EXPWF,SC 111 106.875WF,LC 37 35.625DF,SC 34 35.625DF,LC 8 11.875Chi-square = Sum of (O-E)2/E = 1.55, with df = 3, P-value > .20 since Table 8 (http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables) gives chi-square_{.20} = 4.64. We do not reject H0. There is little or no evidence (P-value > .20) to conclude that the model is incorrect; the evidence is consistent with the Mendelian model. 10.6a: n = 1000 OBS EXP DIFFBOY 520 500 20GIRL 480 500 -20Chi-square = 1.6. With df = 1, Table 9 (http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables) shows P-value > .2010.6b: n = 5000 OBS EXP DIFFBOY 2600 2500 100GIRL 2400 2500 -100Chi-sq = 8. With df = 1, Table 9 (http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables) shows 0.001 < P-value < 0.01.10.6c: n = 10000 OBS EXP DIFFBOY 5200 5000 200GIRL 4800 5000 -200Chi-sq = 16. With df = 1, Table 9 http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables shows . P-value < .0001.10.11: The hypotheses areH0: The men are guessing (Pr(correct) = 1/3) Ha: The men have some ability to detect their partners (Pr(correct) > 1/3)Observed ExpectedCorrect 18 12Wrong 18 24 Total 36 36Chi-Square statistic = 4.5. With df = 1, Table 9 http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tablesgives .01 < P-value < .025 and we reject H0. Note that no alpha level was specified, but a P-valueless than 0.025 is generally considered to be small. 10.17: table is striped all red alive 65 (70.31) 23 (17.69) TOTAL = 88 dead 98 (92.69) 18 (23.31) TOTAL = 116 TOTAL = 163 TOTAL = 41 TOTAL = 204 Null is that there is no difference in the survival rates for the two types, and alternative is that the mimic form (all red) survives more than the striped kind. Test stat is chi-sq = [(65 - 70.31)2/70.31] + [(98 - 92.69)2/92.69] + [(23 - 17.69)2/17.69] + [(18 - 23.31)2/23.31] = 0.40 + 0.30 + 1.59 + 1.21 = 3.50 Again, since alternative is one-tailed, we half to get p-values: 0.025 < P-value < 0.05. Since P-value < , we conclude that the mimic form of P. cinereus seem to survive more successfully that the red-striped. (df = 1) 10.22a: H0: E. coli had no effect on tumor incidences.Ha: E. coli increased tumor incidences. H0: p1 = p2 Ha: p2 > p1 = .05Df = 1Germ-free E. coliTumors 19 (21.34) 8 (5.66) 27No tumors 30 (27.66) 5 (7.34) 35Total 49 13 62chi-sq = 2.17chi-sq_.20 = 1.64 and chi-sq_.10 = 2.71Multiply by half because Ha is directional: therefore, .10 > P > .05We do not reject H0.There is insufficient evidence (.10 > P > .05) to conclude that E. coli increases the number of tumors in mice.10.22b: If the percentages stay the same but the sample sizes double, then the O (Observed) and E (Expected) values double. Also (O-E) doubles, which means that (O-E)2 is four times larger. But when divided by a doubled E, we get that (O-E)2 / E is doubled. So the Chi-square statistic is doubled. Then H0 is rejected because .01 < P-value < .025.Similarly, if the samples were to triple, then the Chi-square statistic would triple. Then .005 < P-value < .01 and, of course, H0 is rejected.This makes sense. If you toss a coin 4 times and get 3 (75%) heads, that is not unusual (z = 1). But if you tossed a coin 100 times and got 75 (75%) heads, then that would be very unusual (z = 5). 10.35: p1 = Pr{HP / MP} and p2 = Pr{HP / MA}. Null is that p1 = p2, and alternative is that p1 and p2 differ. From table 8, http://socr.stat.ucla.edu/Applets.dir/OnlineResources.html#Tables0.001 < P-value < 0.01, and, since P-value < alpha, reject the null and conclude that there is an association (dependence) between the species. Data suggests repulsion. 47.3 % = p1-hat < p2-hat = 70.8%. (df = 1) 10.37a:Pr {Yes|A} : 111/513 = 0.21637 = 21.637% Pr {Yes|B} : 74/515 = 0.1437 = 14.37%10.37b: Pr {A|Yes} : 111/185 = 0.60 = 60%Pr {A|No} : 402/843 = 0.4767 = 47.67%10.73: Let p denote the probability that the uninfected mouse in a cage becomes dominant.H0: Infection has no effect on development of dominant behavior (p = 1/3)HA: Infection tends to inhibit development of dominant behavior (p > 1/3)Uninfected mouseDominant NotDominant15(10) 15(20)Chi-square statistic = 3.75. With df = 1, we get 0.025 = 0.05/2 < P-value < 0.10/2 = 0.05 and we reject Ho. There is sufficient evidence (0.025 < P-value < 0.05) to conclude that infection tends toinhibit development of dominant behavior. 10.87: The hypotheses are H0: Type of treatment does not affect survivalHA: Type of treatment affects survival table is Zidovudine Didanosine Both Total Died 17 (11.29) 7 (11.50) 10(11.21) 34 Survived 259(264.71) 274(269.50)


View Full Document

UCLA STATS 13 - Final Exam Review Problem Solutions

Documents in this Course
lab8

lab8

3 pages

lecture2

lecture2

78 pages

Lecture 3

Lecture 3

117 pages

lecture14

lecture14

113 pages

Lab 3

Lab 3

3 pages

Boost

Boost

101 pages

Noise

Noise

97 pages

lecture10

lecture10

10 pages

teach

teach

100 pages

ch11

ch11

8 pages

ch07

ch07

12 pages

ch04

ch04

10 pages

ch07

ch07

12 pages

ch03

ch03

5 pages

ch01

ch01

7 pages

ch10

ch10

7 pages

Lecture

Lecture

2 pages

ch06

ch06

11 pages

ch08

ch08

5 pages

ch11

ch11

9 pages

lecture16

lecture16

101 pages

lab4

lab4

4 pages

ch01

ch01

7 pages

ch08

ch08

5 pages

lecture05

lecture05

13 pages

Load more
Download Final Exam Review Problem Solutions
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Final Exam Review Problem Solutions and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Final Exam Review Problem Solutions 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?