Coin-Tossing Lab s2

SOLUTIONS TO FINAL EXAM VERSION 1

1) A) The link between IQ and the given variables is essentially nonexistent. The t-statistics for all variables are nearly zero. The corresponding p- values are very large. The F-statistic is not significant (p=.866), indicating that the model is useless. The R-squared is 4.3%, indicating that these variables hardly explain any of the variability in IQ.

B) We estimate that each additional cubic centimeter of brain volume results in an expected decrease of .01874 in the IQ score, assuming that the model is correct, and that head size and body weight are held fixed.

A) The CI is –0.01874  tα 2 0.03017. Using df=16 and =.05, we get tα 2 = 2.120. The CI is therefore (–0.08270 , 0.04522).

B) Using the regression equation, the predicted value is 28.6–0.01874(1000)+1.686(50) –0.0144(60)=93.296.

3) A) The boxplots seem to indicate an advantage for the second-born twin. However, the t-test reveals virtually no relationship between IQ score and birth order. The p-value is .845, a very large number, indicating that natural variability alone would give us such an “interesting” t statistic as we got here 84.5% of the time.

B) Here, we just use an ordinary two-tailed t-test. The t-statistic is 1.2/(8.36/ 10 )=.454. Using 9 degrees of freedom and and =.05, we find a

critical value of tα 2 =2.262. Since the absolute value of our t-statistic does not exceed the critical value, we do not reject the null hypothesis. So once again, we find no evidence of a relationship between IQ and birth order. In principle, this is a different test from the two-sample version considered previously, since here we are using “matched pairs”. In other words, each twin is paired with its sibling. This tends to reduce extraneous sources of variation, since the two twins are (presumably) as alike as two people can be in all respects except birth order.

A) We need to do a right-tailed test. Since the t-statistic on age is positive, we the p-value is half that given by Minitab, that is, .168. Since the p-value is not less than .05, the relationship between salary and age is not significant at level .05. B) Yes, natural variability alone would give us a t-statistic which is at least as large as what we got here in 16.8% of all possible samples.

C) R-squared, 1.6%.

D) F=(MS Regression)/(MS Residual)=45896/48683=.94.

E) S= MS Residual = 48683 =220.6.

5) Since x /SE  (x  ) /(s / n ) has a t-distribution with 9 degrees of freedom, and since the probability that such a t-random variable exceeds 2.262 is .025 (see Table 6), we find that Prob{| x | 2.262 SE }=Prob{|t| > 2.262}=.05.

6) Since the p-value is less than .05, we know that the t-statistic must be greater x  5 than the critical value, which is t.05 =2.132 (df=4). So we have  2.132 2/ 5 . In other words, x  5  (2/ 5)(2.132) , that is, x  6.907. Answer is A.

7) The sample mean, 4, is the midpoint of the confidence interval. Answer is C. The other options given are all wrong. See the handout where we described the proper interpretation of confidence intervals. Answer is C.

8) The mean of X is 2, and the variance of X is (1 2)2 /3  (2  2)2 /3  (3  2)2 /3=2/3. Answer is C.

9) The expectation of the sample mean is 2. The standard error of the sample mean is 2/3 /10  .08165. We get Prob{ x  2.25 }=Prob{Std. Normal  (2.25  2) /.08165}=Prob{Std. Normal > 3.06}=.5–.4989=.0011. Answer is D.

10) We have np=20p=5, so p=.25. The variance of X is npq=(20)(.25)(.75)=3.75. The standard deviation of X is 3.75  1.94. Answer is B.

11) The statement is false (Answer is B), since even if a data point is in the middle in terms of the x-values, it may have an extremely large y-value, which would make it a highly influential point.

SSR / 2 12) We have n=15, k=2, and F   3 so SSR=SSE/2. Since SSE /12 R 2  SSR /SST, we have 1 SST SSR  SSE SSE    1  1 2  3. So R 2  1/3. Answer is B. R 2 SSR SSR SSR

13) Since there are four Kings in the deck, the number of ways of drawing three 4 of them is    4. The probability of drawing three kings is 3 52 4/   4/ 22100  .000181.Answer is A.  3 

14) If A and B are to be independent, we must have P(B | A)  P(B)  P(B | A). If we know that the first card was a King, the probability that the second card will also be a King is P(B | A)  3/51. On the other hand, if we know that the first card was NOT a King, then the probability that the second card will be a King is P(B | A)  4/51. Since 3/51 is not equal to 4/51, the events A and B must be dependent. Answer is D. Note that the events are not mutually exclusive, since it’s possible to get a King on both the first and second cards. They are not complements of each other since the event of getting a King on the first card is not the same as getting a non-King on the second card.

15) The statement is false (Answer is B). In fact, if A, B are independent then P(AB)  P(A)  P(B)  P(A∩B)  P(A)  P(B)  P(A)P(B) , which is not the same thing as P(A)+P(B).