Final Exam Sample Problems Answer Key.             Stat 401, Fall 2010
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ABRAM KAGAN final exam from Fall 2009

1.(i) Notation: let N_{+} = # x_i observations > 0.
Then likelihood is product of  (3x_i^2/2) over all i=1,..,n times
theta^(n-N_{+}) times (2-theta)^(N_{+}).
  (ii) MLE = 2(n-N_{+})/n, and N_{+} ~ Binom(n, (2-theta)/2 ), so
E(2(n-N_{+})/n) = (2/n)(n theta/2) = theta proves unbiased.
  (iii) Var of MLE = thete*(2-theta)/n.

2.(i) 31.8 plus or minus (2.9/sqrt(17))*t_{16,.05} = (30.572, 33.0278)
  (ii) (2.9*sqrt(16/chi^2_{16,.05}), 2.9*sqrt(16/chi^2_{16,.95})) = 
       (2.262, 4.111)

3.(i) H0: mu2 <= mu1, HA: >     [leaving H0 as mu1=mu2 would also be OK]
 (ii) (mean(samp2)-mean(samp1))/sqrt((1/5+1/4)*(3*var(samp1)+4*var(samp2))/7)
[1] 1.197675
### since this is less than t_{7,.05} = 1.895, accept H0.

4. (a) ANOVA F-stat =  (19.74/3)/(24.72/24) = 6.38835 < F_{3,24,.05} = 3.009,
  so accept H0.  

5. (a) line is y=a+bx with (a,b) estimated = (-14.070, 0.15449)
   (b) sample corr = 0.2157

6. (Y(2)+Y(3)+Y(4))/3 ~ N(6.1, .03) , so 
   prob = pnorm(.2/sqrt(.03))-pnorm(-.1/sqrt(.03)) = .594

7. (i) Null hypothesis is that the two-coin result is sum of indep Binom(1,.5) 
and Binom(1,theta) random variables, so p0(theta) = .5*(1-theta), 
p1(theta) = .5, and p2(theta)= .5*theta.
   (ii) MLE of theta is  n2/(n0+n2).
   (iii) chisq goodness of fit statistic with MLE substituted for theta is .72 
with 1 df, Accept H0 that the model in (i) is right.


LIST OF SAMPLE PROBLEMS ADAPTED FROM 95 SAMPLE EXAM

1. (a) S_1^2 ~ chi-squared 2 d.f. ~ Gamma(1,1/2) = Expon(1/2)
so Prob(S_1^2 > 3) = exp(-1.5) = .2231. (b) 9.2* 10^(-6)

2. Method of moments estimator of theta is 1.5*Xbar, which has variance
 = 1.5^2* (theta^2/18)/100 = .005

3.(a) Based on samples 1 and 3 only, interval is 
10.7-11.8 +- sqrt(2/10)*sqrt((9*(4.2+4.9)/18)*t_{18,.025}
 = (-3.104, 0.904); you could also make an interval based 
on a pooled variance using all three samples:
10.7-11.8 +- sqrt(2/10)*sqrt((9*(4.2+4.9+3.5)/27)*t_{27,.025}
 = (-2.9805, 0.7805)

(b) F_{2,27} test statistic = 19.39 statistic, p-value < 1.e-5

4. 99*SY^2/4 ~ chi^2_{99} = Gamma(99/2,1/2) is approximately 
N(99, 198) by the CLT. So probability is approx. 
1-pnorm((4.47-4)/sqrt(8/99))+pnorm((3.53-4)/sqrt(8/99)) = .0983

5. Multinomial goodness of fit test with prob-vector for outcomes 
0,1,2 respectively (1-p)^2, 2p(1-p), p^2. Based on the data, 
MLE for p is (2*58+118)/(2*1388) = .0843, and the expected numbers 
developing 0,1,or 2 ovaries are 1163.8, 214.3, 9.9. The goodness 
of fit chisq (with 1 df) is 279.0, extremely significant so reject H0.

6. (a) For mu:  87.3 +- sqrt(14.7/21))*t_{20,.025} = (85.55,89.05)
 For sigma^2:  (14.7*20/chi^2_{20,.025}, 14.7*20/chi^2_{20,.975}) 
  =  (9.36, 27.09)

7. MLE is (n-N3)/(3n) with variance (1-3p)*(3p)/n

8. (a) t_{28}-statistic is (20.5-23.5)/sqrt((2/15)*(3.3^2+2.5^2)/2)
 = -2.8065 which is significant because abs is > t_{28,.025} = 2.048
  (b) F_{1,28} ANOVA test statistic = 7.8763 = 2.8065^2 with exactly 
 same  p-value as in (a) !
  (c) Same assumptions; but (a) could have been re-done using the 
Satterthwaite approximation [with essentially the aame result] 
WITHOUT assuming variance was the same in both samples.

9. MLE = 1/Xbar, and by CLT  lambda*Xbar*n ~ Gamma(n,1) approx N(n,n)
So approx CI = (1/Xbar)*(1 +- 1.96/sqrt(n))

10. phat = (2*9+25)/(2*64) = .3359, with expected counts (28.23,28.55,7.22)
So 1 df chisq goodness of fit statistic is .99, not significant, 
so accept H0.

11. These are exact binom probs which can be approximated 
by CLT in (a) and Poisson in (b): 
(a) Exact = .966, approx (continuity-corrected = .968
(b) Exact = .972, approx  =  .968