STAT 410 96-2 Assignment 4 Solutions



  1. The estimate is simply N times the proportion of the sample in favour which is 2000 x 120/200 = 1200. The estimated standard error is 2000 x sqrt(120/200 x 80/200) x sqrt(1-200/2000) / sqrt(199) (or you may use sqrt(200) in place of sqrt(199)). This estimated standard error comes to 65.9 so the confidence interval is 1200+-1.96x65.9. Cochran suggests a slightly improved normal approximation which increases the margin by about 5.
  2. A majority means 1001 or more and the test statistic for the null hypothesis A <=1000 is (1200-1000)/65.9 = 3.04 leading to a one sided P value of around .1% so the evidence for majority support is very strong.
  3. The estimated proportion of renters having exclusive use of a toilet is p=109/(109+34) and the estimated standard error is given by sqrt(p(1-p)) x sqrt(1-f) / sqrt(143) where f must be estimated using 290/14828 since the the value N_rented is unknown.
    To estimate the number of renters not having such exclusive use you multiply 34/143 by an estimate of the number of renters namely 14828 x 143/290. In other words you just let P be the fraction of all households which are rented and where the household does not have exclusive use of a toilet. You estimate P by p=34/290 and the corresponding population number A by 14828 x 34 / 290. The estimated standard error of this estimate is 14828 x sqrt(p(1-p)) x sqrt(1-290/14828) / sqrt(289). (This is just formula 3.28.)
  4. In this case you can use 7526 x 34/143 which has the estimated standard error 7526 x sqrt(p(1-p)) x sqrt(1-143/7526) / sqrt(142). This is 3.25 and 3.26. Here p=34/143. In this problem and the last you are free to neglect the fpc and to use sqrt(n) instead of sqrt(n-1) wherever you want.
  5. I intended you to do 3.13 which is a cluster sampling problem and uses 3.31 and 3.34. The total number seeing a dentist is 22 out of a total of 104 people so phat = 22/104. The formula 3.32 gives (ignoring the fpc) (38 - 2 x (22/104) x 87 + 394) / (30 x 29 x (104/30)^2) .
  6. To do question 3.14 you simply say that the event that the mth rare case is seen on the nth trial is the same as the intersection of the 2 events: "there are m-1 rare cases in the first n-1 trials" and "the nth trial is a rare case". These two events are independent and the first has probability n-1 choose m-1 x P^(m-1) x Q^(n-1-(m-1)) while the second has probability P. Multiplying these two probabilities to find the probability of the intersection gives the answer sought. If you were sampling without replacement and there are a total of M rare cases (M > m) in a population of N then the first event mentioned has probability ( M choose m-1) x (N-M choose n-1-(m-1))/(N choose n-1) while the second (which is not now independent of the first) has conditional probability given that the first has happened equal to (M-(m-1))/(N-(n-1)). You multiply these together and do algebra.
The questions.