STAT 410 96-2 Assignment 5 Solutions



  1. 4.1: The standard error of p is smaller for P in the range 5% to 10% than for P in the range 45% to 55%. The largest possible se for P in the latter range occurs when P=50% so this is the value you plug in to 4.1 with t=1, since this corresponds to 1 se.
  2. 4.2: In 4.4 replace rYbar by 1000/676 and use t=1.96 to get a confidence level of 95% = 19/20.
  3. 4.3: For a: First the coefficient of variation of Np is the same as the coefficient of variation of p. So set cv(P) = 0.2 = se(p)/P. Solving this equation boils down to 4.1 with t=1 and d=.2P. Since N is unknown we must assume it is large enough to neglect the fpc and take the denominator in 4.1 to be 1.
    For b: If you take a sample of n people you will get m=n/2 of them of each sex, more or less. So you will need m to be the answer from a and n to be twice that much.
  4. 4.5: Ignoring the fpc the cv of an estimate is a population cv divided by the square root of the sample size. Setting this equal to either 2.5% or 5% gives 5 values for n of the form n = (pop cv / target cv)^2 and you take the largest. Then you compute 317/sqrtt(n) to get the cv for estimated number of unemployed.
  5. Use the sample you drew from stat village to suggest a sample size needed to estimate the mean household government income to within 5%: This is formula 4.4 with r=0.05. You will have to use s, ybar from the sample for S, Ybar.
  6. Again for the sample you drew using stat village give and estimate and an estimated standard error for the proportion of people who are 65 years of age or over: You have to use the formulas 3.31 and 3.32 on page 66.
  7. Use your sample to see how much evidence there is that the variable TENURH (owner occupied or not) is independent of variable 22 (HMSEX: Sex of Primary Household Maintainer): This is a contingency table problem. You will have a 2 by 2 table with 4 entries adding up to 36 and do a chi-squared test.
The questions.