Mathematical Statistics and Data Analysis - Solutions

(a)

We have:

Total Population, $\, N=12000 \,$

Sample size, $\, n=200 \,$

Sample proportion, $\, \hat p = 0.18 \,$

To compute estimated standard error, $\, \sigma_{\hat p} \,$, we use $\, {\sigma_{\hat p}}^2 = \frac {\hat p(1-\hat p)} {n-1} \Prn{1 - \frac n N} \,$. Thus we get $\, {\sigma_{\hat p}^2} = \frac {0.18(1-0.18)} {200-1} \Prn{1-\frac {200} {12000}} = 0.0007417085 \,$. Thus $\, \sigma_{\hat p} = \sqrt {0.0007417085} = 0.027 \,$.

The confidence interval can be computed using $\, \hat p \pm z(\alpha/2) = \hat p \pm \Phi^{-1}(1-\alpha/2)\sigma_{\hat p} \,$ where $\, \alpha = 1-0.90 = 0.1 \,$. Thus the confidence interval becomes $\, 0.18 \pm \Phi^{-1}(1-0.05) 0.027 = 0.18 \pm 1.645 \times 0.027 = 0.18 \pm 0.044415 \,$, or $\, (0.135585,0.224415) \,$.

(b)

We are given two samples from two populations - one from the example quoted and another given in the problem itself.

Sample 1:(from example)

Sample 2:(from problem)

and $\, \hat p_2 = 0.18 \,$ and $\, \hat d = \hat p_1 - \hat p_2 \,$.

Since $\, \hat p_1 \,$ and $\, \hat p_2 \,$ are independent random variables, it follows that $\, \Var(\hat d) = \Var(\hat p_1 - \hat p_2) = \Var(\hat p_1) + {-1}^2 \hat p_2) \,$.

Now putting $\, \Var(\hat p_1) = \frac {\hat p_1(1-\hat p_1)} {n_1-1} \Prn{1 - \frac {n_1} {N_1} } \,$ and $\, \Var(\hat p_2) = \frac {\hat p_2(1-\hat p_2)} {n_2-1} \Prn{1 - \frac {n_2} {N_2} } \,$, we get:

$\, \Var(\hat d) = \frac {\hat p_1(1-\hat p_1)} {n_1-1} \Prn{1 - \frac {n_1} {N_1} } + \frac {\hat p_2(1-\hat p_2)} {n_2-1} \Prn{1 - \frac {n_2} {N_2} } \,$.

Using the standard error computed in example, we have $\, \Var(\hat p_1) = {0.03}^2 = 0.0009 \,$. Similarly from part-a, we have $\, \Var(\hat p_2) = 0.0007417085 \,$. Thus $\, \Var(\hat d) = 0.0009 + 0.0007417085 =0.001642 \,$. Thus $\, \sigma_{\hat d} = \sqrt {0.001642} = 0.0405215 \,$.

(c)

The confidence interval is given by $\, \hat d \pm z(\alpha/2) \sigma_{\hat d} \,$.

We have $\, \hat d = \hat p_1 - \hat p_2 = -0.06 \,$.

For $\, 99\% \,$ confidence interval, $\, \alpha = 1-0.99 = 0.01 \,$. Thus $\, z(\alpha/2) = \Phi^{-1}(1-\alpha/2) = \Phi^{-1}(1-0.01/2) = 2.575 \,$. Thus the confidence interval is $\, -0.06 \pm 2.575 \times 0.0405215 = -0.06 \pm 0.104343 = ( -0.164343, 0.044343) \,$

Similarly, for $\, 95\% \,$, we have $\, z(\alpha/2) = 1.96 \,$. Thus $\, -0.06 \pm 1.96 \times 0.0405215 = -0.06 \pm 0.079422 = (-0.139422,0.019422) \,$

And for $\, 90\% \,$, we get $\, z(\alpha/2) = 1.645 \,$. Thus $\, -0.06 \pm 1,645 \times 0.0405215 = -0.06 \pm 0.066658 = (-0.126658, 0.006658) \,$