Mathematical Statistics and Data Analysis - Solutions

Chapter 7, Survey Sampling

(a)

$$P( \vert \; \bar p - p \; \vert \; \geq \delta ) = 1 - P( \vert \; \bar p - p \; \vert \; \leq \delta )$$
$$= 1 - P( { - \delta } \leq \bar p - p \leq \delta )$$

Dividing by, $\sigma_{\bar p}$, standard error of the sample mean:

$$= 1 - P( \frac { - \delta } { \sigma_{\bar p} } \leq \frac {\bar p - p} {\sigma_{\bar p} } \leq \frac { \delta } { \sigma_{\bar p} } )$$

Now we can approximate it by standard normal using Central Limit Theorem:

$$\approx 1 - \left( \Phi \left( \frac { \delta } { \sigma_{\bar p} } \right) - \Phi \left( \frac { - \delta } { \sigma_{\bar p} } \right) \right)$$
$$= 1 - \left( \Phi \left( \frac { \delta } { \sigma_{\bar p} } \right) - \left( 1 - \Phi \left( \frac { \delta } { \sigma_{\bar p} } \right) \right) \right)$$
$$= 2 \left( 1 - \Phi \left( \frac { \delta } { \sigma_{\bar p} } \right) \right)$$

Since we are given that $P( \vert \; \bar p - p \; \vert \; \geq \delta ) = 0.025$

Thus we get $\delta = \sigma_{\bar p} \times \Phi^{-1} \left( 1 - { \frac {0.025} 2 } \right) = \sigma_{\bar p} \times 2.4 \text{ ( by looking up normal table ) }$

Now since this is a proportion, we know for dichotmous case, and also ignoring the finite population correction:

$$\sigma^2_{\bar p} = \frac { p(1 - p) } n = \frac 1 5 \times \frac 4 5 \times \frac 1 { 100 } = \frac 1 { 625 }$$

which gives us:

$$\delta = 2.4 \times \sigma_{\bar p} = { 2.4 } \times \sqrt{ \frac 1 { 625 } } = 0.096$$

Thus the value of delta, which makes the probability of the difference between sample mean $\bar p$ and the population mean $p$ more than $\delta$ apart equals to $0.025$, is $0.0384$.

(b)

The 95% confidence interval is $\bar p \pm 1.96 \sigma_{\bar p}$.

Ignoring the finite population correction, $\sigma^2_{\bar p} = \frac { \sigma^2 } n$, where $\sigma^2$ is the population variance, which in this case(dichotomous) is $p(1-p) = \frac 1 5 \times \frac 4 5$. Thus $\sigma^2_{\bar p} = \frac 1 5 \times \frac 4 5 \times \frac 1 {100} = \frac 1 {625}$.

Thus the 95% confidence interval becomes $0.25 \pm 1.96 \times \sqrt { \frac 1 {625} } = 0.25 \pm 0.0784$, Or $(0.1716, 0.3284)$
Clearly the interval contains $0.20$

Note: In both parts, $\sigma^2_{\bar p}$ is used instead of $s^2_{\bar p}$ because we can compute it using the population variance $\sigma^2 = p(1-p)$. Thus we need not to use $s^2_{\bar p}$ as an approxiamtion for $\sigma^2_{\bar p}$. In other words, use $s^2_{\bar p}$ only when we can not compute $\sigma^2_{\bar p}$.

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