# Mathematical Statistics and Data Analysis - Solutions

### Chapter 7, Survey Sampling

#### Solution 19

For the first part, we need to find the constant $\, k \,$ such that the interval $\, (-\infty, \bar X + ks_{\bar X}) \,$ is a 90% confidence interval for $\, \mu \,$.

We shall use the same function here $\, z(\alpha) \,$ as used in the book. Also let $\, Z \,$ denotes the standard normal function.

Thus we have:

$$\, P(Z < z(\alpha)) = \Phi(z(\alpha)) \,$$
\, \begin{align*} &\Rightarrow P\left( \frac{\bar X - \mu}{s_{\bar X} } < z(\alpha)\right) = 1 - \alpha && \text{Using CLT in LHS and def. of z in RHS} \\ &\Rightarrow P(\bar X - \mu < z(\alpha)s_{\bar X}) = 1-\alpha \\ &\Rightarrow P(\mu - \bar X > -z(\alpha)s_{\bar X}) = 1-\alpha && \text{Reversing the sign of inequality} \\ &\Rightarrow P(\mu > \bar X - z(\alpha)s_{\bar X}) = 1-\alpha \\ &\Rightarrow P(\mu > \bar X + z(1-\alpha)s_{\bar X}) = 1-\alpha && -z(\alpha) = z(1-\alpha) \text{ By its definition} \\ &\Rightarrow P(\mu < \bar X + z(1 - \alpha)s_{\bar X}) = 1 - (1-\alpha) && \text{Sum of total probability is 1} \\ &\Rightarrow P(\mu < \bar X + z(1 - \alpha)s_{\bar X}) = \alpha \end{align*} \,

Thus for 90% confidence interval we need $\, \alpha = 0.9 \,$. And $\, k = z(1-\alpha) \,$. By the definition of $\, z \,$, we have $\, z(\alpha) = \Phi^{-1}(1-\alpha) \,$. Thus $\, k= \Phi^{-1}(1-(1-\alpha)) = \Phi^{-1}(0.9) \,$

Now we just need to lookup in the standard normal table to find $\, k = 1.28 \,$.

Second part is also similar where we need to find the value of $\, k \,$ for the one sided confidence interval $\, (\bar X - ks_{\bar X}) \,$.

We can proceed from the last result:

\, \begin{align*} &\Rightarrow P(\mu < \bar X + z(1 - \alpha)s_{\bar X}) = \alpha \\ &\Rightarrow P(\mu < \bar X - z(\alpha)s_{\bar X}) = \alpha && \text{Since here k is negative, we will use -z(\alpha) = z(1-\alpha)} \\ &\Rightarrow P(\mu > \bar X - z(\alpha)s_{\bar X}) = 1-\alpha \end{align*} \,

Thus for 95% confidence interval we need $\, \alpha = 1-0.95=0.05 \,$. And $\, k = z(\alpha) \,$. By the definition of $\, z \,$, we have $\, z(\alpha) = \Phi^{-1}(1-\alpha) \,$. Thus $\, k= \Phi^{-1}(1-\alpha) = \Phi^{-1}(0.95) = 1.645 \,$.

$$\tag*{\blacksquare}$$