Chapter 18, Problem 18.49E

(a)

To determine

To find: The z statistic in terms of the sample mean $\overline{x}$.

To find: The value of z does the two-sided test reject $H_0$ at the 5% level of significance.

Answer to Problem 18.49E

The value of z statistic in terms of the sample mean $\overline{x}$ is $0.894\overline{x}-8.944$.

The two-sided test will reject $H_0$ if $z>1.96$ or $z<-1.96$.

Explanation of Solution

Given info:

The population of all measurements is normal with standard deviation $\sigma =2.5$.

Calculation:

z statistic:

Formula for z statistic is $z=\frac{\overline{x}-{\mu }_0}{\left(\frac{\sigma }{\sqrt{n}}\right)}$.

Substitute 10 for ${\mu }_0$, 2.5 for $\sigma$ and 5 for n,

$\begin{array}{c}z=\frac{\overline{x}-10}{\left(\frac{2.5}{\sqrt{5}}\right)}\\ =\frac{\overline{x}}{1.118}-\frac{10}{1.118}\\ =0.894\overline{x}-8.944\end{array}$

The value of z statistic is $0.894\overline{x}-8.944$.

Rejection rule for two-sided hypothesis:

If $\left|z\right|>z^{*}$, then reject the null hypothesis $\left(H_0\right)$.

From table C: t Distribution Critical Values, the value of $z^{*}$ corresponding with $\alpha =0.05$ and two-sided hypothesis is 1.96. That is, $z^{*}=1.96$.

Thus, the two-sided test will reject $H_0$ if $z>1.96$ or $z<-1.96$.

(b)

To determine

To find: The values of $\overline{x}$ lead to rejection of $H_0$.

Answer to Problem 18.49E

The values $\overline{x}>12.197$ or $\overline{x}<7.812$ will lead to rejection of $H_0$.

Explanation of Solution

Calculation:

Here, the two-sided test will reject $H_0$ if $z>1.96$ or $z<-1.96$.

Consider,

$z>1.96$.

Substitute $0.894\overline{x}-8.944$ for z,

$\begin{array}{c}z>1.96\\ 0.894\overline{x}-8.944>1.96\\ 0.894\overline{x}>10.904\\ \overline{x}>12.197\end{array}$

Consider,

$z<-1.96$.

Substitute $0.894\overline{x}-8.944$ for z,

$\begin{array}{c}z<-1.96\\ 0.894\overline{x}-8.944<-1.96\\ 0.894\overline{x}<6.984\\ \overline{x}<7.812\end{array}$

Thus, the values $\overline{x}>12.197$ or $\overline{x}<7.812$ will lead to rejection of $H_0$.

(c)

To determine

To find: The probability of observing $\overline{x}$ that leads to rejection of $H_0$.

Answer to Problem 18.49E

The probability of observing $\overline{x}$ that leads to rejection of $H_0$ is 0.4286.

Explanation of Solution

Given info:

The value of $\mu$ is 12.

Calculation:

Power:

Formula for power is,

$\begin{array}{c}\text{Power}=P\left(\overline{x}>12.197\right)+P\left(\overline{x}<7.812\right)\\ =P\left(\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}>\frac{12.197-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\right)+P\left(\frac{\overline{x}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<\frac{7.812-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\right)\\ =P\left(Z>\frac{12.197-12}{\left(\frac{2.5}{\sqrt{5}}\right)}\right)+P\left(Z<\frac{7.812-12}{\left(\frac{2.5}{\sqrt{5}}\right)}\right)\\ =P\left(Z>\frac{0.197}{1.118}\right)+P\left(Z<\frac{-4.188}{1.118}\right)\end{array}$

              $\begin{array}{l}=P\left(Z>0.18\right)+P\left(Z<-3.75\right)\\ =1-P\left(Z\le 0.18\right)+P\left(Z<-3.75\right)\end{array}$

From Table A: Standard Normal Cumulative Proportions, the value of $P\left(Z\le 0.18\right)$ is 0.5714 and the value of $P\left(Z<-3.75\right)$ is 0.0000.

Therefore,

$\begin{array}{c}\text{Power}=1-P\left(Z\le 0.18\right)+P\left(Z<-3.75\right)\\ =1-0.5714+0.0000\\ =0.4286\end{array}$

Thus, the probability of observing $\overline{x}$ that leads to rejection of $H_0$ is 0.4286.