Chapter 18, Problem 18.49E

a.

To determine

The z test statistics in terms of the sample mean $\overline{x}$.

The value of z for which two sided test reject $H_0$ at 5% level of significance.

Answer to Problem 18.49E

The null hypothesis will be rejected for $\underset{\_}{Z\le -1.96}\text{or }\underset{\_}{Z\ge 1.96}$ at 5% level of significance.

Explanation of Solution

Given info:

Random sample of size 5 is as follows:

6.47

7.51

10.10

13.63

9.91

The Null and Alternative hypothesis is given as:

The null hypothesis:

${\text{H}}_0:\mu =8$

The alternative hypothesis:

${\text{H}}_a:\mu \ne 8$

Simple random sample of size n is 5 and standard deviation is $\sigma =2$.

Calculation:

The formula for sample mean is,

$\overline{X}=\frac{{\displaystyle \sum _i{X}_i}}{n}$

Where,

• ${\displaystyle \sum \text{X}}$ is the sum of all sample values.
• n is the sample size.
• $\overline{\text{X}}$ is the sample mean.

The formula for the z statistic is,

$z=\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$

Where,

• $\overline{X}$ is the sample mean.
• $\mu$ is the population mean.
• $\sigma$ is the population standard deviation.
• n is the sample size.

Add all the sample observations to find ${\displaystyle \sum _i{X}_i}$.

$\begin{array}{c}{\displaystyle \sum _i{X}_i}=\text{6}\text{.47+7}\text{.51+10}\text{.10+13}\text{.63+9}\text{.91}\\ =47.62\end{array}$

Substitute 47.62 for ${\displaystyle \sum _i{X}_i}$ and 5 for n in equation for sample mean to find $\overline{X}$.

$\begin{array}{c}\overline{X}=\frac{47.62}{5}\\ =9.524\end{array}$

Substitute 8 for $\mu$, 9.524 for $\overline{X}$, 2 for $\sigma$ and 5 for n in equation for z statistic to find z statistic.

$\begin{array}{c}z=\frac{9.524-8}{\frac{2}{\sqrt{5}}}\\ =\frac{1.524}{0.894}\\ =1.740\end{array}$

The value of z statistic is 1.740.

At a 5% level of significance, the critical value of z for two tailed test is 1.96.

From the table of “The standard Normal Cumulative Proportions”, the critical value for the left tail test corresponding to an area of 0.25 is $-1.96$ and for the right tail test corresponding to an area of 0.975 $\left(1-0.975=0.25\right)$ is 1.96.

Decision rule:

If absolute value of test statistic is less than critical value then there is not enough evidence to reject the null hypothesis.

The test statistic value 1.740 is less than that of the critical value 1.96, hence there is enough evidence to reject the null hypothesis.

Thus, null hypothesis will be rejected for $Z\le -1.96\text{or }Z\ge 1.96$.

b.

To determine

The value of sample mean lead to reject the null hypothesis.

Answer to Problem 18.49E

$H_0$ is rejected when $\underset{\_}{\overline{X}\le 6.247}\text{and}\underset{\_}{\overline{X}\ge 9.753}$.

Explanation of Solution

Calculation:

The formula for the z statistic is given by;

$z=\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}$ (2)

Where,

• $\overline{X}$ is the sample mean.
• $\mu$ is the population mean.
• $\sigma$ is the population standard deviation.
• n is the sample size.

Substitute 8 for $\mu$, 9.524, for $\sigma$ and 5 for n in equation.

$\begin{array}{c}z=\frac{\overline{X}-8}{\frac{2}{\sqrt{5}}}\\ =\frac{\sqrt{5}}{2}\times \left(\overline{X}-8\right)\\ =1.118\times \left(\overline{X}-8\right)\end{array}$

Reject the null hypothesis $H_0$ when,

$\begin{array}{l}=1.118\left(\overline{X}-8\right)\ge 1.96\\ =\left(\overline{X}-8\right)\ge \frac{1.96}{1.118}\\ =\overline{X}\ge 1.753+8\\ =\overline{X}\ge 9.753\end{array}$

Reject the null hypothesis ${\text{H}}_0$ when,

$\begin{array}{l}=1.118\left(\overline{X}-8\right)\le -1.96\\ =\left(\overline{X}-8\right)\le \frac{-1.96}{1.118}\\ =\overline{X}\le -1.753+8\\ =\overline{X}\le 6.247\end{array}$

Thus, at 5% level of significance level ${\text{H}}_0$ is rejected when $\overline{X}\le 6.247\text{and}\overline{X}\ge 9.753$.

c.

To determine

The power of the test.

Answer to Problem 18.49E

The power of the test is about 39%.

Explanation of Solution

Calculation:

Power:

Power of a test of hypothesis is defined as the probability of rejecting null hypothesis when the alternative is true.

The sample mean $\overline{\text{X}}$ follows normal distribution with population mean $\mu =10$  and standard deviation $\frac{\sigma }{\sqrt{n}}=\frac{2}{\sqrt{5}}$ is defined as:

$\overline{X}\sim N\left(10,\frac{2}{\sqrt{5}}\right)$

Therefore,

$\begin{array}{l}P\left(\overline{X}\le 6.247\right)+\left(\overline{X}\ge 9.753\right)=P\left(6.247\le \overline{X}\le 9.753\right)\\ =P\left(\frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\le z\le \frac{\overline{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\right)\\ =P\left(\frac{6.247-10}{\frac{2}{\sqrt{5}}}\le z\le \frac{9.753-10}{\frac{2}{\sqrt{5}}}\right)\\ =P\left(\frac{-3.753}{0.8944}\le z\le \frac{-0.247}{0.8944}\right)\end{array}$

$\begin{array}{l}=\text{P}\left(-4.20\le z\le -0.28\right)\\ =\text{P}\left(z\le -0.28\right)-\text{P}\left(z\le -4.20\right)\\ =\text{P}\left(z>0.28\right)-\text{P}\left(z>4.20\right)(\text{By}\text{symmetric})\\ =0.5-\text{P}\left(z<0.28\right)-\left(0.5-\text{P}\left(z<4.20\right)\right)\end{array}$

$\begin{array}{l}=\text{P}\left(z<4.20\right)-\text{P}\left(z<0.28\right)\\ =0.5-0.1103\left(\text{Using}\text{Standard}\text{Normal}\text{Table}\text{A}\right)\\ =0.3897\end{array}$

$\approx 0.39$

Hence, the power of the test is about 39%.