Chapter 9.3, Problem 9.8E

i.

To determine

Critical value of $\overline{x}$ used for rejecting $H_0$ .

Answer to Problem 9.8E

Since$z\left(2.0400\right)>z_a\left(1.645\right)$and a null hypothesis is rejected.

Explanation of Solution

Given:

1. The size of the sample is $n=35$ .
2. The mean $\overline{x}=2.4$ .
3. The standard deviation is $s=0.29$ .

Formula Used:

Test statistic:

  $z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Calculation:

Null hypothesis is $H_0:\mu =2.3$

Alternative hypothesis is $H_0:\mu >2.3$ (one-tailed)

So, for determining the tests statistics under the null hypothesis the test statistics is defined as below:

  $\begin{array}{l}z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}\\ =\frac{2.4-2.3}{\frac{0.2}{\sqrt{35}}}\\ =2.0400\end{array}$

Thus, critical value as the significance level is $\alpha =0.05$ and the test is one- tailed test, the rejection region is determined by a critical value with tail area equals to a $\alpha =0.05$ .

So, $H_0$ can be rejected if $z>1.645$ .

Conclusion:

Since $z\left(2.0400\right)>z_a\left(1.645\right)$ and a null hypothesis is rejected.

ii.

To determine

  $\beta =P$

  $\left(Accept{H}_{0}When\mu =2.4\right)$

Answer to Problem 9.8E

  $\beta =P$ is $0.3416$ .

Explanation of Solution

Given:

  $\mu =2.4$

Formula Used:

Test statistic:

  $z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Calculation:

The critical value at $5\%$ level of significance is $1.645$ .

  $z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}>1.645$

So, it implies that the acceptance region is,

  $\begin{array}{l}\frac{\overline{x}-2.4}{\frac{0.29}{\sqrt{35}}}<1.645\\ \overline{x}<2.38\\ \beta =P\left(Accept{H}_{0}When\mu =2.4\right)\\ =P\left(\overline{x}<2.38/\mu =2.4\right)\\ =P\left(z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}<\frac{2.38-2.4}{\frac{0.29}{\sqrt{35}}}\right)\\ =P\left(z<0.4080\right)\\ =0.3416\end{array}$

The power of the test is,

  $\begin{array}{l}1-\beta =1-0.3416\\ =0.6584\end{array}$

Conclusion:

Thus, $\beta =P=0.3416$ .

iii.

To determine

  $\beta$ when $\mu =2.3,2.5,2.6$

Answer to Problem 9.8E

  $\beta$ when $\mu =2.3$ is $0.9487$ , $\beta$ when $\mu =2.5$ is $0.007$ , $\beta$ when $\mu =2.6$ is $0.000004$

Explanation of Solution

Given:

  $\mu =2.3,2.5,2.6$

Formula Used:

Test statistic:

  $z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}$

Calculation:

Initially, to find out $\beta$ $\left(Accept{H}_{0}When\mu =2.3\right)$

The critical value at $5\%$ level of significance is $1.645$ .

  $z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}>1.645$

So, it implies that the acceptance region is,

  $\begin{array}{l}\frac{\overline{x}-2.3}{\frac{0.29}{\sqrt{35}}}<1.645\\ \overline{x}<2.38\\ \beta =P\left(Accept{H}_{0}When\mu =2.3\right)\\ =P\left(\overline{x}<2.38/\mu =2.3\right)\\ =P\left(z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}<\frac{2.38-2.3}{\frac{0.29}{\sqrt{35}}}\right)\\ =P\left(z<1.6320\right)\\ =0.9487\end{array}$

The power of the test is,

  $\begin{array}{l}1-\beta =1-0.9487\\ =0.0513\end{array}$

Now, to find out $\beta$ $\left(Accept{H}_{0}When\mu =2.5\right)$

The critical value at $5\%$ level of significance is $1.645$ .

  $z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}>1.645$

So, it implies that the acceptance region is,

  $\begin{array}{l}\frac{\overline{x}-2.5}{\frac{0.29}{\sqrt{35}}}<1.645\\ \overline{x}<2.38\\ \beta =P\left(Accept{H}_{0}When\mu =2.5\right)\\ =P\left(\overline{x}<2.38/\mu =2.5\right)\\ =P\left(z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}<\frac{2.38-2.5}{\frac{0.29}{\sqrt{35}}}\right)\\ =P\left(z<2.4480\right)\\ =0.007\end{array}$

The power of the test is,

  $\begin{array}{l}1-\beta =1-0.007\\ =0.9928\end{array}$

So, to find out $\beta$ $\left(Accept{H}_{0}When\mu =2.6\right)$

The critical value at $5\%$ level of significance is $1.645$ .

  $z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}>1.645$

So, it implies that the acceptance region is,

  $\begin{array}{l}\frac{\overline{x}-2.6}{\frac{0.29}{\sqrt{35}}}<1.645\\ \overline{x}<2.38\\ \beta =P\left(Accept{H}_{0}When\mu =2.6\right)\\ =P\left(\overline{x}<2.38/\mu =2.6\right)\\ =P\left(z=\frac{\overline{x}-\mu }{\frac{s}{\sqrt{n}}}<\frac{2.38-2.6}{\frac{0.29}{\sqrt{35}}}\right)\\ =P\left(z<4.48806\right)\\ =0.000004\end{array}$

The power of the test is,

  $\begin{array}{l}1-\beta =1-0.000004\\ =0.9999\end{array}$

Conclusion:

Therefore, $\beta$ when $\mu =2.3$ is $0.9487$ , $\beta$ when $\mu =2.5$ is $0.007$ , $\beta$ when $\mu =2.6$ is $0.000004$

iv.

To determine

To draw: Graph the power curve for the test using values of $\beta$ from part (ii) and part (iii).

Answer to Problem 9.8E

Power curve for the test is shown as below:

  

Explanation of Solution

Given:

Values of $\beta$ ,

1. $\beta =0.3416$
2. $\beta =0.9487$
3. $\beta =0.007$
4. $\beta =0.000004$

Calculation:

By considering the following table,

True mean $\left(\mu \right)$	Power of the test$\left(1-\beta \right)$
$2.3$	$0.0513$
$2.4$	$0.6584$
$2.5$	$0.9928$
$2.6$	$0.9999$

Power curve for the test is shown as below:

  

Conclusion:

Therefore, Power curve for the test is shown as below: