Chapter 6, Problem 12SE

Resistors for use in a certain application are supposed to have a mean resistance μ greater than 100 Ω. Assume that the standard deviation of the resistances is 5 Ω. Resistances will be measured for a sample of resistors, and a test of the hypothesis H₀: μ ≤ 100 versus H₁: μ > 100 will be made. Assume that in fact the true mean resistance is 101 Ω.

1. a. If 100 resistors are sampled, what is the power of a test made at the 5% level?
2. b. How many resistors must be sampled so that a 5% level test has power 0.95?
3. c. If 100 resistors are sampled, at what level must the test be made so that the power is 0.90?
4. d. If 100 resistors are sampled, and the rejection region is $\bar{X}>100.5$, what is the power of the test?

To determine

Find the power of the test made at the 5% level, if the 100 resistors are samples.

Answer to Problem 12SE

The power is 0.6406.

Explanation of Solution

Given info:

The hypotheses are: $H_0:\mu \le 100$ versus $H_1:\mu >100$.

The standard deviation of the resistance is 5Ω. The true mean resistance is 101 Ω.

Calculation:

Under $H_0$, the sample mean is approximately normal with mean and standard deviation is 100 and 0.05 $\left(=\frac{5}{\sqrt{100}}\right)$, respectively.

The 5% rejection region is $\overline{X}>x_5$ because the alternative hypothesis is of the form $\mu >{\mu }_0$.

The upper boundary of 5% is calculated as follows:

Software Procedure:

Step-by-step procedure to obtain the 5^th percentile using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose Probability Value and Right Tail for the region of the curve to shade.
• Enter the Probability value as 0.5.
• Click OK.

Output using the MINITAB software is given below:

From the output, the z-score corresponding to the 5^th percentile is +1.645.

The value of $x_5$ is calculated as follows:

$\begin{array}{c}{x}_5=100+1.645\left(0.5\right)\\ =100+\text{0}\text{.8225}\\ =\text{100}\text{.8225}\end{array}$

Therefore, the rejection region is $\overline{X}\ge \text{100}\text{.8225}$.

Power:

$\begin{array}{c}\text{Power}=P\left(\overline{X}\ge \text{100}\text{.8225}\right)\\ =P\left(z\ge \frac{\text{100}\text{.8225}-101}{0.5}\right)\\ =P\left(z\ge \frac{-\text{0}\text{.1775}}{0.5}\right)\\ =P\left(z\ge -0.36\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\ge -0.36\right)$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the data value as -0.36.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\ge -0.36\right)=0.6406$.

Thus, the power is 0.6406.

To determine

Find the number of resistors that must be sampled so that a 5% level test has power 0.95.

Answer to Problem 12SE

About 271 resistors must be sampled so that a 5% level test has power 0.95.

Explanation of Solution

Calculation:

The 5% rejection region is $\overline{X}\ge x_5$ because the alternative hypothesis is of the form $\mu >{\mu }_0$.

From part a., the z-score corresponding to the 5^th percentile is +1.645.

The value of $x_0$ is calculated as follows:

$x_0=100+1.645\left(\frac{5}{\sqrt{n}}\right)$

The rejection region is $\overline{X}\ge x_0$ if power 0.95. That is $P\left(\overline{X}\ge x_0\right)=0.95$.

Software Procedure:

Step-by-step procedure to obtain the 95^th percentile using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the data value as 0.95.
• Click OK.

Output using the MINITAB software is given below:

The value of $x_0$ is calculated as follows:

$x_0=101-1.645\left(\frac{5}{\sqrt{n}}\right)$

Therefore:

$\begin{array}{c}100+1.645\left(\frac{5}{\sqrt{n}}\right)=101-1.645\left(\frac{5}{\sqrt{n}}\right)\\ 1.645\left(\frac{5}{\sqrt{n}}\right)+1.645\left(\frac{5}{\sqrt{n}}\right)=101-100\\ \left(\frac{\text{8}\text{.225}}{\sqrt{n}}\right)+\left(\frac{\text{8}\text{.225}}{\sqrt{n}}\right)=1\\ \frac{\text{8}\text{.225}+\text{8}\text{.225}}{\sqrt{n}}=1\end{array}$

                       $\begin{array}{c}\sqrt{n}=\text{16}\text{.45}\\ =\text{270}\text{.6025}\\ \approx \text{271}\end{array}$

Thus, the required sample size is 271.

To determine

Find the level of the test.

Answer to Problem 12SE

The level is 0.2358.

Explanation of Solution

Calculation:

Type-1 error: Rejecting the null hypothesis $\left(H_0\right)$ when it is true. It is denoted by α.

$\begin{array}{c}\alpha \text{ = P( type I error) }\\ \text{= P( rejecting }{H}_0\text{when }{H}_0\text{is true)}\end{array}$

Software Procedure:

Step-by-step procedure to obtain the 10^th percentile using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Left Tail for the region of the curve to shade.
• Enter the data value as 0.10.
• Click OK.

Output using the MINITAB software is given below:

Thus, the z-score corresponding to the 10^th percentile is –1.28.

The value of $x_0$ is calculated as follows:

$\begin{array}{c}{x}_0=101-1.28\left(\frac{5}{\sqrt{100}}\right)\\ =101-\text{0}\text{.64}\\ =100.36\end{array}$

The level is calculated as follows:

$\begin{array}{c}\text{Level}=P\left(\overline{X}\ge 100.36\right)\\ =P\left(z\ge \frac{100.36-100}{0.5}\right)\\ =P\left(z\ge \frac{0.36}{0.5}\right)\\ =P\left(z\ge 0.72\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\ge 0.72\right)$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the data value as 0.72.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\ge 0.72\right)=0.2358$.

Therefore, the level is 0.2358.

To determine

Find the power of the test, if the 100 resistors are sampled and the rejection region is $\overline{X}>100.5$.

Answer to Problem 12SE

The power is 0.8413.

Explanation of Solution

Calculation:

Power:

$\begin{array}{c}\text{Power}=P\left(\overline{X}\ge 100.5\right)\\ =P\left(z\ge \frac{\text{100}\text{.5}-101}{0.5}\right)\\ =P\left(z\ge \frac{-0.5}{0.5}\right)\\ =P\left(z\ge -1\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\ge -1\right)$ using the MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Right Tail for the region of the curve to shade.
• Enter the data value as -1.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\ge -1\right)=0.8413$.

Thus, the power is 0.8413.