Chapter 6, Problem 11SE

A machine that grinds valves is set to produce valves whose lengths have mean 100 mm and standard deviation 0.1 mm. The machine is moved to a new location. It is thought that the move may have upset the calibration for the mean length, but that it is unlikely to have changed the standard deviation. Let μ represent the mean length of valves produced after the move. To test the calibration, a sample of 100 valves will be ground, their lengths will be measured, and a test will be made of the hypotheses H₀: μ = 100 versus H₁: μ ≠ 100.

1. a. Find the rejection region if the test is made at the 5% level.
2. b. Find the rejection region if the test is made at the 10% level.
3. c. If the sample mean length is 99.97 mm, will H₀ be rejected at the 5% level?
4. d. If the sample mean length is 100.01 mm, will H₀ be rejected at the 10% level?
5. e. A critical point is 100.015 mm. What is the level of the test?

To determine

Find the rejection region, if the test is made at 5% level.

Answer to Problem 11SE

The null hypothesis will be rejected, if $\overline{X}\le \text{99}\text{.9804}$ or $\overline{X}\ge \text{100}\text{.0196}$.

Explanation of Solution

Given info:

The hypotheses are: $H_0:\mu =100$ versus $H_1:\mu \ne 100$.

The standard deviation is 0.1 mm and the sample size is 100.

Calculation:

Under $H_0$, the sample mean is approximately normal with mean and standard deviation is 100 and 0.01 $\left(=\frac{0.1}{\sqrt{100}}\right)$, respectively.

The rejection region consists of lower and upper 2.5% of the null distribution because the alternative hypothesis is of the form is $\mu \ne {\mu }_0$.

The lower and upper boundary 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 Both Tails for the region of the curve to shade.
• Enter the Probability value as 0.05.
• Click OK.

Output using the MINITAB software is given below:

Therefore, the z-sore corresponding to the lower and upper 2.5% is –1.96 and +1.96, respectively.

The boundaries are calculated as follows:

Lower boundary:

$\begin{array}{c}100-1.96\left(0.01\right)=100-0.0196\\ =\text{99}\text{.9804}\end{array}$

Upper boundary:

$\begin{array}{c}100+1.96\left(0.01\right)=100+0.0196\\ =\text{100}\text{.0196}\end{array}$

Therefore, the rejection region is $\overline{X}\le \text{99}\text{.9804}$ and $\overline{X}\ge \text{100}\text{.0196}$

If $\overline{X}\le \text{99}\text{.9804}$ or $\overline{X}\ge \text{100}\text{.0196}$, the null hypothesis  will be rejected.

To determine

Find the rejection region, if the test is made at 10% level.

Answer to Problem 11SE

The null hypothesis will be rejected, if $\overline{X}\ge \text{100}\text{.01645}$ or $\overline{X}\le \text{99}\text{.98355}$.

Explanation of Solution

Calculation:

The rejection region consists of lower and upper 5% of the null distribution because the alternative hypothesis is of the form is $\mu \ne {\mu }_0$.

The lower and upper boundary is calculated as follows:

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 Probability Value and Both Tails for the region of the curve to shade.
• Enter the Probability value as 0.10.
• Click OK.

Output using the MINITAB software is given below:

Therefore, the z-sore corresponding to the lower and upper 5% is –1.645 and +1.645, respectively.

The boundaries are calculated as follows:

Lower boundary:

$\begin{array}{c}100-1.645\left(0.01\right)=100-\text{0}\text{.01645}\\ =\text{99}\text{.98355}\end{array}$

Upper boundary:

$\begin{array}{c}100+1.645\left(0.01\right)=100+\text{0}\text{.01645}\\ =\text{100}\text{.01645}\end{array}$

Therefore, the rejection region is $\overline{X}\le \text{99}\text{.98355}$ and $\overline{X}\ge \text{100}\text{.01645}$

If $\overline{X}\ge \text{100}\text{.01645}$ or $\overline{X}\le \text{99}\text{.98355}$, the null hypothesis  will be rejected.

To determine

Check whether the null hypothesis will be rejected at the 5% level, if the sample mean length is 99.97 mm.

Answer to Problem 11SE

Yes, the null hypothesis will be rejected at 5% level.

Explanation of Solution

Calculation:

From part a., the null hypothesis will be rejected, if $\overline{X}\le \text{99}\text{.9804}$ or $\overline{X}\ge \text{100}\text{.0196}$.

Here, the sample mean is 99.97 mm.

Thus, the null hypothesis will be rejected at 5% level because $\overline{X}\left(=99.97\right)<\text{99}\text{.9804}$.

To determine

Check whether the null hypothesis will be rejected at the 10% level, if the sample mean length is 100.01 mm.

Answer to Problem 11SE

No, the null hypothesis will not be rejected at 10% level.

Explanation of Solution

Calculation:

From part b., the null hypothesis will be rejected, if $\overline{X}\ge \text{100}\text{.01645}$ or $\overline{X}\le \text{99}\text{.98355}$.

Here, the sample mean is 100.01 mm.

Therefore, the null hypothesis will not be rejected at 10% level because $\overline{X}\left(=100.01\right)<\text{100}\text{.01645}$.

To determine

Find the level of the test, if the critical point is 100.015 mm.

Answer to Problem 11SE

The level is 0.1336.

Explanation of Solution

Calculation:

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

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

The alternative hypothesis is of the form is $\mu \ne {\mu }_0$. This indicates that the test is two tailed test. Therefore, there are two critical points.

The given critical point is 100.015 mm and the other critical point is $100-0.015=99.985$.

The level of the test is, $P\left(\overline{X}\le 99.985\right)+P\left(\overline{X}\ge 100.015\right)$.

The z-score of 99.985 is calculated below:

$\begin{array}{c}P\left(\overline{X}\le 99.985\right)=P\left(z\le \frac{99.985-100}{0.01}\right)\\ =P\left(z\le \frac{-0.015}{0.01}\right)\\ =P\left(z\le -1.5\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\le -1.5\right)$ 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 Left Tail for the region of the curve to shade.
• Enter the data value as -1.5.
• Click OK.

Output using the MINITAB software is given below:

Thus, $P\left(z\le -1.5\right)=0.06681$.

The z-score of 100.015 is calculated below:

$\begin{array}{c}P\left(\overline{X}\ge 100.015\right)=P\left(z\ge \frac{100.015-100}{0.01}\right)\\ =P\left(z\ge \frac{0.015}{0.01}\right)\\ =P\left(z\ge 1.5\right)\end{array}$

Software Procedure:

Step-by-step procedure to obtain the $P\left(z\ge 1.5\right)$ 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 1.5.
• Click OK.

Output using the MINITAB software is given below:

From the output, $P\left(z\ge 1.5\right)=0.06681$.

Therefore, the level is given below:

$\begin{array}{c}P\left(\overline{X}\le 99.985\right)+P\left(\overline{X}\ge 100.015\right)=0.06681+0.06681\\ =\text{0}\text{.1336}\end{array}$

Thus, the level is 0.1336.