Chapter 10.4, Problem 46E

a.

To determine

State the appropriate conclusion for the given information.

Answer to Problem 46E

The conclusion is that there is no convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.

Explanation of Solution

The given information states the sample of 13 ball bearings. It is also known that the test statistic is $t\text{ = 1}\text{.6}$ and the significance level is $\alpha =0.05$.

The test hypotheses are given below:

Null hypothesis: $H_0:\mu =0.5$

That is, the mean diameter of ball bearings of a particular type is 0.5.

Alternative hypothesis: $H_a:\mu \ne 0.5$

That is, the mean diameter of ball bearings of a particular type is different from 0.5.

Here, the direction of the test is two-tailed.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =13-1\\ =12\end{array}$

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot > choose View Probability> OK.
• From Distribution, choose t.
• Enter df as 12.
• Choose X value and Both Tail for the region of the curve to shade.
• Enter the X value as 1.6.
• Click OK.

Output using the MINITAB software is given below:

From the output, the P-value of the one-tailed test is 0.06779.

The P-value of the test statistic is obtained as follows:

$\begin{array}{c}P\text{-value}=2\times P\left(t>1.6\right)\\ =2\times 0.06779\\ \approx 0.136\end{array}$

Decision rule:

• If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.
• If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ 0.136 is greater than the significance level 0.05.

That is, $P\text{-value}\left(=0.136\right)>\text{Significance level}\left(=0.05\right)$.

The decision is that the null hypothesis is not rejected at 0.05 significance level.

Therefore, there is convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.

b.

To determine

State the appropriate conclusion for the given information.

Answer to Problem 46E

The conclusion is that there is no convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.

Explanation of Solution

The given information states the sample of 13 ball bearings. It is also known that the test statistic is $t\text{ = }-\text{1}\text{.6}$ and the significance level is $\alpha =0.05$.

Here, the direction of the test is two-tailed.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =13-1\\ =12\end{array}$

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot > choose View Probability> OK.
• From Distribution, choose t.
• Enter df as 12.
• Choose X value and Both Tail for the region of the curve to shade.
• Enter the X value as –1.6.
• Click OK.

Output using the MINITAB software is given below:

From the output, the P-value of the one-tailed test is 0.06779.

The P-value of the test statistic is obtained as follows:

$\begin{array}{c}P\text{-value}=2\times P\left(t<-1.6\right)\\ =2\times 0.06779\\ =0.136\end{array}$

Decision rule:

• If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.
• If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ 0.136 is greater than the significance level 0.05.

That is, $P\text{-value}\left(=0.136\right)>\text{Significance level}\left(=0.05\right)$.

The decision is that the null hypothesis is not rejected at 0.05 significance level.

Therefore, there is no convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.

c.

To determine

State the appropriate conclusion for the given information.

Answer to Problem 46E

The conclusion is that there is no convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.

Explanation of Solution

The given information states the sample of 13 ball bearings. It is also known that the test statistic is $t\text{ = }-\text{2}\text{.6}$ and the significance level is $\alpha =0.01$.

Here, the direction of the test is two-tailed.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =25-1\\ =24\end{array}$

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot > choose View Probability> OK.
• From Distribution, choose t.
• Enter df as 24.
• Choose X value and Both Tail for the region of the curve to shade.
• Enter the X value as –2.6.
• Click OK.

Output using the MINITAB software is given below:

From the output, the P-value of the one-tailed test is 0.008.

The P-value of the test statistic is obtained as follows:

$\begin{array}{c}P\text{-value}=2\times P\left(t<-2.6\right)\\ =2\times 0.008\\ =0.016\end{array}$

Decision rule:

• If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.
• If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ 0.016 is greater than the significance level 0.01.

That is, $P\text{-value}\left(=0.016\right)>\text{Significance level}\left(=0.01\right)$.

The decision is that the null hypothesis is not rejected at 0.01 significance level.

Therefore, there is no convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.

d.

To determine

State the appropriate conclusion for the given information.

Answer to Problem 46E

The conclusion is that there is no convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.

Explanation of Solution

The given information states the sample of 13 ball bearings. It is also known that the test statistic is $t\text{ = }-\text{3}\text{.6}$.

Here, the significance level $\alpha$ is not known.

Assume that the significance level is $\alpha =0.05$.

Here, the direction of the test is two-tailed.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =25-1\\ =24\end{array}$

P-value:

Software procedure:

Step-by-step procedure to obtain the P-value using the MINITAB software:

• Choose Graph > Probability Distribution Plot > choose View Probability> OK.
• From Distribution, choose t.
• Enter df as 24.
• Choose X value and Both Tail for the region of the curve to shade.
• Enter the X value as –3.6.
• Click OK.

Output obtained using the MINITAB software is given below:

From the output, the P-value of the one-tailed test is 0.0007.

The P-value of the test statistic is obtained as follows:

$\begin{array}{c}P\text{-value}=2\times P\left(t<-3.6\right)\\ =2\times 0.0007\\ =0.0014\end{array}$

Decision rule:

• If $P\text{-value}\le \text{Significance level}$, then reject the null hypothesis $H_0$.
• If $P\text{-value}>\text{Significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ 0.0014 is less than the significance level 0.05.

That is, $P\text{-value}\left(=0.0014\right)<\text{Significance level}\left(=0.05\right)$.

The decision is that the null hypothesis is rejected at 0.05 significance level.

Therefore, there is convincing evidence that the mean diameter of ball bearings of a particular type is different from 0.5.