Chapter 9, Problem 72SE

The following summary data on bending strength (lb-in/in) of joints is taken from the article “Bending Strength of Corner Joints Constructed with Injection Molded Splines” (Forest Products J., April, 1997: 89–92).

	Sample	Sample	Sample
Type	Size	Mean	SD
Without side coating	10	80.95	9.59
With side coating	10	63.23	5.96

a. Calculate a 95% lower confidence bound for true average strength of joints with a side coating.

b. Calculate a 95% lower prediction bound for the strength of a single joint with a side coating. c. Calculate an interval that, with 95% confidence, includes the strength values for at least 95% of the population of all joints with side coatings.

d. Calculate a 95% confidence interval for the difference between true average strengths for the two types of joints.

To determine

Calculate the 95% lower confidence bound for the true average strength of the joints with a side coating.

Answer to Problem 72SE

The 95% lower confidence bound for the true average bending strength of the joints with a side coatingis at least59.78(lb-in/in).

Explanation of Solution

Given info:

The information is based on the bending strength (lb-in/in) of the joints:

Let $\overline{x}$ denotes the sample mean bending strength of the joints with a side coating

$\overline{x}=63.23$,$s=5.96$ and $n=10$.

Calculation:

Lower Confidence interval:

The Lower Confidence level can be defined is,

$\overline{x}-t_{\frac{\alpha }{2}.n-1}\frac{s}{\sqrt{n}}$ ,

Where $\overline{x}$ is the sample mean,$s$ is the standard deviation and n be the sample size.

Lower Confidence interval:

Step-by-step procedure to obtain the confidence interval using the MINITAB software:

• Choose Stat > Basic Statistics > One sample t.
• Choose Summarized columns. Enter the sample mean as 63.23 and standard deviation as 5.96.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select Greater than.
• Click OK in all the dialog boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the lower confidence bound is 59.78.

Interpretation:

Thus, The 95% lower confidence bound for the true average bending strength of the joints with a side coating is at least 59.78(lb-in/in).

To determine

Calculate the 95% lower prediction bound for the true average strength of the single joint with a side coating.

Answer to Problem 72SE

The 95% lower prediction bound for the true average bending strength of the single joint with a side coating will be at least 51.77(lb-in/in).

Explanation of Solution

Given info:

Let $\overline{x}$ denotes the sample mean bending strength of the joints with a side coating

$\overline{x}=63.23$, $s=5.96$ and $n=10$.

Calculation:

Lower Prediction interval:

$\overline{x}-t_{\left(\frac{\alpha }{2},n-1\right)}\left[s\left(\sqrt{1+\frac{1}{n}}\right)\right]$,

Where $\overline{x}$ is the sample mean,$s$ is the standard deviation of and n be the sample size.

Degrees of freedom:

The degrees of freedom is $df=n-1$, for sample size n.

Thus, for sample of size 10,

$\begin{array}{c}n-1=10-1\\ =9\end{array}$

Hence, the degrees of freedom are 9.

Level of significance:

It is given that the level of significance is 0.05.

The lower prediction bound for the true average bending strength of the single joint with a side coating is obtained as given below:

$\overline{x}-t_{\left(\frac{\alpha }{2},n-1\right)}\left[s_D\left(\sqrt{1+\frac{1}{n}}\right)\right]=63.23-t_{\left(0.025,9\right)}\left[5.96\left(\sqrt{1+\frac{1}{10}}\right)\right]$

From Appendix Tables, “Table A.5 Critical Values for t Distribution”, the critical value for the 9df  with level of significance 0.025 is $t_{0.05,9}=1.833$.

The lower prediction bound is given by;

$\begin{array}{c}\overline{x}-t_{\left(\frac{\alpha }{2},n-1\right)}\left[s_D\left(\sqrt{1+\frac{1}{n}}\right)\right]=63.23-t_{\left(0.025,9\right)}\left[5.96\left(\sqrt{1+\frac{1}{10}}\right)\right]\\ =63.23-\left(1.833\right)\left[5.96\left(\sqrt{1+0.1}\right)\right]\\ =63.23-1.833\left[5.96\left(\sqrt{1.1}\right)\right]\\ =63.23-1.833\left(6.251\right)\end{array}$

$\begin{array}{c}=63.23-11.46\\ =51.77\end{array}$

Thus, the 95% lower prediction bound is 51.77.

Thus, the 95% lower prediction bound for the true average bending strength of the single joint with a side coating will be at least 51.77(lb-in/in).

To determine

Calculate the 95% confidence interval that includes the strength values for at least 95% of the population of all joints with side coatings.

Answer to Problem 72SE

The 95% confidence interval that includes the strength values for at least 95% of the population of all joints with side coatings lies between 43.09(lb-in/in) and 83.37(lb-in/in).

Explanation of Solution

Given info:

Let $\overline{x}$ denotes the sample mean bending strength of the joints with a side coating

$\overline{x}=63.23$, $s=5.96$ and $n=10$.

Calculation:

To find thestrength values for at least 95%, use the method of tolerance interval.

Tolerance interval:

$\overline{x}\pm \left(\begin{array}{l}\text{Tolerance critical value}\\ \text{for 100}\left(\alpha -1\right)\text{with}n\end{array}\right)s$

Where $\overline{x}$ is the sample mean, $s$ is the standard deviation and n be the sample size.

Degrees of freedom:

The degrees of freedom is $df=n-1$, for sample size n.

Thus, for sample of size 10,

$\begin{array}{c}n-1=10-1\\ =9\end{array}$

Hence, the degrees of freedom are 9.

Level of significance:

It is given that the level of significance is 0.05.

The tolerance intervalthat includes the strength values for at least 95% of the population of all joints with side coatings is obtained as given below:

$\overline{x}\pm \left(\begin{array}{l}\text{Tolerance critical value}\\ \text{for }95\%\text{with}n=10\end{array}\right)s=63.23\pm \left(\begin{array}{l}\text{Tolerance critical value}\\ \text{for }95\%\text{with}n=10\end{array}\right)5.96$

From Appendix Tables, “Table A.6 Tolerance Critical Values for Normal population Distribution”, the tolerance critical value for the n = 10 with confidence level 95% for 95% population is $3.379$.

The interval is given by;

$\begin{array}{c}\overline{x}\pm \left(\begin{array}{l}\text{Tolerance critical value}\\ \text{for }95\%\text{with}n=10\end{array}\right)s=63.23\pm \left(\begin{array}{l}\text{Tolerance critical value}\\ \text{for }95\%\text{with}n=10\end{array}\right)5.96\\ =63.23\pm 3.379\left(5.96\right)\\ =63.23\pm 20.14\\ =\left(43.09,83.37\right)\end{array}$

Thus, the 95% tolerance interval is (43.09, 83.37).

Thus, the 95% tolerance interval that includes the strength values for at least 95% of the population of all joints with side coatings lies between 43.09(lb-in/in) and 83.37(lb-in/in).

Interpretation:

Hence it is highly confident that at least 95% of the population of all joints with side coatings have the bend strength between 43.09(lb-in/in) and 83.37(lb-in/in).

To determine

Find the 95% confidence interval for the difference between two average strengths for the two types of joints.

Answer to Problem 72SE

The interpretation is, there is 95% confidence that the average strength for the joints without side coating is greater than that of strength for the joints with side coating by between10.11(lb-in/in) and 25.33(lb-in/in).

Explanation of Solution

Given info:

Let $\overline{x}$ denotes the sample mean bending strength of the joints without a side coating and $\overline{y}$ denotes the sample mean bending strength of the joints with a side coating.

$\overline{x}=80.95$,$\overline{y}=63.23$, $s_1=9.59$, $s_2=5.96$. The respective sample sizes are $m=10$, and $n=10$.

Calculation:

Let ${\mu }_1$ represent the average of bending strength of the joints without a side coating and ${\mu }_2$ represents the average of bending strength of the joints with a side coating.

Requirements for a two sample t-test:

• The sample X and Y taken from the population is selected at random.
• The samples X and Y are dependent of each other.
• Samples must be distributed to normal.

Here, the samples selected from the strength without coating and with coating were selected at random and dependent. Moreover, the sample size is assumed to be normally distributed. Hence, the assumptions are satisfied.

Thus, the condition is satisfied.

Confidence interval:

The confidence interval is given as,

$\left(\overline{x}-\overline{y}\right)\pm t_{\frac{\alpha }{2},n-1}\sqrt{\frac{s_x}{n_1}+\frac{s_y}{n_2}}$ ,

Where $\overline{x}$ and $\overline{y}$ are the sample means of first and second sample respectively,$s_x$ and $s_y$ are the sample standard deviation of first and second sample respectively, and $n_1$ and $n_2$  are the sample size of first and second sample respectively.

Confidence interval:

Step-by-step procedure to obtain the confidence interval using the MINITAB software:

• Choose Stat > Basic Statistics > 2-sample t test.
• Choose Summarized data.
• In first, enter Sample size as 10, Mean as 80.95, Standard deviation as 9.59.
• In second, enter Sample size as 10, Mean as 63.23, Standard deviation as 5.96.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select Not equal.
• Click OK in all the dialog boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the confidence interval is (10.11, 25.33).

Thus, there is 95% confidence that the average strength for the joints without side coating is greater than that of strength for the joints with side coating by between 10.11(lb-in/in).and 25.33(lb-in/in).