Chapter 9, Problem 67SE

The accompanying summary data on compression strength (lb) for 12 × 10 × 8 in. boxes appeared in the article “Compression of Single-Wall Corrugated Shipping Containers Using Fixed and Floating Test Platens” (J. Testing and Evaluation, 1992: 318–320). The authors stated that “the difference between the compression strength using fixed and floating platen method was found to be small compared to normal variation in compression strength between identical boxes.” Do you agree? Is your analysis predicated on any assumptions?

	Sample	Sample	Sample
Method	Size	Mean	SD
Fixed	10	807	27
Floating	10	757	41

To determine

Check whether the claim that the “difference between compression strength using fixed method and floating platen is smaller than the normal variation in compression strength between identical boxes” is appropriate.

Check whether the analysis is based on any assumptions.

Answer to Problem 67SE

The authors claim that “difference between compression strength using fixed method and floating platen is smaller than the normal variation in compression strength between identical boxes” is not agreed.

Yes, the analysis is based on certain assumptions.

Explanation of Solution

Given info:

Let $\overline{x}$ denotes the sample mean compression strength of fixed method and $\overline{y}$ denotes the sample mean compression strength of floating platen method.

$\overline{x}=807$, $\overline{y}=757$, $s_1=27$, $s_2=41$. The level of significance is $\alpha =0.05$. The respective sample sizes are $m=10$, and $n=10$.

Calculation:

Here, ${\mu }_1$ represents the mean compression strength of fixed method and ${\mu }_2$ represents the mean compression strength of floating platen method.

The test hypotheses are,

Null hypothesis:

$H_0:{\mu }_1-{\mu }_2=0$

That is, the mean compression strength fixed method is different from the floating platen method.

Alternative hypothesis:

$H_a:{\mu }_1-{\mu }_2\ne 0$

That is, there is evidence that the mean compression strength fixed method is different from the floating platen method.

Assumption for the two sample t-test:

• The samples X and Y are selected from the population at random.
• The samples X and Y are independent of each other.
• Samples must be distributed normally.

Here, the samples selected from the fixed method and floating method were selected at random and independently. Moreover, the sample size is large and distributed normally. Hence, the assumptions are satisfied.

Conduct the two-sample t-test to test the hypotheses.

Test statistic:

Step-by-step procedure to obtain the test statistic using the MINITAB software:

• Choose Stat > Basic Statistics > 2-Sample t.
• Choose Sample from the columns.
• In first, enter Sample size as 10, Mean as 807, Standard deviation as 27.
• In second, enter Sample size as 10, Mean as 757, Standard deviation as 41.
• 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 output, the test statistic is 3.22 and the P- value is 0.006.

Rejection rule:

If $P\text{-value}<\alpha$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

Here, the P-value is less than the level of significance.

That is,$P\text{-value}\left(=0.006\right)<\alpha \left(=0.05\right)$.

Therefore, the decision is “reject the null hypothesis”.

Thus, it can be concluded that there is evidence that the mean compression strength fixed method is different from the floating platen method.

The authors claim that “difference between compression strength using fixed method and floating platen is smaller than the normal variation in compression strength between identical boxes” is not agreed.

Here, the sample size of the compression strength data is small and it is assumed to be distributed to normal. Also, it is assumed that the data were selected at random and independent.