Question

A mechanical engineer wishes to compare strength properties of steel beams with similar beams made with a particular alloy. The same number of beams, \(n\), of each type will be tested. Each beam will be set in a horizontal position with a support on each end, a force of \(2500lb\) will be applied at the center, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for both types of beam is \(.05in\). Because the alloy is more expensive, the engineer wishes to test at level \(.01\) whether it has smaller average deflection than the steel beam. What value of n is appropriate if the desired type II error probability is \(.05\) when the difference in true average deflection favors the alloy by \(.04in\).?

Short answer

the solution is

\(n = 50\)

Step-by-step solution

standard normal distribution

The null hypothesis is \({H_0}:{\mu _1} - {\mu _2} = 0\), \({\mu _2}\) with representing the true average of alloy and \({\mu _1}\)being the true average of steel beam. The type II error \(\beta \) for \({\mu _1} - {\mu _2} = {\Delta ^'}\) vary depending on the alternative hypothesis. The alternative hypothesis is \({H_a}:{\mu _1} - {\mu _2} > 0\) and \({\Delta ^'} = 0.4, \), which indicates that the type II error is present

\(\beta \left( {{\Delta ^'}} \right) = \Phi \left( {{z_\alpha } - \frac{{{\Delta ^'} - {\Delta _0}}}{\sigma }} \right)\)

Where

\(\sigma = \sqrt {\frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}} \)

\({\sigma _1} = {\sigma _2} = 0.05\)and \(m = n\). The significance level is \(\alpha = 0.01,{z_\alpha } = {z_{0.01}} = 2.33\). Using the fact

\(\Phi (1.645) = 0.05\)

The following holds true, as determined by the table in the book's appendix.

\(\begin{array}{l}\beta \left( {{\Delta ^'}} \right) = \Phi \left( {{z_\alpha } - \frac{{{\Delta ^'} - {\Delta _0}}}{\sigma }} \right)\\0.05 = \Phi \left( {{z_\alpha } - \frac{{{\Delta ^'} - {\Delta _0}}}{\sigma }} \right) and 0.05 = \Phi (1.645)\\\Phi (1.645) = \Phi \left( {{z_\alpha } - \frac{{{\Delta ^'} - {\Delta _0}}}{\sigma }} \right)\end{array}\)

The following is true because \(\Phi \) is a cdf of standard normal distribution.

\(\begin{array}{l}{z_\alpha } - \frac{{{\Delta ^'} - {\Delta _0}}}{\sigma } = 1.645\\2.33 - \frac{{0.04 - 0}}{\sigma } = 1.28\\{\sigma ^2} = \frac{{0.0{4^2}}}{{{{(1.645 + 2.33)}^2}}}\end{array}\)

find n value

remember that

\(\sigma = \sqrt {\frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}} ,\)

and

\({\sigma ^2} = \frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n} = \frac{{0.0{5^2}}}{n} + \frac{{0.0{5^2}}}{n} = \frac{{0.0025 + 0.0025}}{n}\)

by substituting this, the following is true

\(\begin{array}{l}\frac{{0.0025 + 0.0025}}{n} = \frac{{0.0016}}{{{{(1.645 + 2.33)}^2}}}\\n = \frac{{(0.0025 + 0.0025) \times {{(2.33 + 1.645)}^2}}}{{0.0016}}\\n = 49.38\end{array}\)

However, because \(n\) must be an integer number (sample size), the required value of \(n\) is

\(n = 50\)

conclusion

\(n = 50\)