Question

An experiment was performed to compare the fracture toughness of high-purity \(18Ni\) maraging steel with commercial-purity steel of the same type (Corrosion Science, 1971: 723–736). For \(m = 32\)specimens, the sample average toughness was \(\overline x = 65.6\) for the high purity steel, whereas for \(n = 38\)specimens of commercial steel \(\overline y = 59.8\). Because the high-purity steel is more expensive, its use for a certain application can be justified only if its fracture toughness exceeds that of commercial purity steel by more than 5. Suppose that both toughness distributions are normal.

a. Assuming that \({\sigma _1} = 1.2\) and \({\sigma _2} = 1.1\), test the relevant hypotheses using \(\alpha = .001\).

b. Compute \(\beta \) for the test conducted in part (a) when \({\mu _1} - {\mu _2} = 6.\)

Short answer

the solution is

a)do not reject null hypothesis

b)\(\beta (6) = 0.095\)

Step-by-step solution

find the value of the test statistic

\(\begin{array}{*{20}{l}}{\bar x = 65.6;}&{m = 32}\\{\bar y = 59.8;}&{n = 38.}\end{array}\)

Both toughness distribution are normal.

a)

\({\sigma _1} = 1.2\)for the first sample and \({\sigma _2} = 1.1. \)for the second sample.

When there is a null hypothesis.

\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)

The test statistic value is based on the assumption of two normal population distributions with known \({\sigma _1}\) and \({\sigma _2}\) values, as well as two independent random samples.

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}} }}\)

The P value for the acceptable alternative hypothesis is derived as the sufficient area under the standard normal curve.

The hypothesis of interest are \({H_0}:{\mu _1} - {\mu _2} = 5\) versus \({H_a}:{\mu _1} - {\mu _2} > 5\)

The value of the test statistic is

\(z = \frac{{65.6 - 59.8 - 5}}{{\sqrt {1.{2^2}/32 + 1.{1^2}/38} }} = 2.89\)

The area under the standard normal curve to the right of \(z\) can be used to get the \(P\) value. The \(P\) value can be calculated using the table in the appendix or software.

\(P = P(Z > 2.89) = 1 - P(Z \le 2.89) = 1 - 0.9981 = 0.0019.\)

Since

\(P = 0.0019 > 0.001 = \alpha \)

The null hypothesis should not be reject

At level \(0.001\).

Compute \(\beta \)

b)

The type II error \(\beta \) for \({\mu _1} - {\mu _2} = {\Delta ^'}\) vary depending on the alternative hypothesis. \({H_a}:{\mu _1} - {\mu _2} > 5\) and \({\Delta ^'} = 1\) is the alternative hypothesis, in which case the type II error is.

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

Where

The type II error \(\beta \) when \({\mu _1} - {\mu _2} = 6\) is

\(\begin{array}{l}\beta (6) = \Phi \left( {3.09 - \frac{{6 - 5}}{{\sqrt {1.{2^2}/32 + 1.{1^2}/38} }}} \right)\\ = \Phi \left( {3.09 - \frac{1}{{0.2272}}} \right)\\ = \Phi ( - 1.311) = 0.095.\end{array}\)

The table in the appendix was used to determine values of \(\Phi \) and \({z_\alpha } = {z_{0.001}} = 3.09.\)

conclusion

a)do not reject null hypothesis

b)\(\beta (6) = 0.095\)