Question

To decide whether two different types of steel have the same true average fracture toughness values, n specimens of each type are tested, yielding the following results:

\(\begin{array}{l}\underline {\begin{array}{*{20}{c}}{ Type }&{ Sample Average }&{ Sample SD }\\1&{60.1}&{1.0}\\2&{59.9}&{1.0}\\{}&{}&{}\end{array}} \\\end{array}\)

Calculate the P-value for the appropriate two-sample \(z\) test, assuming that the data was based on \(n = 100\). Then repeat the calculation for \(n = 400\). Is the small P-value for \(n = 400\) indicative of a difference that has practical significance? Would you have been satisfied with just a report of the P-value? Comment briefly.a

Short answer

the solution is

The slight difference in the genuine average has been amplified statistically due to the huge sample size \(n = 400\). In practice, the little difference in sample average values would lead one to believe that there is no difference between them. The P- value would be insufficient - it would be unsatisfactory

Step-by-step solution

calculate p value

when the null hypothesis is

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

the test statistic value

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

The \(P\) value is derived for the acceptable alternative hypothesis by calculating the adequate area under the standard normal curve. When both \(m > 40 and n > 40\), this test can be employed.

The sample size is the same and exceeds \(40(100,400 > 40)\). When \(n = 100\), the value of the test statistic is.

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{60.1 - 59.9}}{{\sqrt {\frac{1}{{100}} + \frac{1}{{100}}} }} = 1.41.\)

calculated phi

the alternative hypothesis is two sided \(\left( {{H_a}:{\mu _1} - {\mu _2} \ne 0} \right)\) the p value

\(\begin{array}{l}2.P(Z > 1.41) = 2 \times (1 - P(Z \le 1.41) = 2.(1 - \Phi (1.41))\\ = 0.1586\end{array}\)

where \(\Phi (1.41)\) was calculated using the appendix table.

When \(n = 400\)

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{60.1 - 59.9}}{{\sqrt {\frac{1}{{400}} + \frac{1}{{400}}} }} = 2.83.\)

The alternative hypothesis is two sided \(\left( {{H_a}:{\mu _1} - {\mu _2} \ne 0} \right)\), the \(P\) value is

\(\begin{array}{l}2.P(Z > 2.83) = 2 \times (1 - P(Z \le 2.83) = 2 \times (1 - \Phi (2.83))\\ = 0.0046\end{array}\)

where \(\Phi (2.83)\) was calculated using the appendix table.

The slight difference in the genuine average has been amplified statistically due to the huge sample size \(n = 400\). In practice, the little difference in sample average values would lead one to believe that there is no difference between them. The P- value would be insufficient - it would be unsatisfactory.

conclusion

The slight difference in the genuine average has been amplified statistically due to the huge sample size\(n = 400\). In practice, the little difference in sample average values would lead one to believe that there is no difference between them. The P- value would be insufficient - it would be unsatisfactory