91影视

given

\(\begin{array}{*{20}{l}}{\bar x = 18.12kgf/c{m^2};}&{m = 40}\\{\bar y = 16.87kgf/c{m^2};}&{n = 32}\end{array}\)

Assume that the bond strength distributions are normal.

a)

the null hypothesis is

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

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 is used to determine the sufficient area under the standard normal curve for the adequate alternative hypothesis.

The standard deviations' values are known.

\(\begin{array}{l}{\sigma _1} = 1.6\\{\sigma _2} = 1.4\end{array}\)

The \({\Delta _0} = 0\) and the alternative hypothesis is \({H_a}:{\mu _1} - {\mu _2} > 0\). The value of the test statistic is

\(z = \frac{{18.12 - 16.87}}{{\sqrt {\frac{{1.{6^2}}}{{40}} + \frac{{1.{4^2}}}{{32}}} }} = 3.53\)

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

\(P = P(Z > 3.53) = 1 - P(Z \le 3.53) = 1 - 0.9998 = 0.0002\)

Since

\(P = 0.0002 < 0.01 = \alpha \)

The null hypothesis should be reject.

depending on alternative hypothesis the type II error\(\beta \)when \({\mu _1} - {\mu _2} = \Delta '\)

differs. The alternative hypothesis is \({H_a}:{\mu _1} - {\mu _2} > 0\) and \(\Delta ' = 1\), in this case the type II error is

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

Where

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

The significance level is \(\alpha = 0.01,{z_\alpha } = {z_{0.01}} = 2.33, \)

\(\sigma = \sqrt {\frac{{1.{6^2}}}{{40}} + \frac{{1.{4^2}}}{{32}}} = 0.3539.\)

Finally, the type II error when \({\mu _1} - {\mu _2} = 1\) is

\(\beta (1) = \Phi \left( {2.33 - \frac{{1 - 0}}{{0.3539}}} \right) = \Phi ( - 0.4) = 0.3085\)

similarly as in (b), from

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

For \(\Delta ' = 1,\beta (1) = 0.1{z_\alpha } = {z_{0.05}} = 1.645.\)using the fact

\(\Phi (1.28) = 0.1\)

Which was obtained from the table in the appendix in the book, the following holds

\(\begin{array}{l}\beta \left( {\Delta '} \right) = \Phi \left( {{z_\alpha } - \frac{{\Delta ' - {\Delta _0}}}{\sigma }} \right)\\0.1 = \Phi \left( {{z_\alpha } - \frac{{\Delta ' - {\Delta _0}}}{\sigma }} \right) and 0.1 = \Phi (1.28)\\\Phi (1.28) = \Phi \left( {{z_\alpha } - \frac{{\Delta ' - {\Delta _0}}}{\sigma }} \right)\end{array}\)

Because \(\Phi \)is cdf of standard normal distribution, the following is true

\(\begin{array}{l}{z_\alpha } - \frac{{\Delta ' - {\Delta _0}}}{\sigma } = 1.28\\2.33 - \frac{{1 - 0}}{\sigma } = 1.28\\{\sigma ^2} = \frac{1}{{{{(1.645 + 1.28)}^2}}}\end{array}\)

Remember that

\(\begin{array}{l}\sigma = \sqrt {\frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n}} \\{\sigma ^2} = \frac{{\sigma _1^2}}{m} + \frac{{\sigma _2^2}}{n} = \frac{{1.{6^2}}}{{40}} + \frac{{1.{4^2}}}{n}\end{array}\)

by substituting this, the following is true

\(\begin{array}{l}\frac{{1.{6^2}}}{{40}} + \frac{{1.{4^2}}}{n} = \frac{1}{{{{(1.645 + 1.28)}^2}}}\\\frac{{1.{6^2}}}{{40}} + \frac{{1.{4^2}}}{n} = 0.1169\\\frac{{1.96}}{n} = 0.0529\end{array}\)

\(\begin{array}{l}n = 37.06\\{\rm{n = 38}}\end{array}\)

d)

The sample size \(n = 32\) is insufficient for the large sample test to be used. This question is a "bait" to get you to say:

When there is a null hypothesis.

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

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 \(m > 40 and n > 40\) are true, this test can be applied.

However, because \(n\) is insufficient, the \(t\) test from section \(9.2\) should be applied.

\(\begin{array}{l}b. \beta (1) = 0.3085;\\c. n = 38; \end{array}\)

d. could not use the same test