Question

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar x = 18.12kgf/c{m^2}\)for the modified mortar \((m = 40)\) and \(\bar y = 16.87kgf/c{m^2}\) for the unmodified mortar \((n = 32)\). Let \({\mu _1}\) and \({\mu _2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal.

a. Assuming that \({\sigma _1} = 1.6\) and \({\sigma _2} = 1.4\), test \({H_0}:{\mu _1} - {\mu _2} = 0\) versus \({H_a}:{\mu _1} - {\mu _2} > 0\) at level . \(01.\)

b. Compute the probability of a type II error for the test of part (a) when \({\mu _1} - {\mu _2} = 1\).

c. Suppose the investigator decided to use a level \(.05\) test and wished \(\beta = .10\) when \({\mu _1} - {\mu _2} = 1. \). If \(m = 40\), what value of n is necessary?

d. How would the analysis and conclusion of part (a) change if \({\sigma _1}\) and \({\sigma _2}\) were unknown but \({s_1} = 1.6\)and \({s_7} = 1.4?\)

Short answer

the solution is

a.Reject null hypothesis

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

d. could not use the same test

Step-by-step solution

calculate p value

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.

compute probability of type II error

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

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

use the table in the appendix

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

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

conclusion of part (a)

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.

conclusion

1. Reject null hypothesis

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

d. could not use the same test