Chapter 9.1, Problem 6E

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 $\bar{x}$= 18.12 kgf/cm² for the modified mortar (m = 40) and y $\bar{y}$= 16.87 kgf/cm² for the unmodified mortar (n = 32). Let µ₁ and µ₂ 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 σ₁ = 1.6 and σ₂ = 1.4, test H₀: µ₁ − µ₂ = 0 versus H_a: µ₁ − µ₂ > 0 at level .01.

b. Compute the probability of a type II error for the test of part (a) when µ₁ − µ₂ = 1.

c. Suppose the investigator decided to use a level .05 test and wished b 5 .10 when µ₁ − µ₂ = 1. If m = 40, what value of n is necessary?

d. How would the analysis and conclusion of part (a) change if σ₁ and σ₂ were unknown but s₁ = 1.6 and s₂ = 1.4?

To determine

Test the hypotheses using $\alpha =0.01$

Answer to Problem 6E

The conclusion is that, there is enough evidence to infer that the true average tension bond strength for modified mortar is higher than that of true average tension bond strength for unmodified mortar.

Explanation of Solution

Given info:

${\overline{x}}_1=18.12$, $\overline{y}=16.87$, ${\sigma }_1=1.6$, ${\sigma }_2=1.4$, $\alpha =0.01$, $m=40$, and $n=32$.

Calculation:

Here, the aim of the hypothesis test is to obtain the true average tension bond strength for modified mortar is higher than that of true average tension bond strength for unmodified mortar and it is appropriate to use right tailed test.

Here, ${\mu }_1$ represents the mean tension bond strength for modified mortar and ${\mu }_2$ represents the mean tension bond strength for unmodified mortar.

The test hypotheses are,

Null hypothesis:

$H_0:{\mu }_1-{\mu }_2=0$

That is, the true average tension bond strength for modified mortar is same as the true average tension bond strength for unmodified mortar.

Alternative hypothesis:

$H_a:{\mu }_1-{\mu }_2>0$

That is, the true average tension bond strength for modified mortar is higher than that of true average tension bond strength for unmodified mortar.

Test statistic:

Here, the sample size is assumed to be large. That is, $m>40$ and $n>40$. Hence, the test statistic for the large samples is obtained as:

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

The difference between the average $\Delta =0$

The test statistic value is obtained below:

$\begin{array}{c}z=\frac{\left(18.12-16.87\right)-0}{\sqrt{\frac{{1.6}^2}{40}+\frac{{1.4}^2}{32}}}\\ =\frac{1.25}{\sqrt{\frac{2.56}{40}+\frac{1.96}{32}}}\\ =\frac{1.25}{0.3539}\\ =3.53\end{array}$

Thus, the test statistic value is 3.53.

The corresponding P-value for the left-tailed test is obtained as:

$P\left(\left|z\right|\ge 3.53\right)=\left[1-\Phi \left(3.53\right)\right]$

From the Appendix “Table A.3”, the critical value $\left(z_{0.01}\right)$ at level of significance $\alpha =0.01$ is $\underset{\_}{0.9998}$.

$\begin{array}{c}P\left(\left|z\right|\ge 3.53\right)=\left[1-\Phi \left(3.53\right)\right]\\ =1-0.9998\\ =0.0002\end{array}$

Thus, the P-value is 0.0002.

Decision rule:

Rejection region based on P-value approach:

If $P\text{-value}>\alpha$, then fail to reject the null hypothesis $\left(H_0\right)$.

If $P\text{-value}<\alpha$, then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

The level of significance is $\alpha =0.01$ and the P- value is $\underset{\_}{0.0002}$.

Here, the P-value is less than the level of significance.

That is,$P\text{-value}\left(=0.0002\right)<\alpha \left(=0.01\right)$.

Thus, the decision is “reject the null hypothesis”.

Thus, there is enough evidence to infer that the true average tension bond strength for modified mortar is higher than that of true average tension bond strength for unmodified mortar.

To determine

Calculate the value of $\beta$ when ${\mu }_1-{\mu }_2=1$.

Answer to Problem 6E

The value of $\beta$ is 0.3085.

Explanation of Solution

Calculation:

The type-II error $\beta$ is defined as the probability of accepting the null hypothesis when it is true. That is,

$\beta =P\left(\text{not rejecting }{H}_0|H_1\text{ is true}\right)$

When $H_a:{\mu }_1-{\mu }_2>{\Delta }_0$,

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

Where $\sigma =\sqrt{\frac{{\sigma }_1^2}{m}+\frac{{\sigma }_2^2}{n}}$

The value of $\beta$ is obtained as shown below:

From Appendix “Table A.3-Standard Normal Curves Areas”, the $z_{0.01}$ is 2.33.

$\begin{array}{c}\beta =\Phi \left(z_{0.01}-\frac{1-0}{\sqrt{\frac{{1.6}^2}{40}+\frac{{1.4}^2}{32}}}\right)\\ =\Phi \left(2.33-\frac{1}{0.3539}\right)\\ =\Phi \left(2.33-2.83\right)\\ =\Phi \left(-0.50\right)\end{array}$

From Appendix “Table A.3-Standard Normal Curves Areas”, the standard normal value of –0.50  is 0.3085.

Thus, the value of $\beta$ when ${\mu }_1-{\mu }_2=1$ is 0.3085.

To determine

Find the value of n.

Answer to Problem 6E

The value of n is 37.

Explanation of Solution

Calculation:

The value of n is obtained as shown below:

$\frac{{\sigma }_1^2}{m}+\frac{{\sigma }_2^2}{n}=\frac{1}{{\left(Z_{\alpha }+Z_{\beta }\right)}^2}$

Substitute ${\sigma }_1=1.6,{\sigma }_2=1.4,m=40,\alpha =0.05\text{and}\beta =0.1$

$\frac{{1.6}^2}{40}+\frac{{1.4}^2}{n}=\frac{1}{{\left(Z_{0.05}+Z_{0.10}\right)}^2}$

From Appendix “Table A.3-Standard Normal Curves Areas”, the $z_{0.05}$ is 1.645 and the $z_{0.1}$ is 1.28.

The value of n is given below:

$\begin{array}{c}\frac{2.56}{40}+\frac{1.96}{n}=\frac{1}{{\left(1.645+1.28\right)}^2}\\ 0.064+\frac{1.96}{n}=0.1169\\ \frac{1.96}{n}=0.1169-0.064\\ \frac{1.96}{n}=0.0529\end{array}$

$\begin{array}{c}1.96=0.0529n\\ n=\frac{1.96}{0.0529}\\ n=37.05\end{array}$

Thus, the value of n is round as 37.

To determine

Explain the change happen in the part (a), when ${\sigma }_1\text{and }{\sigma }_2$ is unknown.

Explanation of Solution

Justification:

The sample size n is not a large sample and it is not appropriate to use the large sample z test. When the ${\sigma }_1\text{and }{\sigma }_2$ is unknown, small sample t procedure will be more appropriate. Moreover, the value of test statistic does not change for both procedure. Thus, null hypothesis $H_0$ will be rejected at $\alpha =0.01$ for both procedure.