Chapter 9.4, Problem 23SE

a.

To determine

Write the two-tailed decision rule to detect quality control violations using the 5 percent level of significance.

Answer to Problem 23SE

Two-tailed decision rule to detect quality control violations using the 5 percent level of significance is “reject the null hypothesis if $z_{\text{calc}}>+1.96$ or $z_{\text{calc}}<-1.96$”.

Explanation of Solution

Calculation:

The given information is that, the filling process follows a normal distribution with a mean of 10 ounces and a standard deviation of 0.07 ounce. That is, $\mu =10$ and $\sigma =0.07$. The data represents the weights (ounces) of 12 bottles.

The test hypotheses are given below:

Null hypothesis:

$H_0:\mu =10\text{ounces}$

That is, the bottles of Galena Spring Water contain an average of 10 ounces.

Alternative hypothesis:

$H_1:\mu \ne 10\text{ounces}$

That is, the bottles of Galena Spring Water do not contain an average of 10 ounces.

Critical value:

For two tailed test,

$\begin{array}{c}1-\frac{\alpha }{2}=1-\frac{0.05}{2}\\ =1-0.025\\ =0.975\end{array}$

Critical value:

Software procedure:

Step-by-step software procedure to obtain critical value using EXCEL is as follows:

• Open an EXCEL file.
• In cell A1, enter the formula “=NORM.S.INV(0.975)”
• Output using Excel software is given below:

From the output, the critical value of z is $z_{\frac{0.05}{2}}=\pm 1.96$.

Decision rule for two-tailed test at $\alpha =0.05$:

If $z_{\text{calc}}>+1.96$, then reject the null hypothesis.

If $z_{\text{calc}}<-1.96$, then reject the null hypothesis.

b.

To determine

Find the test statistic and make the decision.

Answer to Problem 23SE

The test statistic is 0.78.

The decision is “null hypothesis is not rejected”.

Explanation of Solution

Calculation:

Mean:

$\begin{array}{c}\overline{x}=\frac{{\displaystyle \sum x_i}}{n}\\ =\frac{\left(\begin{array}{l}10.02+9.95+10.11+10.10+10.08+10.04+\\ 10.06+10.03+9.98+10.01+9.92+9.89\end{array}\right)}{12}\\ =\frac{120.19}{12}\\ =10.0158\end{array}$

Standard deviation:

The formula for standard deviation is,

$\sigma =\sqrt{\frac{{{\displaystyle \sum \left(x_i-\overline{x}\right)}}^2}{N}}$

The value of ${{\displaystyle \sum \left(x_i-\overline{x}\right)}}^2$ is calculated as follows:

$x$	${\left(x_i-\overline{x}\right)}^2$
10.02	0.00001764
9.95	0.00432964
10.11	0.00887364
10.1	0.00708964
10.08	0.00412164
10.04	0.00058564
10.06	0.00195364
10.03	0.00020164
9.98	0.00128164
10.01	0.00003364
9.92	0.00917764
9.89	0.01582564
	${{\displaystyle \sum \left(x_i-\overline{x}\right)}}^2=0.05349$

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{\frac{0.05349}{12}}\\ =\sqrt{0.0044575}\\ =0.07\end{array}$

The formula for test statistic is,

$z_{\text{calc}}=\frac{\overline{x}-{\mu }_0}{\frac{\sigma }{\sqrt{n}}}$

Where $\overline{x}$ is the sample mean, ${\mu }_0$ is the population mean, $\sigma$ is the population standard deviation and n is the sample size.

Substitute $\overline{x}=10.0158$, ${\mu }_0=10$, $\sigma =0.07$ and $n=12$ in the test statistic formula.

$\begin{array}{c}{z}_{\text{calc}}=\frac{10.0158-10}{\frac{0.07}{\sqrt{12}}}\\ =\frac{0.0158}{\left(\frac{0.07}{3.4641}\right)}\\ =\frac{0.0158}{0.0202}\\ =0.78\end{array}$

Thus, the test statistic is 0.78.

Conclusion:

Here, the test statistic is less than the critical value. That is, $z_{\text{calc}}\left(=0.78\right)<+1.96$.

Therefore, the null hypothesis is not rejected.

c.

To determine

Describe the assumptions for given hypotheses tests.

Explanation of Solution

The assumption necessary to validate the given test is the population is normally distributed.