Chapter 9, Problem 109CE

a.

To determine

Check whether the mean brightness exceeds the specification or not.

Answer to Problem 109CE

There is evidence to infer that the mean brightness exceeds the specification.

Explanation of Solution

Calculation:

The given information is that, the data represents the sample of 24 test sheets from a day’s production run.

State the hypotheses:

Null hypothesis:

$H_0:\mu \le 106$

That is, the mean brightness is not greater than 106.

Alternative hypothesis:

$H_1:\mu >106$

That is, the mean brightness is greater than 106.

Critical value:

For right tailed test,

$\begin{array}{c}1-\alpha =1-0.005\\ =0.995\end{array}$

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =24-1\\ =23\end{array}$

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 “=T.INV(0.995,23)”
• Output using Excel software is given below:
• 

From the output, the critical value is 2.807.

Decision rule for right-tailed test at $\alpha =0.005$:

If $t_{\text{calc}}>+2.807$, then reject the null hypothesis.

• Sample mean and variance:
• The formula for mean is,
• $\overline{x}=\frac{{\displaystyle \sum x_i}}{n}$
• The formula for standard deviation is,
• $s=\sqrt{\frac{{\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}}{n-1}}$
• The value of ${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}$ and mean is calculated as follows:

Brightness	$\left(x_i-\overline{x}\right)$	${\left(x_i-\overline{x}\right)}^2$
106.98	–0.01670	0.000279
107.02	0.02330	0.000543
106.99	–0.00670	0.0000449
106.98	–0.01670	0.000279
107.06	0.06330	0.004007
107.05	0.05330	0.002841
107.03	0.03330	0.001109
107.04	0.04330	0.001875
107.01	0.01330	0.000177
107.00	0.00330	0.0000109
107.02	0.02330	0.000543
107.04	0.04330	0.001875
107.00	0.00330	0.0000109
106.98	–0.01670	0.000279
106.91	–0.08670	0.007517
106.93	–0.06670	0.004449
107.01	0.01330	0.000177
106.98	–0.01670	0.000279
106.97	–0.02670	0.000713
106.99	–0.00670	0.0000449
106.94	–0.05670	0.003215
106.98	–0.01670	0.000279
107.03	0.03330	0.001109
106.98	–0.01670	0.000279
${\displaystyle \sum x_i}=2,567.92$		${\displaystyle \sum {\left(x_i-\overline{x}\right)}^2}=0.031933$
$\overline{x}=106.9967$

• The standard deviation is,
• $\begin{array}{c}s=\sqrt{\frac{0.031933}{24-1}}\\ =\sqrt{0.001388}\\ =0.0373\end{array}$
• Thus, the standard deviation is 0.0373.
• Test statistic:

The formula for test statistic is,

$t_{\text{calc}}=\frac{\overline{x}-{\mu }_0}{\frac{s}{\sqrt{n}}}$

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

Substitute $\overline{x}=106.9967$, ${\mu }_0=106$, $s=0.0373$ and $n=24$ in the test statistic formula.

$\begin{array}{c}{t}_{\text{calc}}=\frac{106.9967-106}{\frac{0.0373}{\sqrt{24}}}\\ =\frac{0.9967}{\left(\frac{0.0373}{4.8990}\right)}\\ =\frac{0.9967}{0.007614}\\ =130.9\end{array}$

Thus, the test statistic is 130.9.

Conclusion for critical value method:

Here, the test statistic is greater than the critical value.

That is, $t_{\text{calc}}\left(=130.9\right)>+2.807$

Therefore, the null hypothesis is rejected.

Hence, there is evidence to infer that the mean brightness exceeds the specification.

b.

To determine

Check whether the sample shows that ${\sigma }^2<0.0025$.

Answer to Problem 109CE

The sample does not show that ${\sigma }^2<0.0025$.

Explanation of Solution

Calculation:

Here, the hypotheses test is left-tailed test.

The test hypotheses are given below:

Null hypothesis:

$H_0:{\sigma }^2\ge 0.0025$

That is, the true variance is greater than or equal to 0.0025.

Alternative hypothesis:

$H_1:{\sigma }^2<0.0025$

That is, the true variance is less than 0.0025.

Degrees of freedom:

$\begin{array}{c}df=n-1\\ =24-1\\ =23\end{array}$

Critical value:

From “Appendix E: CHI-SQAURE CRITICAL VALUES”,

• Locate the value 23 in the first column of the table.
• Locate the value 0.995 in the first row of the table.
• The intersecting value of row and column is 9.26.

Thus, the lower critical value is 9.26.

Decision rule for left-tailed test:

• If ${\chi }_{\text{calc}}^2<{\chi }_{\text{lower}}^2\left(=9.26\right)$, then reject the null hypothesis.
• Otherwise, do not reject the null hypothesis
• Test statistic:
• $\begin{array}{c}{\chi }_{\text{calc}}^2=\frac{\left(n-1\right)s^2}{{\sigma }_0^2}\\ =\frac{\left(24-1\right){\left(0.0373\right)}^2}{0.0025}\\ =\frac{0.0320}{0.0025}\\ =12.8\end{array}$
• Thus, the test statistic is 12.8.

Conclusion for critical value method:

Here, the test statistic is greater than critical value.

That is, ${\chi }_{\text{calc}}^2\left(=12.8\right)>{\chi }_{\text{lower}}^2\left(=9.26\right)$

Therefore, the null hypothesis is not rejected.

Hence, the sample does not show that ${\sigma }^2<0.0025$.