Chapter 9, Problem 80SE

Soap Production Process. H₀: m = 120 and H_a: μ ≠ 120 are used to test whether a bath soap production process is meeting the standard output of 120 bars per batch. Use a .05 level of significance for the test and a planning value of 5 for the standard deviation.

1. a. If the mean output drops to 117 bars per batch, the firm wants to have a 98% chance of concluding that the standard production output is not being met. How large a sample should be selected?
2. b. With your sample size from part (a), what is the probability of concluding that the process is operating satisfactorily for each of the following actual mean outputs: 117, 118, 119, 121, 122, and 123 bars per batch? That is, what is the probability of a Type II error in each case?

To determine

Find the sample size.

Answer to Problem 80SE

Welcome123

The sample size is 45.

Explanation of Solution

Calculation:

The given information is that $\beta =0.02$, $\text{Power}=0.98\left(=1-0.02\right)$, $\alpha =0.05$, $\sigma =5$, ${\mu }_a=117$, and ${\mu }_0=120$.

The null and alternative hypotheses are given below:

Null hypothesis:

$H_0:\mu =120$

Alternative hypothesis:

$H_a:\mu \ne 120$

Sample size:

The formula for finding the sample size is as follows:

$n=\frac{{\left(z_{\frac{\alpha }{2}}+z_{\beta }\right)}^2{\sigma }^2}{{\left({\mu }_0-{\mu }_a\right)}^2}$

Here, $z_{\frac{\alpha }{2}}$ represents the z-value providing an area of $\frac{\alpha }{2}$ in the upper tail of a standard normal distribution, $z_{\beta }$ represents the z-value providing an area of $\beta$ in the upper tail of a standard normal distribution, $\sigma$ represents the population standard deviation, ${\mu }_0$ represents the value of the population mean in the null hypothesis, and ${\mu }_a$ represents the value of the population mean used for the type II error.

For $z_{\frac{\alpha }{2}}$:

It is given that $\alpha =0.05$. Therefore, $\frac{\alpha }{2}=0.025$.

The cumulative area to the left is as follows:

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.025\\ =0.975\end{array}$

Critical value:

Use Table 1: Cumulative probabilities for the standard Normal Distribution to find $z_{\frac{\alpha }{2}}$.

• Locate the area of 0.975 in the body of Table 1.
• Move left until the first column and note the values as 1.9.
• Move upward until the top row is reached and note the value as 0.06.

Thus, the critical value $z_{\frac{\alpha }{2}}$ is 1.96

For $z_{\beta }$:

It is given that $\beta =0.02$.

The cumulative area to the left is as follows:

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.02\\ =0.98\end{array}$

Critical value:

Use Table 1: Cumulative probabilities for the standard Normal Distribution to find $z_{\beta }$.

• Locate the area of 0.98 in the body of Table 1.
• Move left until the first column and note the values as 2.0.
• Move upward until the top row is reached and note the value as 0.05.

Thus, the critical value $z_{\beta }$ is 2.05.

Substitute 1.96 for $z_{\frac{\alpha }{2}}$, 2.05 for $z_{\beta }$, 5 for $\sigma$, 120 for ${\mu }_0$, and 117 for ${\mu }_a$ in n formula.

$\begin{array}{c}n=\frac{{\left(1.96+2.05\right)}^2{\left(5\right)}^2}{{\left(120-117\right)}^2}\\ =\frac{16.0801\times 25}{9}\\ \approx 45\end{array}$

Thus, the sample size is 45.

To determine

Find the probability of making a Type II error.

Answer to Problem 80SE

The probabilities of making a Type II error for $\mu$ equals to 117, 118, 119, 121, 122, and 123 are tabulated below:

$\mu$	$\beta$
117	0.0192
118	0.2358
119	0.7324
121	0.7324
122	0.2358
123	0.0192

Explanation of Solution

Calculation:

The given information is that $n=45$, $\mu =117,118,119,121\text{ and 123}$, $\sigma =5$, and $\alpha =0.05$.

The rejection rule for the two-tailed test using the critical value is as follows:

If $z\ge z_{\frac{\alpha }{2}}$ or $z\le -z_{\frac{\alpha }{2}}$, reject the null hypothesis

Critical value:

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

The cumulative area to the left is as follows:

$\begin{array}{c}\text{Area to the left}=1-\text{Area to the right}\\ =1-0.025\\ =0.975\end{array}$

Use Table 1: Cumulative probabilities for the standard Normal Distribution to find $z_{\frac{\alpha }{2}}$.

• Locate the area of 0.975 in the body of Table 1.
• Move left until the first column and note the values as 1.9.
• Move upward until the top row is reached and note the value as 0.06.

Thus, the critical values $z_{\frac{\alpha }{2}}$ and $-z_{\frac{\alpha }{2}}$ are 1.96 and –1.96.

Rejection rule:

If $z\ge 1.96$ or $z\le -1.96$, reject the null hypothesis.

The value ${\overline{x}}_1$ is obtained below:

$\begin{array}{c}{\overline{x}}_1=\mu -z\left(\frac{\sigma }{\sqrt{n}}\right)\\ =120-1.96\left(\frac{5}{\sqrt{45}}\right)\\ =\text{118}\text{.54}\end{array}$

The value ${\overline{x}}_2$ is obtained below:

$\begin{array}{c}{\overline{x}}_2=\mu +z\left(\frac{\sigma }{\sqrt{n}}\right)\\ =120+1.96\left(\frac{5}{\sqrt{45}}\right)\\ =\text{121}\text{.46}\end{array}$

The rejection rule based on the value of $\overline{x}$ for the test is as follows:

• If $\overline{x}\le 118.54$ or $\overline{x}\ge 121.46$, reject the null hypothesis.
• If $118.54<\overline{x}<121.46$, do not reject the null hypothesis.

For $\mu =117$:

The value of z when $\mu =117$ isas follows:

$\begin{array}{c}z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}\\ =\frac{118.54-117}{\frac{5}{\sqrt{45}}}\\ =2.07\end{array}$

Use Table 1: Cumulative probabilities for the standard Normal Distribution to find probability.

• Locate the value 2.0 in the first column.
• Locate the value 0.07 in the first row.
• The intersecting value that corresponds to 2.07 is 0.9808.

Thus, the area value is 0.9808.

The probability of making a Type II error if $\mu =117$ is as follows:

$\begin{array}{c}\beta =1-0.9808\\ =\text{0}\text{.0192}\end{array}$

Thus, the probability of making a Type II error if $\mu =117$ is 0.0192.

Similarly, the probabilities of making a Type II error for the remaining $\mu$ values are tabulated below:

$\mu$	$z=\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$	Area value	$\beta$
117	$\frac{118.54-117}{\frac{5}{\sqrt{45}}}=2.07$	0.9808	0.0192
118	$\frac{118.54-118}{\frac{5}{\sqrt{45}}}=0.72$	0.7642	0.2358
119	$\frac{118.54-119}{\frac{5}{\sqrt{45}}}=-0.62$	0.2676	0.7324
121	$\frac{121.46-121}{\frac{5}{\sqrt{45}}}=0.62$	0.2676	0.7324
122	$\frac{121.46-122}{\frac{5}{\sqrt{45}}}=-0.72$	0.7642	0.2358
123	$\frac{121.46-123}{\frac{5}{\sqrt{45}}}=-2.07$	0.9808	0.0192