Chapter 11.4, Problem 13E

Sugar content: A broth used to manufacture a pharmaceutical product has its sugar content, in milligrams per milliliter, measured several times on each of three successive days.

Day 1: 5.0 4.8 5.1 5.1 4.8 5.1 4.8 4.8 5.0 5.2 4.9 4.9 5.0

Day 2: 5.8 4.7 4.7 4.9 5.1 4.9 5.4 5.3 4.8 5.7 5.1 5.7

Day 3: 6.3 4.7 5.1 5.9 5.1 5.9 4.7 6.0 5.3 4.9 5.7 5.3 5.6

1. Can you conclude that the variability of the process is greater on the second day than on the first day? Use the $\alpha =0.05$ level of significance.
2. Can you conclude that the variability of the process is greater on the third day than on the second day? Use the $\alpha =0.01$ level of significance.

To determine

To find:

Whether the variability of the process is greater on the second day than on the first day.

Answer to Problem 13E

The variability of the process is greater on the second day than on the first day.

Explanation of Solution

Given Information:

Sugar content: A broth used to manufacture a pharmaceutical product has its sugar content, in milligrams per milliliter, measured several times on each of three successive days.

Day   $1$	$5.04.85.15.14.85.14.84.85.05.24.94.95.0$
Day   $2$	$5.84.74.74.95.14.95.45.35.34.85.75.15.7$
Day   $3$	$6.34.75.15.95.15.94.76.05.34.95.75.35.6$

Formula used:

To test the hypothesis is that the variability of tile process is higher than on second day and first day at $5\%$ level of significance.

First, give hypothesis, compute test-statistics, p-value then give conclusion based on p-value.

The null and alternative hypothesis is.

  $\begin{array}{l}{H}_0:{\sigma }_1^2={\sigma }_2^2\\ H_a:{\sigma }_1^2>{\sigma }_2^2\end{array}$

Calculation:

The F-test statistics is.

By using MINITAB, find F-test statistics with the help of following steps is:

1. Import the data.
2. Select the Stat and choose the Basic Statistics option
3. Select the $2$Variances and choose variable option and put Sample in different columns
4. Select Confidence level,
5. Click 0k

Test and $CI$ for two variances: Day $2$ , Day $1$ .

Method

  $\begin{array}{l}\text{Null hypothesis Variance}\left(\text{Day 2}\right)/\text{Variance}\left(\text{Day 1}\right)=:\\ \text{Alternative hypothesis Variance}\left(\text{Day 2}\right)/\text{Variance}\left(\text{Day 1}\right)>:\end{array}$

Statistics

  $\begin{array}{l}\text{Variable N StDev Variance}\\ \text{Day 2 13 0}\text{.385 0}\text{.148}\\ \text{Day 1 13 0}\text{.139 0}\text{.019}\end{array}$

Tests

  $\begin{array}{l}\text{Method }\text{Test}\\ \text{F Test}\left(\text{normal}\right)\text{DF1}\text{DF2 Statistic p-value}\\ \text{7}\text{.70 12 12}\text{ 7}\text{.70 0}\text{.001}\end{array}$

From the MINITAB output the $F$ test statistics is.

The p-value for this test is,

  $0.001$ From the MINITAB output, the p-value for this test is.

The conclusion is that the p-value in this context is less than $0.05$ which is $0.001$ , so the null hypothesis is rejected at $5\%$ level of significance. There is sufficient evidence to indicate that the variability of the process is higher than on second day and first day. The result is statistically significant.

To determine

To find:

Whether the vanabihty of the process is greater on the third day than on the second day.

Answer to Problem 13E

The vanabihty of the process is not greater on the third day than on the second day.

Explanation of Solution

Given Information:

To test the hypothesis is that the variability of tile process is higher than on third day and second day at $5\%$ level of significance.

First, give hypothesis, compute test-statistics, p-value then give conclusion based on p-value.

The null and alternative hypothesis is.

  $\begin{array}{l}{H}_0:{\sigma }_2^2={\sigma }_3^2\\ H_a:{\sigma }_2^2>{\sigma }_3^2\end{array}$

Formula used:

The F-test statistics is.

By using MINITAB, find F-test statistics with the help of following steps is:

1. Import the data.
2. Select the Stat and choose the Basic Statistics option
3. Select the $2$Variances and choose variable option and put Sample in different columns
4. Select Confidence level,
5. Click 0k

Calculation:

Test and $CI$ for two variances: Day $3$ , Day $2$ .

Method

  $\begin{array}{l}\text{Null hypothesis Variance}\left(\text{Day 3}\right)/\text{Variance}\left(\text{Day 2}\right)=1\\ \text{Alternative hypothesis Variance}\left(\text{Day 3}\right)/\text{Variance}\left(\text{Day 2}\right)>1\\ \text{Significance level Alpha}=0.05\end{array}$

Statistics

  $\begin{array}{l}\text{Variable N StDev Variance}\\ \text{Day 3 13 0}\text{.518 0}\text{.269}\\ \text{Day 2 13 0}\text{.385 0}\text{.148}\end{array}$

Tests

  $\begin{array}{l}\text{Method }\text{Test}\\ \text{F Test}\left(\text{normal}\right)\text{DF1}\text{DF2 Statistic p-value}\\ \text{1}\text{.81 12 12}\text{ 1}\text{.81 0}\text{.158}\end{array}$

From the MINITAB output the $F$ test statistics is.

The p-value for this test is,

  $0.158$

The conclusion is that the p-value in this context is higher than $0.05$ which is $0.158$ , so the null hypothesis is rejected at $5\%$ level of significance. There is sufficient evidence to indicate that the variability of the process is higher than on third day and second day. The result is statistically significant.