91Ó°ÊÓ

A pharmaceutical manufacturer purchases a particular material from two different suppliers. The mean level of impurities in the raw material is approximately the same for both suppliers, but the manufacturer is concerned about the variability of the impurities from shipment to shipment. To compare the variation in percentage impurities for the two suppliers, the manufacturer selects 10 shipments from each of the two suppliers and measures the percentage of impurities in the raw material for each shipment. The sample means and variances are shown in the table. $$ \begin{aligned} &\begin{array}{ll} \text { Supplier } \mathrm{A} & \text { Supplier } \mathrm{B} \\ \hline \bar{x}_{1}=1.89 & \bar{x}_{2}=1.85 \\ s_{1}^{2}=.273 & s_{2}^{2}=.094 \end{array}\\\ &n_{1}=10 \quad n_{2}=10 \end{aligned} $$ a. Do the data provide sufficient evidence to indicate a difference in the variability of the shipment impurity levels for the two suppliers? Test using \(\alpha=.01 .\) Based on the results of your test, what recommendation would you make to the pharmaceutical manufacturer? b. Find a \(99 \%\) confidence interval for \(\sigma_{2}^{2}\) and interpret your results.

Step by step solution

01

Identify the given information

We have been given: - Sample sizes: \(n_1 = n_2 = 10\) - Sample means: \(\bar{x}_1 = 1.89\) and \(\bar{x}_2 = 1.85\) - Sample variances: \(s_1^2 = 0.273\) and \(s_2^2 = 0.094\) - Significance level: \(\alpha = 0.01\)

02

Calculate the F-ratio

Now, we need to calculate the F-ratio. For the F-test, we'll compare the ratio of variances. To do that, we'll divide the larger variance by the smaller one: $$ F = \frac{s_1^2}{s_2^2} = \frac{0.273}{0.094} \approx 2.904 $$

03

Determine the degrees of freedom

The degrees of freedom for each sample are \(n_1 - 1\) and \(n_2 - 1\). So, we have: $$ df_1 = n_1 - 1 = 10 - 1 = 9 $$ $$ df_2 = n_2 - 1 = 10 - 1 = 9 $$

04

Find the F-critical values

Using the F distribution table, or an F-distribution calculator, we'll find the F-critical values at \(\alpha = 0.01\) and the given degrees of freedom, \(df_1 = 9\) and \(df_2 = 9\). We get: $$ F_\text{lower} = 0.166 $$ $$ F_\text{upper} = 6.000 $$

05

Compare the F-ratio with critical values

Now, we'll compare the calculated F-ratio with the critical values: If \(F_\text{lower} < F < F_\text{upper}\), we cannot reject the null hypothesis, meaning that there is not sufficient evidence to indicate a difference in variability. $$ 0.166 < 2.904 < 6.000 $$ Since the F-ratio is within the critical values, we cannot reject the null hypothesis. This means there is not sufficient evidence to indicate a significant difference in variability between the shipments from the two suppliers. The recommendation for the pharmaceutical manufacturer would be that there is no significant difference in the variability of shipment impurity levels between the two suppliers, so both suppliers can be considered as having a similar range of variability concerning impurity levels at the 1% significance level.

06

Part b: 99% confidence interval for Supplier B's variance

In order to find the confidence interval for the variance of supplier B, we need to use the following formula: $$ \frac{(n - 1) s^2}{\chi^2_\text{upper}} < \sigma^2 < \frac{(n - 1) s^2}{\chi^2_\text{lower}} $$ We have: - Sample size: \(n = 10\) - Sample variance: \(s^2 = 0.094\) - Significance level: 1 - 0.99 = 0.01 - Degrees of freedom: \(n - 1 = 9\) Using a chi-square distribution table, we can find the chi-square values: $$ \chi^2_\text{lower} \approx 1.735 $$ $$ \chi^2_\text{upper} \approx 23.589 $$ Now, we can calculate the confidence interval: $$ \frac{(10 - 1) 0.094}{23.589} < \sigma^2 < \frac{(10 - 1) 0.094}{1.735} $$ $$ 0.033 < \sigma^2 < 0.456 $$ So, the 99% confidence interval for supplier B's impurity level variance is (0.033, 0.456). This means that we are 99% confident that the true variance of impurity levels from supplier B shipments lies between 0.033 and 0.456.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

F-test

The F-test is a statistical test used to compare the variances of two samples to determine if they come from populations with the same variance. In the context of the variance analysis for the pharmaceutical manufacturer, the F-test helps us understand if there is a significant difference in the variability of impurity levels between two suppliers.

To perform an F-test, we calculate the F-ratio, which is the ratio of the larger sample variance to the smaller sample variance. This ratio helps assess the difference in variability between the two groups.

In our exercise, we calculate the F-ratio by dividing the variance of supplier A by that of supplier B: \( F = \frac{s_1^2}{s_2^2} = \frac{0.273}{0.094} \approx 2.904 \).

A significant F-ratio indicates that the variances are unlikely to have come from populations with equal variances. If the calculated F-ratio falls outside the range of critical values, we reject the null hypothesis of equal variances.

Confidence Interval

A confidence interval provides a range of values within which we expect the true population parameter to lie with a certain level of confidence. In this exercise, we calculate a 99% confidence interval for the variance of supplier B's shipment impurity levels.

To do this, we use the chi-square distribution, given the variance follows a chi-square distribution when sample variance data is used.

The formula is: \[ \frac{(n - 1) s^2}{\chi^2_{\text{upper}}} < \sigma^2 < \frac{(n - 1) s^2}{\chi^2_{\text{lower}}} \] where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \chi^2_{\text{lower}} \) and \( \chi^2_{\text{upper}} \) are chi-square values.

In our case, we find the confidence interval is approximately from 0.033 to 0.456. This means there's a 99% chance that the true variance of supplier B's impurity levels lies within this range. This helps in making informed decisions about supplier variability.

Degrees of Freedom

Degrees of Freedom (df) are an essential concept in statistics that refers to the number of independent values or quantities which can be assigned to a statistical distribution. They are crucial in calculating accurate test statistics such as the F-test or the t-test.

In the context of our variance analysis, for each supplier, the degrees of freedom is calculated as the sample size minus one: \( df = n - 1 \).

For both suppliers A and B, since \( n = 10 \), the degrees of freedom is calculated as follows: \( df_1 = n_1 - 1 = 10 - 1 = 9 \) and \( df_2 = n_2 - 1 = 10 - 1 = 9 \).

Degrees of freedom are used not only to determine the distribution of the test statistic but also to find critical values from statistical tables like the F-distribution table. This helps ensure the results from these tests are accurate and reliable.

Significance Level

The significance level, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true. It is a threshold set by the researcher to limit the risk of a Type I error, which occurs when a true null hypothesis is incorrectly rejected.

In our exercise, the significance level is set at 1%, or \( \alpha = 0.01 \).

This low significance level indicates a stringent criterion, suggesting the analysis is conservative. In other words, the results need to be very strong to conclude that there is a significant difference in the variances of impurities between the suppliers.

At a 1% significance level, we would only reject the null hypothesis and claim a significant difference in variability if the calculated F-ratio is greater than the critical value obtained from the F-distribution table. This conservative approach helps ensure that any conclusions drawn are robust and not due to random variations in the data.