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A comparison of the precisions of two machines developed for extracting juice from oranges is to be made using the following data: $$\begin{array}{ll}\hline \text { Machine A } & \text { Machine B } \\\\\hline s^{2}=2790 \mathrm{~mL}^{2} & s^{2}=1260 \mathrm{~mL}^{2} \\\n=25 & n=25 \\\\\hline\end{array}$$ a. Is there sufficient evidence to indicate that there is a difference in the precision of the two machines at the 5\% level of significance? b. Find a \(95 \%\) confidence interval for the ratio of the two population variances. Does this interval confirm your conclusion from part a? Explain.

Short Answer

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And what is the 95% confidence interval for the ratio of their population variances?

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\(H_0\)): There is no difference in the precision of the two machines, which means the ratio between the population variances is 1. Mathematically, $$H_0: \frac{\sigma^2_A}{\sigma^2_B} = 1$$ The alternative hypothesis (\(H_1\)): There is a difference in precision between the two machines, so the ratio between the population variances is not 1. $$ H_1: \frac{\sigma^2_A}{\sigma^2_B} \neq 1$$
02

Calculate the test statistic

To calculate the F-statistic, we need to divide the larger sample variance by the smaller sample variance: $$F_{obs} = \frac{s^2_A}{s^2_B} = \frac{2790}{1260} \approx 2.214$$
03

Determine the critical values

We need to determine the critical values for the F-test at a 5% significance level. We have: - Degrees of freedom for numerator (Machine A): \(\nu_1 = n_A - 1 = 25 - 1 = 24\) - Degrees of freedom for denominator (Machine B): \(\nu_2 = n_B - 1 = 25 - 1 = 24\) Using an F-distribution table, we find the critical values for an \(\alpha = 0.05\) significance level with \(24\) degrees of freedom in the numerator and \(24\) degrees of freedom in the denominator: $$F_{\frac{\alpha}{2}} \approx 0.463$$ and $$F_{1-\frac{\alpha}{2}} \approx 2.154$$
04

Compare the test statistic with the critical values

Our observed test statistic value, \(F_{obs} \approx 2.214\), falls outside the critical region between 0.463 and 2.154. This means we reject the null hypothesis at a 5% level of significance. So, there is sufficient evidence to indicate a difference in the precision of the two machines.
05

Calculate the confidence interval for the ratio of variances

We will now find a 95% confidence interval for the ratio of the two population variances using the formula for the F-distribution confidence interval: $$\frac{s^2_A / \nu_1}{s^2_B / \nu_2} \cdot \frac{1}{F_{1-\frac{\alpha}{2}}} \leq \frac{\sigma^2_A}{\sigma^2_B} \leq \frac{s^2_A / \nu_1}{s^2_B / \nu_2} \cdot F_{1-\frac{\alpha}{2}}$$ Plugging in the values, we get: $$\frac{2790 / 24}{1260 / 24} \cdot \frac{1}{2.154} \leq \frac{\sigma^2_A}{\sigma^2_B} \leq \frac{2790 / 24}{1260 / 24} \cdot 2.154$$ $$0.464 \leq \frac{\sigma^2_A}{\sigma^2_B} \leq 2.155$$
06

Interpret the confidence interval

The 95% confidence interval for the ratio of the two population variances is between 0.464 and 2.155. Since the interval does not contain 1, it confirms the conclusion from part a that there is a significant difference in the precision of the two machines.

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is crucial when assessing statistical evidence in experiments and studies. In the context of variance ratio tests, the null hypothesis, denoted as \( H_0 \), asserts that there is no difference between the variances of two populations—or in our specific case, the precision of two machines used for extracting juice from oranges.

When dealing with variance ratio tests, we express the null hypothesis by setting the ratio of the two population variances equal to 1, indicating that any observed difference in sample variances is due to random chance rather than a systematic difference. If the calculated test statistic falls within a certain range, we fail to reject the null hypothesis, suggesting that there's not enough evidence to support the claim of a variance difference.

However, if the test statistics fall outside this range, as in the exercise, where we have an observed value that exceeds the upper critical value, it provides sufficient evidence to reject the null hypothesis and consider that there may be a significant difference between the machine precisions.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \), is the counterpart to the null hypothesis. It represents the outcome that the study or experiment is attempting to support—namely, that there is a real effect or difference present.

In our juice extraction machine example, the alternative hypothesis posits that there is a significant difference in the precision of the two machines, which means the ratio of the population variances is not equal to 1. If evidence suggests rejecting the null hypothesis, you lend support to the alternative hypothesis, prompting further investigation or acceptance of a new theory.

The alternative hypothesis is what motivates researchers to scrutinize the null hypothesis through statistical testing. It embodies the notion of a scientific breakthrough or the confirmation of a suspected effect or difference that could lead to new understanding and advancements in the field.
F-Statistic
The F-statistic is a powerful tool in hypothesis testing, especially when comparing variances. It is used in the Analysis of Variance (ANOVA) and the variance ratio test, as in our example with the juice extraction machines.

To calculate the F-statistic, we divide the sample variances with the assumption that the population variances are equal under the null hypothesis—this means dividing the larger variance by the smaller one to ensure a positive value. In our case, we calculated \( F_{obs} = 2.214 \), by dividing \( s^2_A = 2790 \) by \( s^2_B = 1260 \).

The F-statistic allows us to make decisions about the null hypothesis. By comparing our calculated F-statistic to critical values derived from the F-distribution, we can determine the p-value or the probability of observing such an F-statistic if the null hypothesis were true. A low p-value indicates that such an extreme statistic is unlikely under the null hypothesis, leading to its rejection.
Confidence Interval
Confidence intervals provide a range of values that you can be certain, to a given level of confidence, contains the true population parameter. They are a vital part of inferential statistics and give us not just an estimate of the parameter but also an idea about the precision and reliability of that estimate.

For a variance ratio test, like in our exercise, a 95% confidence interval gives us a range where we expect the true ratio of population variances to fall 95% of the time, had we conducted the experiment an infinite number of times. Calculating this interval helps us understand the variability between the two machines' precision.

In our example, the 95% confidence interval for the ratio of the variances of machine A to machine B was between 0.464 and 2.155. This interval does not include the value 1, which means the ratio of the true variances is very unlikely to be 1, thus reinforcing the conclusion that the machines' precisions differ significantly.

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Most popular questions from this chapter

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