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Chapter 10: Problem 29

A comparison of the precisions of two machines developed for extracting juice from oranges is to be made using the following data: Machine A Machine B $$ s^{2}=3.1 \text { ounces }^{2} \quad s^{2}=1.4 \text { ounces }^{2} $$ \(n=25 \quad n=25\) 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

Expert verified
Question: Based on the given problem, perform a hypothesis test to check if there is a significant difference in the precision of the two juice extracting machines and calculate a 95% confidence interval for the ratio of the population variances. Answer: After performing the F-test, we have enough evidence to reject the null hypothesis, indicating that there is a significant difference in the precision of the two machines at a 5% significance level. The 95% confidence interval for the ratio of the population variances is (1.02, 4.66), which confirms that the population variances are significantly different.

Step by step solution

01

State the null and alternative hypotheses

The null hypothesis (\(H_0\)) states that the population variances of the two machines are equal, while the alternative hypothesis (\(H_1\)) states that they are not equal: $$H_0: \sigma_A^2 = \sigma_B^2$$ $$H_1: \sigma_A^2 \neq \sigma_B^2$$
02

Perform an F-test

We will use the F-test to compare the sample variances using the given sample sizes. The F-test statistic is calculated as the ratio of the larger sample variance to the smaller sample variance: $$F = \frac{s_A^2}{s_B^2}=\frac{3.1}{1.4}$$ Now, we need the degrees of freedom for both machines (which is equal to the sample sizes minus one): $$df_A=25-1=24$$ $$df_B=25-1=24$$ Using these degrees of freedom along with the F-test statistic and the alpha level (\(\alpha = 0.05\)) as the significance level, we will find the critical F-values as: $$F_{crit1} = F_{(0.025, 24, 24)}$$ $$F_{crit2} = F_{(0.975, 24, 24)}$$
03

Compare F-test statistic with critical F-values

If the F-test statistic falls between the critical F-values, we fail to reject the null hypothesis; otherwise, we reject the null hypothesis. After referring to the F-distribution table or using any statistical software, we obtain: $$F_{crit1} \approx 0.4631$$ $$F_{crit2} \approx 2.1608$$ And we have previously calculated $$F \approx 2.2143$$ Since \(F > F_{crit1}\) and \(F>F_{crit2}\), we reject the null hypothesis. So, there is enough evidence to indicate a significant difference in the precision of the two machines at a 5% significance level.
04

Calculate the 95% confidence interval for the ratio of the population variances

Now, we will find the 95% confidence interval for the ratio of the population variances: $$\frac{s_A^2}{s_B^2} \times \frac{1}{F_{(0.975, 24, 24)}} < \frac{\sigma_A^2}{\sigma_B^2} < \frac{s_A^2}{s_B^2} \times \frac{1}{F_{(0.025, 24, 24)}}$$ Substituting the values, we get: $$\frac{3.1}{1.4} \times \frac{1}{2.1608} < \frac{\sigma_A^2}{\sigma_B^2} < \frac{3.1}{1.4} \times \frac{1}{0.4631}$$ $$1.02 < \frac{\sigma_A^2}{\sigma_B^2} < 4.66$$
05

Interpret the confidence interval and answer part b

The 95% confidence interval for the ratio of population variances \((1.02, 4.66)\) does not contain the value 1, which means the population variances are significantly different. This interval confirms our conclusion from the F-test in part a that there is a 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.

Confidence Interval
A confidence interval provides a range of values which is likely to contain the true population parameter with a certain level of confidence. In statistical problems, such as the one concerning the comparison of two machines, confidence intervals are used to estimate the true ratio of two population variances.
  • A 95% confidence interval means that if we were to take numerous samples and calculate a confidence interval for each of them, 95% of these intervals would be expected to contain the true population parameter.
  • In this exercise, a confidence interval is calculated for the ratio of the variances of Machine A and Machine B.
  • The calculated interval was (1.02, 4.66), which gives us a range within which the true ratio of the variances is likely to lie with 95% confidence.
This interval helps statisticians conclude whether there is a significant difference between the precision of two machines by checking if the interval contains the value 1. Since 1 means equal variances, the absence of 1 in the interval indicates a difference.
Null Hypothesis
The null hypothesis is a statement that there is no effect, or no difference, and it serves as a starting point for statistical testing. In terms of hypothesis testing between variances, the null hypothesis specifically posits that there is no difference in the population variances.
  • Mathematically, it is formulated as \(H_0: \sigma_A^2 = \sigma_B^2\).
  • It assumes that any observed difference in sample variances is due to random chance.
  • The null hypothesis is tested using an F-test, which involves comparing the ratio of the sample variances to see if it is significantly different from 1, the theoretical ratio under the null hypothesis.
In our exercise, the null hypothesis is rejected, implying the sample data provides enough evidence to support that there is indeed a difference in precision between the two machines.
Significance Level
The significance level in hypothesis testing represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. It is denoted by \( \alpha \) and is chosen before conducting a test.
  • For this exercise, a significance level of 0.05 was utilized, which implies a 5% risk of committing a Type I error.
  • A smaller \(\alpha\) would reduce the likelihood of falsely rejecting the null hypothesis, requiring more evidence to declare a difference.
  • Conversely, at a significance level of 0.05, there is a 95% confidence in the results being non-random if the null hypothesis is rejected.
The choice of a 5% significance level is common in many statistical tests and means that the researchers are willing to accept a 5% chance of falsely declaring a difference in machine precision.
Population Variance
Population variance is a measure of the dispersion or spread of a set of data points in a population. It reflects how much the values in a population deviate from the mean of the population.
  • Variance is calculated as the average of the squared differences from the mean.
  • In the context of our exercise, population variance refers to the variance of the juice output precision for each machine, which would theoretically describe the entire population of machine outputs.
  • The F-test used in the solution compares the sample variances as estimates of these population variances to determine whether there is a statistically significant difference between them.
Understanding population variance helps provide insight into the consistency and reliability of a process or phenomenon, in this case, the precision of juice extraction by the two machines. Lower variance equates to higher precision and consistency.

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Most popular questions from this chapter

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