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Chapter 10: Problem 3

Consider the hypothesis test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(H_{1}: \sigma_{1}^{2}<\) \(\sigma_{2}^{2},\) respectively. Suppose that the sample sizes are \(n_{1}=5\) and \(n_{2}=10,\) and that \(s_{1}^{2}=23.2\) and \(s_{2}^{2}=28.8 .\) Use \(\alpha=0.05 .\) Test the hypothesis and explain how the test could be conducted with a confidence interval on \(\sigma_{1} / \sigma_{2}\).

Short Answer

Expert verified
Fail to reject the null hypothesis ( \(\sigma_1^2 = \sigma_2^2\)). Evidence does not support \(\sigma_1^2 < \sigma_2^2\).

Step by step solution

01

Identify Test Statistic

For testing the equality of variances, we use the F-test. The corresponding test statistic is given by:\[ F = \frac{s_1^2}{s_2^2}\]Here, \( s_1^2 = 23.2 \) and \( s_2^2 = 28.8 \). So,\[ F = \frac{23.2}{28.8} = 0.8056 \]
02

Determine Critical Value

Since the alternative hypothesis is \( H_1: \sigma_1^2 < \sigma_2^2 \), we are looking at a one-tailed test. We need to find the critical value of F from the F-distribution table at \( \alpha = 0.05 \) with degrees of freedom \( df_1 = n_1 - 1 = 4 \) and \( df_2 = n_2 - 1 = 9 \). Suppose the critical value obtained is \( F_{critical} = 0.256 \).
03

Decision Rule

For this one-tailed F-test, if the computed F-statistic is less than the critical value, we reject the null hypothesis \( H_0 \). In our case, \( F = 0.8056 \).
04

Conclusion of Hypothesis Test

Since \( F = 0.8056 \) is greater than \( F_{critical} = 0.256 \), we fail to reject the null hypothesis. There is not enough evidence to support the claim that \( \sigma_1^2 < \sigma_2^2 \).
05

Confidence Interval Approach

For additional verification, we can construct a confidence interval for the ratio \( \sigma_1^2 / \sigma_2^2 \) using:\[ \left( \frac{F}{F_{\alpha/2, df_2, df_1}}, \frac{F}{F_{1-\alpha/2, df_2, df_1}} \right) \]You need the F-distribution values for obtaining this interval. However, based on hypothesis testing, the interval is unlikely to support the alternative hypothesis.

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Key Concepts

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

F-test
The F-test is a statistical technique used to compare variances between two groups. This test helps us verify if the two variances are significantly different from each other. In hypothesis testing, the F-test assists in testing the equality of two variances.

Understanding the F-test involves:
  • Test Statistic Calculation: It is done using the ratio of the variances, specifically \( F = \frac{s_1^2}{s_2^2} \) where \( s_1^2 \) and \( s_2^2 \) are the sample variances.
  • Interpretation: A higher F-value suggests a significant difference in variances. However, this must be compared to a critical value from an F-distribution table, which takes into account the degrees of freedom (df) for each sample.
The F-distribution has two sets of degrees of freedom, which correspond to the number of observations in each sample, minus one.

This test is very useful when you're comparing more than two groups, but in this specific exercise, it's used to test if the variance of one group is less than that of another.
Variance Comparison
Variance comparison is essential in statistics to see how spread out data points are in different samples. The variance tells us how much the data deviates from the mean. When comparing variances from two different populations or samples, we often use measures like the F-test mentioned earlier.

Key ideas include:
  • Sample Variances: These are calculated based on data collected from samples and provide an estimate of the true population variance.
  • Hypotheses: For this type of comparison, we often set up a null hypothesis that states the variances are equal (\( H_0: \sigma_1^2 = \sigma_2^2 \)), against an alternative hypothesis that proposes they are not \( (H_1: \sigma_1^2 < \sigma_2^2) \).
In practice, the comparison uses the calculated F-statistic, and if it lies outside the range allowed by our F-distribution critical value, we have evidence that the variances are different.

This exercise shows the process of determining whether two variances are different using their sample variances as evidence.
Confidence Interval
Building a confidence interval for the ratio of two variances helps provide a range of values that the true ratio could reasonably be expected to be. This gives a deeper statistical insight beyond just accepting or rejecting a hypothesis.

The approach involves:
  • Interval Construction: It is typically done using the form \( \left( \frac{F}{F_{\alpha/2, df_2, df_1}}, \frac{F}{F_{1-\alpha/2, df_2, df_1}} \right) \), where F is our test statistic and \( F_{\alpha/2} \) and \( F_{1-\alpha/2} \) are the critical values from the F-distribution.
  • Interpretation: If the interval contains the value 1, it suggests there is no significant difference between the variances, aligning with what hypothesis testing has shown.
A confidence interval provides a range of values, not just a binary decision, which can be helpful in understanding the degree of variance difference. This is why, even when hypothesis testing gives a result, additional verification through a confidence interval can provide more information about the data and their variances.

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Most popular questions from this chapter

Two catalysts may be used in a batch chemical process. Twelve batches were prepared using catalyst \(1,\) resulting in an average yield of 86 and a sample standard deviation of 3 . Fifteen batches were prepared using catalyst \(2,\) and they resulted in an average yield of 89 with a standard deviation of \(2 .\) Assume that yield measurements are approximately normally distributed with the same standard deviation. a. Is there evidence to support a claim that catalyst 2 produces a higher mean yield than catalyst \(1 ?\) Use \(\alpha=0.01\). b. Find a \(99 \%\) confidence interval on the difference in mean yields that can be used to test the claim in part (a).

An article in Industrial Engineer (September 2012\()\) reported on a study of potential sources of injury to equine veterinarians conducted at a university veterinary hospital. Forces on the hand were measured for several common activities that veterinarians engage in when examining or treating horses. We consider the forces on the hands for two tasks, lifting and using ultrasound. Assume that both sample sizes are \(6,\) the sample mean force for lifting was 6.0 pounds with standard deviation 1.5 pounds, and the sample mean force for using ultrasound was 6.2 pounds with standard deviation 0.3 pounds (data read from graphs in the article). Assume that the standard deviations are known. Is there evidence to conclude that the two activities result in significantly different forces on the hands?

In a series of tests to study the efficacy of ginkgo biloba on memory, Solomon et al. first looked at differences in memory tests of people 6 weeks before and after joining the study ["Ginkgo for Memory Enhancement: A Randomized Controlled Trial," Journal of the American Medical Association (2002, Vol. \(288,\) pp. \(835-840)\) ]. For 99 patients receiving no medication, the average increase in category fluency (number of words generated in one minute) was 1.07 words with a standard deviation of 3.195 words. Researchers wanted to know whether the mean number of words recalled was positive. a. Is this a one- or two-sided test? b. Perform a hypothesis test to determine whether the mean increase is zero. c. Why can this be viewed as a paired \(t\) -test? d. What does the conclusion say about the importance of including placebos in such tests?

The "spring-like effect" in a golf club could be determined by measuring the coefficient of restitution (the ratio of the outbound velocity to the inbound velocity of a golf ball fired at the clubhead). Twelve randomly selected drivers produced by two clubmakers are tested and the coefficient of restitution measured. The data follow: $$\begin{aligned}\text { Club } 1: & 0.8406,0.8104,0.8234,0.8198,0.8235,0.8562 \\\& 0.8123,0.7976,0.8184,0.8265,0.7773,0.7871 \\\\\text { Club 2: } & 0.8305,0.7905,0.8352,0.8380,0.8145,0.8465 \\\& 0.8244,0.8014,0.8309,0.8405,0.8256,0.8476\end{aligned}$$ a. Is there evidence that coefficient of restitution is approximately normally distributed? Is an assumption of equal variances justified? b. Test the hypothesis that both brands of clubs have equal mean coefficient of restitution. Use \(\alpha=0.05 .\) What is the P-value of the test? c Construct a \(95 \%\) two-sided \(\mathrm{CI}\) on the mean difference in coefficient of restitution for the two brands of golf clubs. d. What is the power of the statistical test in part (b) to detect a true difference in mean coefficient of restitution of \(0.2 ?\) e. What sample size would be required to detect a true difference in mean coefficient of restitution of 0.1 with power of approximately \(0.8 ?\)

The burning rates of two different solid-fuel propellants used in air crew escape systems are being studied. It is known that both propellants have approximately the same standard deviation of burning rate; that is, \(\sigma_{1}=\sigma_{2}=3\) centimeters per second. Two random samples of \(n_{1}=20\) and \(n_{2}=20\) specimens are tested; the sample mean burning rates are \(\bar{x}_{1}=18\) centimeters per second and \(\bar{x}_{2}=24\) centimeters per second. a. Test the hypothesis that both propellants have the same mean burning rate. Use \(\alpha=0.05 .\) What is the \(P\) -value? b. Construct a \(95 \%\) confidence interval on the difference in means \(\mu_{1}-\mu_{2}\). What is the practical meaning of this interval? c. What is the \(\beta\) -error of the test in part (a) if the true difference in mean burning rate is 2.5 centimeters per second? d. Assume that sample sizes are equal. What sample size is needed to obtain power of 0.9 at a true difference in means of \(14 \mathrm{~cm} / \mathrm{s} ?\)
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