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Consider the hypothesis test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(H_{1}: \sigma_{1}^{2} \neq \sigma_{2}^{2} .\) Suppose that the sample sizes are \(n_{1}=15\) and \(n_{2}=15,\) and the sample variances are \(s_{1}^{2}=2.3\) and \(s_{2}^{2}=1.9 .\) Use \(\alpha=0.05\) (a) Test the hypothesis and explain how the test could be conducted with a confidence interval on \(\sigma_{1} / \sigma_{2}\) (b) What is the power of the test in part (a) if \(\sigma_{1}\) is twice as large as \(\sigma_{2} ?\) (c) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if the \(\sigma_{2}\) is half of \(\sigma_{1} ?\)

Short Answer

Expert verified
(a) Fail to reject \( H_0 \) at \( \alpha = 0.05 \). (b) Calculate power for \( \sigma_1 = 2\sigma_2 \) using F-tests. (c) Determine \( n \) for \( \beta = 0.05 \), typically requiring larger samples.

Step by step solution

01

Identify Relevant Information

We are given two samples with sizes \( n_1 = 15 \) and \( n_2 = 15 \), and sample variances \( s_1^2 = 2.3 \) and \( s_2^2 = 1.9 \). The null hypothesis is \( H_0: \sigma_1^2 = \sigma_2^2 \) against the alternative hypothesis \( H_1: \sigma_1^2 eq \sigma_2^2 \). The significance level is \( \alpha = 0.05 \).
02

Conduct F-Test

The test statistic for comparing two variances is the F-ratio, calculated as \( F = \frac{s_1^2}{s_2^2} \). Substitute the given values: \( F = \frac{2.3}{1.9} \approx 1.21 \).
03

Determine Critical Value

Since \( \alpha = 0.05 \) and the test is two-tailed, find the critical values for \( F \) with \( df_1 = n_1 - 1 = 14 \) and \( df_2 = n_2 - 1 = 14 \). Use an F-distribution table or calculator to find \( F_{upper} = F_{0.975, 14, 14} \) and \( F_{lower} = 1/F_{upper} \).
04

Conclusion of F-Test

Since the calculated \( F \approx 1.21 \) falls between the critical values \( F_{lower} \) and \( F_{upper} \), we fail to reject the null hypothesis \( H_0 \). This indicates no significant difference between the two variances at \( \alpha = 0.05 \).
05

Confidence Interval Approach

To construct a confidence interval for \( \frac{\sigma_1^2}{\sigma_2^2} \), use the calculated F-statistic and critical values. The interval provides the range in which the true variance ratio \( \sigma_1 / \sigma_2 \) likely falls, affirming the test decision.
06

Power Calculation

If \( \sigma_1 = 2\sigma_2 \), calculate power using relevant non-central F-distribution or power calculation methods specific to F-tests. Hence, power involves determining the probability of detecting a true variance ratio \( 4:1 \).
07

Determine Sample Size for \( \beta = 0.05 \)

To achieve a power \( 1 - \beta = 0.95 \), compute requisite sample sizes. Consider variance ratio \( (\sigma_2 / \sigma_1)^2 = 0.25 \) and use statistical software or tables to find \( n \) ensuring \( \beta = 0.05 \).

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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 method used to compare two variances to determine if they are significantly different. In hypothesis testing, this is adeptly applied when assessing assumptions about populations based on sample data. The key underlying principle of an F-test is the F-ratio, which is the test statistic calculated to evaluate the null hypothesis.

In our example, the null hypothesis is that the variances from two distinct samples are equal, i.e., \(H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}\). The alternative hypothesis, conversely, suggests these variances are not equal. To derive the F-ratio, you compute it as the ratio of the variances, typically like this: \(F = \frac{s_{1}^{2}}{s_{2}^{2}}\). Given the sample variances of 2.3 and 1.9 in this exercise, the F-ratio is approximately 1.21.

The resulting F-statistic, when compared against critical values derived from the F-distribution, informs whether the null hypothesis can be rejected. A key component to note is that the F-distribution requires degrees of freedom, based on sample sizes, for determining the critical values.
Variance Comparison
Variance comparison is crucial in statistics because it helps us understand the variability within datasets. When two groups are being examined for similarities or differences, variance comparison allows us to investigate the spread or variability of their data points.

In this context, variance comparison hinges upon the hypothesis that the variances of two independent samples are equal. By conducting an F-test, we aim to establish whether the observed differences in sample variances could have occurred by random chance, or if they are statistically significant. For instance, in our problem, we're comparing sample variances of 2.3 and 1.9 with equal degrees of freedom (since both sample sizes are 15, thus, 14 degrees of freedom).

Ultimately, if the calculated F-statistic falls within a critical value range, it suggests the variances are not significantly different, thereby favoring the null hypothesis. Thus, variance comparison via an F-test is an insightful way of determining the reliability and consistency of the underlying data.
Critical Value
A critical value is a threshold in hypothesis testing that determines whether the null hypothesis should be rejected. In practice, it is a predefined point on the statistical distribution that outlines the significance level's boundaries. It plays a fundamental role in identifying unusual test statistic values deemed significant either way on a two-tailed test.

For the assignment at hand, with a significance level \(\alpha = 0.05\), we are conducting a two-tailed test. The degrees of freedom are determined by the sample sizes as \(df_1 = 14\) and \(df_2 = 14\). Using an F-distribution table or calculator, you can find two critical values: one falls to the right tail of the distribution, and the other on the left. To compute these, we use the notation: \(F_{\alpha/2, df_1, df_2}\) for the upper and its reciprocal for the lower critical value.

The calculated F-statistic, 1.21 in our example, will be assessed against these critical points. Because it falls between the critical values, the conclusion is to not reject the null hypothesis. Essentially, the critical value marks the boundary for decision-making in hypothesis testing.
Sample Size Calculation
Determining an adequate sample size is vital to testing accuracy and reliability. In hypothesis testing, the objective is to ensure the sample size is sufficient for the power of the test, which is the probability of correctly rejecting a false null hypothesis. An inadequately sized sample can result in errors like false negatives (Type II errors).

Consider the stated problem if \(\sigma_{1}\) is twice \(\sigma_{2}\), we are tasked with achieving a power (1 - \(\beta\)) of 0.95. This informs us that the probability of making a Type II error should not exceed 0.05. Accordingly, the requisite sample size can be derived using specialized statistical software or pre-calculated tables that factor in expected effect size and the desired significance level.

In simpler terms, sample size calculation ensures that our findings in hypothesis testing are robust and conclusive. By defining the expected variance ratio and acceptable error rates, researchers and statisticians can determine the number of observations needed, hence minimizing the likelihood of incorrect inferences.

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Most popular questions from this chapter

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: Club 1: 0.8406,0.8104,0.8234,0.8198,0.8235,0.8562 $$0.8123,0.7976,0.8184,0.8265,0.7773,0.7871$$ 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 (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 \(\mathrm{Cl}\) 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 \(0.8 ?\)

Two chemical companies can supply a raw material. The concentration of a particular element in this material is important. The mean concentration for both suppliers is the same, but you suspect that the variability in concentration may differ for the two companies. The standard deviation of concentration in a random sample of \(n_{1}=10\) batches produced by company 1 is \(s_{1}=4.7\) grams per liter, and for company \(2,\) a random sample of \(n_{2}=16\) batches yields \(s_{2}=5.8\) grams per liter. Is there sufficient evidence to conclude that the two population variances differ? Use \(\alpha=0.05\).

The brcaking strength of yarn supplicd by two manufacturers is being investigated. You know from experience with the manufacturers' processes that \(\sigma_{1}=5\) psi and \(\sigma_{2}=4\) psi. A random sample of 20 test specimens from each manufacturer results in \(\bar{x}_{1}=88\) psi and \(\bar{x}_{2}=91\) psi, respectively. (a) Using a \(90 \%\) confidence interval on the difference in mean breaking strength, comment on whether or not there is cvidence to support the claim that manufacturer 2 produces yarn with higher mean breaking strength. (b) Using a \(98 \%\) confidence interval on the difference in mean breaking strength, comment on whether or not there is evidence to support the claim that manufacturer 2 produces yarn with higher mean breaking strength. (c) Comment on why the results from parts (a) and (b) are different or the same. Which would you choose to make your decision and why?

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2} .\) Suppose that sample sizes \(n_{1}=10\) and \(n_{2}=10,\) that \(\bar{x}_{1}=7.8\) and \(\bar{x}_{2}=5.6,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=9 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\) (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) if \(\mu_{1}\) is 3 units greater than \(\mu_{2} ?\) (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 3 units greater than \(\mu_{2}\) ? Assume that \(\alpha=0.05 .\)

Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2} .\) Suppose that sample sizes are \(n_{1}=15\) and \(n_{2}=15\) that \(\bar{x}_{1}=4.7\) and \(\bar{x}_{2}=7.8,\) and that \(s_{1}^{2}=4\) and \(s_{2}^{2}=6.25 .\) Assume that \(\sigma_{1}^{2}=\sigma_{2}^{2}\) and that the data are drawn from normal distributions. Use \(\alpha=0.05\). (a) Test the hypothesis and find the \(P\) -value. (b) Explain how the test could be conducted with a confidence interval. (c) What is the power of the test in part (a) for a true difference in means of \(3 ?\) (d) Assume that sample sizes are equal. What sample size should be used to obtain \(\beta=0.05\) if the true difference in means is -2 ? Assume that \(\alpha=0.05\).

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