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Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1}>\mu_{2}\) with known variances \(\sigma_{1}=10\) and \(\sigma_{2}=5\) Suppose that sample sizes \(n_{1}=10\) and \(n_{2}=15\) and that \(\bar{x}_{1}=24.5\) and \(\bar{x}_{2}=21.3 .\) Use \(\alpha=0.01\). (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 2 units greater than \( \mu_{2}\)? (d) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if \(\mu_{1}\) is 2 units greater than \(\mu_{2} ?\) Assume that \(\alpha=0.05 .\)

Short Answer

Expert verified
(a) Fail to reject \( H_0 \); P-value \( \approx 0.1736. \) (b) Confidence interval [-5.71, 12.71] includes zero. (c) Power \( \approx 0.64. \) (d) Sample size \( \approx 44. \)

Step by step solution

01

State the Null and Alternate Hypotheses

The given hypotheses for the problem are\( H_0: \mu_1 = \mu_2 \) (the means are equal) and \( H_1: \mu_1 > \mu_2 \) (the mean of the first population is greater).
02

Calculate the Test Statistic

The test statistic for comparing two means with known variances is given by:\[ Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Substituting the given values:\( \bar{x}_1 = 24.5, \bar{x}_2 = 21.3, \sigma_1 = 10, \sigma_2 = 5, n_1 = 10, n_2 = 15 \), we have:\[ Z = \frac{24.5 - 21.3}{\sqrt{\frac{10^2}{10} + \frac{5^2}{15}}} = \frac{3.2}{\sqrt{10 + \frac{25}{15}}} = \frac{3.2}{\sqrt{11.67}} \approx 0.94 \]
03

Determine the Critical Value and P-value

Given \( \alpha = 0.01 \) for a one-tailed test, look up the critical value for \( Z \) in a standard normal distribution table. The critical value is approximately 2.33. Since the calculated \( Z \approx 0.94 \) does not exceed 2.33, we fail to reject \( H_0 \). The P-value for \( Z = 0.94 \) is approximately 0.1736.
04

Interpret the Hypothesis Test Result

Since the P-value (0.1736) is greater than \( \alpha = 0.01 \), we do not reject the null hypothesis. There is not enough evidence to support that \( \mu_1 > \mu_2 \).
05

Explain Test with Confidence Interval

To conduct a test with a confidence interval, we calculate the interval for \( (\bar{x}_1 - \bar{x}_2) \) at a confidence level of 99%. The interval is given by:\[ (\bar{x}_1 - \bar{x}_2) \pm Z_{\alpha/2} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]For \( Z_{\alpha/2} = 2.58 \), the interval is:\[ 3.2 \pm 2.58 \times \sqrt{11.67} \approx 3.2 \pm 8.91 \]This results in the interval [-5.71, 12.71], which includes zero, indicating no significant difference at the 1% level.
06

Compute the Power of Test

To find the power when \( \mu_1 = \mu_2 + 2 \), adjust the test statistic:\[ Z = \frac{(\bar{x}_1 - \bar{x}_2) - 2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Using \( 3.2 - 2 \approx 1.2 \), the new \( Z \) is:\[ Z = \frac{1.2}{\sqrt{11.67}} \approx 0.35 \]This results in a power of approximately 0.64, meaning a 64% chance to detect a true difference of 2.
07

Calculate Required Sample Size for Given \( \beta \)

To find the sample size necessary for \( \beta = 0.05 \) with the same difference of 2, set power (1-β) = 0.95. Use the formula:\[ n = \left( \frac{Z_{\alpha} + Z_{1-\beta}}{\delta / \sqrt{\frac{\sigma_1^2 + \sigma_2^2}{2}}} \right)^2 \]Where \( \delta = 2, Z_{0.05} \approx 1.645, Z_{0.95} \approx 1.645 \). The sample size is:\[ n = \left( \frac{1.645 + 1.645}{2 / \sqrt{\frac{100 + 25}{2}}} \right)^2 \approx 44 \] per group.

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Key Concepts

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

Test Statistic
In hypothesis testing, the test statistic is a crucial element. It helps determine whether to reject the null hypothesis. When comparing two means with known variances, the test statistic is often calculated using the formula:\[Z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]This formula combines the sample means (\(\bar{x}_1\) and \(\bar{x}_2\)), the known variances (\(\sigma_1\) and \(\sigma_2\)), and the sizes of the samples (\(n_1\) and \(n_2\)). By substituting the given values, we determine if there's enough statistical evidence to support our alternate hypothesis.The result tells us how many standard deviations our observed difference is from the expected difference under the null hypothesis. This guides us in making decisions regarding our hypotheses.
P-Value
The p-value is a probability measure that helps us interpret the result of a hypothesis test. It quantifies the evidence against the null hypothesis.- A small p-value (typically \(\leq\) 0.05) indicates strong evidence against the null hypothesis, suggesting it is unlikely to be true.- A large p-value (\(\geq\) 0.05) suggests weak evidence against the null hypothesis, meaning we fail to reject it.In our test, with a p-value of approximately 0.1736, this indicates there's not enough evidence to reject the null hypothesis using an alpha level of 0.01. We interpret this to mean that there isn't significant evidence to conclude that the first population mean is greater than the second.
Confidence Interval
Confidence intervals provide a range of values within which we believe the true parameter lies, with a certain level of confidence.To create a confidence interval for the difference between two means, we use:\[(\bar{x}_1 - \bar{x}_2) \pm Z_{\alpha/2} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]In this case, for a 99% confidence level, we find an interval of \([-5.71, 12.71]\). Since this interval includes zero, it implies no significant difference between the means at the 1% level. Confidence intervals give us a practical way of considering hypothesis testing, offering a 'range' rather than a simple "accept/reject" decision.
Sample Size Calculation
Calculating an appropriate sample size is essential for ensuring our test has sufficient power to detect a true effect.To compute the sample size based on desired power and type I and II errors, we use:\[n = \left( \frac{Z_{\alpha} + Z_{1-\beta}}{\delta / \sqrt{\frac{\sigma_1^2 + \sigma_2^2}{2}}} \right)^2\]This formula considers the effect size \(\delta\), the critical values for \(\alpha\) (type I error) and \(\beta\) (type II error). It ensures a balance between finding true effects and minimizing false negatives or positives.For our scenario, to achieve a power of 0.95 (\(\beta = 0.05\)), the required sample size per group is about 44. Such calculations help design effective experiments that address the research questions properly with adequate power to detect true differences when they exist.

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Most popular questions from this chapter

A polymer is manufactured in a batch chemical process. Viscosity measurements are normally made on each batch, and long experience with the process has indicated that the variability in the process is fairly stable with \(\sigma=20\). Fifteen batch viscosity measurements are given as follows: $$724,718,776,760,745,759,795,756,742,740,761,749, 739,747,742$$ A process change is made which involves switching the type of catalyst used in the process. Following the process change, eight batch viscosity measurements are taken: $$735,775,729,755,783,760,738,780$$ Assume that process variability is unaffected by the catalyst change. If the difference in mean batch viscosity is 10 or less, the manufacturer would like to detect it with a high probability. (a) Formulate and test an appropriate hypothesis using \(\alpha=\) \(0.10 .\) What are your conclusions? Find the \(P\) -value. (b) Find a \(90 \%\) confidence interval on the difference in mean batch viscosity resulting from the process change. (c) Compare the results of parts (a) and (b) and discuss your findings.

An electrical engineer must design a circuit to deliver the maximum amount of current to a display tube to achieve sufficient image brightness. Within her allowable design constraints, she has developed two candidate circuits and tests prototypes of each. The resulting data (in microamperes) are as follows: $$\begin{array}{l|l}\text { Circuit } 1: & 251,255,258,257,250,251,254,250,248 \\\\\hline \text { Circuit } 2: & 250.253 .249 .256 .259 .252 .260 .251\end{array}$$ (a) Use the Wilcoxon rank-sum test to test \(H_{0}: \mu_{1}=\mu_{2}\) against the alternative \(H_{1}: \mu_{1}>\mu_{2} .\) Use \(\alpha=0.025 .\) (b) Use the normal approximation for the Wilcoxon rank-sum test. Assume that \(\alpha=0.05 .\) Find the approximate \(P\) -value for this test statistic.

The melting points of two alloys used in formulating solder were investigated by melting 21 samples of each material. The sample mean and standard deviation for alloy 1 was \(\bar{x}_{1}=420^{\circ} \mathrm{F}\) and \(s_{1}=4^{\circ} \mathrm{F},\) while for alloy 2 they were \(\bar{x}_{2}=426^{\circ} \mathrm{F}\) and \(s_{2}=3^{\circ} \mathrm{F}\). (a) Do the sample data support the claim that both alloys have the same melting point? Use \(\alpha=0.05\) and assume that both populations are normally distributed and have the same standard deviation. Find the \(P\) -value for the test. (b) Suppose that the true mean difference in melting points is \(3^{\circ} \mathrm{F}\). How large a sample would be required to detect this difference using an \(\alpha=0.05\) level test with probability at least 0.9 ? Use \(\sigma_{1}=\sigma_{2}=4\) as an initial estimate of the common standard deviation.

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