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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) Assume that sample sizes are equal. 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
The hypothesis test does not reject \(H_{0}\) as P-value = 0.1757. The confidence interval (-4.79, 11.19) includes zero, confirming this. For power 0.0401 when \(\mu_1 = \mu_2 + 2\), sample size should be 32 for \(\beta = 0.05\) and equal groups.

Step by step solution

01

Calculate the Test Statistic

We begin by calculating the test statistic for the hypothesis test. Since the variances are known, we use the formula for the two-sample Z-test: \[ Z = \frac{(\bar{x}_{1} - \bar{x}_{2}) - (\mu_{1} - \mu_{2})}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}}} \]Plugging in the values, we get:\[ Z = \frac{(24.5 - 21.3) - 0}{\sqrt{\frac{10^{2}}{10} + \frac{5^{2}}{15}}} = \frac{3.2}{\sqrt{10 + \frac{25}{15}}} = \frac{3.2}{\sqrt{11.67}} \approx 0.935 \]Thus, the Z-test statistic is approximately 0.935.
02

Determine P-value from Z-Test

The P-value for a right-tailed test where \(Z = 0.935\) is found using the standard normal distribution table. The cumulative probability for Z=0.935 is approximately 0.8243. Since we have a right-tailed test, the P-value is:\[ P = 1 - 0.8243 = 0.1757 \]Thus, the P-value is approximately 0.1757.
03

Compare P-value with Significance Level

Compare the computed P-value (0.1757) with the significance level (\(\alpha = 0.01\)). Because the P-value (0.1757) is greater than \(\alpha\), we do not reject the null hypothesis \(H_{0}\). The conclusion is that there is not enough evidence to support the alternative hypothesis \(\mu_{1} > \mu_{2}\) at the 1% level.
04

Conduct Test with Confidence Interval

To use a confidence interval, calculate the critical value \(Z_{\alpha}\) for \(\alpha = 0.01\), which is approximately 2.33 for a right-tail test. The confidence interval for the difference in means is:\[ \hat{\mu_{1}} - \hat{\mu_{2}} \pm Z_{\alpha} \times \sqrt{\frac{\sigma_{1}^{2}}{n_{1}} + \frac{\sigma_{2}^{2}}{n_{2}}} \]\[ 3.2 \pm 2.33 \times \sqrt{11.67} = 3.2 \pm 7.99 \]Interval: \([-4.79, 11.19]\). Since 0 lies within this interval, it supports our decision not to reject \(H_{0}\).
05

Calculate the Power of the Test

If \(\mu_{1} = \mu_{2} + 2\), the test's power is calculated using the non-centrality parameter \(\delta\) in the test statistic:\[ \delta = \frac{2 - 0}{\sqrt{11.67}} \approx 0.58 \]To calculate the power, determine the Z-score at which we would reject the null hypothesis:\[ Z = Z_{\alpha} - \delta \approx 2.33 - 0.58 = 1.75 \]Thus, the power is the probability \(P(Z > 1.75)\), approximately 0.0401 after using normal distribution tables.
06

Find Required Sample Size for Equal Sizes

For equal sample sizes \(n\), to achieve \(\beta = 0.05\) and \(\alpha = 0.05\): 1. \(Z_{\alpha} \approx 1.645\) (for two-tailed \(\alpha = 0.05\)) and \(Z_{\beta} \approx 1.645\) (for \(\beta = 0.05\)).2. Use the formula: \[ n = \left( \frac{(Z_{\alpha} + Z_{\beta}) \times \sqrt{\sigma_{1}^{2} + \sigma_{2}^{2}}}{\text{Effect Size}} \right)^{2} \]3. Plug \(\sigma_{1}^{2}=10, \sigma_{2}^{2}=5, \text{Effect Size} = 2\).\[ n = \left( \frac{(1.645 + 1.645) \times 3.87}{2} \right)^{2} \approx 32 \]Thus, the sample size needed for equal groups is approximately 32.

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Key Concepts

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

Z-test
A Z-test is a type of statistical test used to determine whether there are meaningful differences between two population means. Considered a parametric test, it requires that certain conditions, like normal distribution of the data and known variances, are met.
For this exercise, we're dealing with two sample means: 24.5 and 21.3, with known variances. The calculation of the Z-test statistic involves computing the difference between sample means and dividing by the standard error of the difference.
  • This critical value allows us to determine whether the observed differences between means are statistically significant, at a given level of significance (in the example, 0.01).
The Z-test is emphasizing normal distribution and knowledge of variance, which makes it specific, albeit powerful, under the right conditions.
Confidence Intervals
Confidence intervals provide a range of values which are believed to contain the true difference between two population means. They give us an intuitive sense of how much the sample means could vary if the experiment were repeated.
In our exercise, the confidence interval for the difference is calculated as 3.2, adjusted by the critical value times the standard error. This forms the boundaries -
  • Lower Boundary: 3.2 − 7.99 = -4.79
  • Upper Boundary: 3.2 + 7.99 = 11.19
The interval [-4.79, 11.19] encompasses zero, which suggests that there isn't enough evidence to indicate a significant difference at our 1% significance level. Confidence intervals provide additional insights beyond hypothesis testing, helping substantiate or challenge conclusions drawn from statistical tests.
P-value
The P-value indicates the probability of observing results as extreme as, or more so than, the actual results (*assuming the null hypothesis is true*). It determines how surprised we'd be if the test result were observed again.
For a Z-test with a statistic of 0.935, we compute the P-value using the standard normal distribution table.
  • P-value = 1 - 0.8243 = 0.1757
Because our P-value (0.1757) exceeds the significance threshold (0.01), we lack sufficient evidence to reject the null hypothesis. The P-value serves as a crucial indicator, providing a clear measure of statistical significance versus arbitrary cutoffs.
Power of a Test
The power of a test reflects the likelihood of correctly rejecting a false null hypothesis. It indicates how well the test can detect effects when they genuinely exist. High power is always desirable as it implies greater reliability of test results.In our example, if the true difference in means (\( u \)) is 2, we calculate the power by first computing the non-centrality parameter and then determining the probability of obtaining a relevant Z-score:
  • Non-centrality parameter *\( \delta \)* = *\( 0.58 \)*
  • Power is the probability of *Z* > *1.75*, approximately *0.0401*.
Our calculated power here indicates a low probability of detecting a true difference, suggesting improvement or adjustments might be needed for better test reliability.
Sample Size Calculation
Determining the proper sample size is vital, influencing the balance between statistical power and practical limitations. It ensures that tests carried out are both efficient and accurate.For equal sample sizes with desired power (1-\( \beta \)) of 0.05:
  • Use *\( Z_{\alpha} \approx 1.645 \)* and *\( Z_{\beta} \approx 1.645 \)*
  • Calculate using: \[ n = \left( \frac{(Z_{\alpha} + Z_{\beta}) \times \sqrt{2 \times 5}}{2} \right)^2 \]
  • Result is approximately 32 samples per group.
An appropriately sized sample enables more effective testing, offering stronger confidence in results and minimizing risks of false conclusions due to sampling variability.

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Most popular questions from this chapter

The manufacturer of a hot tub is interested in testing two different heating elements for its product. The element that produces the maximum heat gain after 15 minutes would be preferable. The manufacturer obtains 10 samples of each heating unit and tests each one. The heat gain after 15 minutes (in \({ }^{\circ} \mathrm{F}\) ) follows. $$\begin{array}{l|l}\text { Unit 1: } & 25,27,29,31,30,26,24,32,33,38 \\\\\hline \text { Unit 2: } & 31,33,32,35,34,29,38,35,37,30\end{array}$$ (a) Is there any reason to suspect that one unit is superior to the other? Use \(\alpha=0.05\) and the Wilcoxon rank sum test. (b) Use the normal approximation for the Wilcoxon rank sum test. Assume that \(\alpha=0.05 .\) What is the approximate \(P\) -value for this test statistic?

Consider the hypothesis test \(H_{0}: \sigma_{1}^{2}=\sigma_{2}^{2}\) against \(H_{1}: \sigma_{1}^{2}>\sigma_{2}^{2},\) Suppose that the sample sizes are \(n_{1}=20\) and \(n_{2}=8,\) and that \(s_{1}^{2}=4.5\) and \(s_{2}^{2}=2.3 .\) Use \(\alpha=0.01 .\) Test the hypothesis and explain how the test could be conducted with a confidence interval on \(\sigma_{1} / \sigma_{2}\),

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 \(\alpha=0.05 .\)

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?

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 six 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?

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