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Consider the hypothesis test \(H_{0}: \mu_{1}=\mu_{2}\) against \(H_{1}: \mu_{1} \neq \mu_{2}\) with known variances \(\sigma_{1}=10\) and \(\sigma_{2}=5\) Suppose that sample sizes \(n_{1}=10\) and \(n_{2}=15\) and that \(\bar{x}_{1}=4.7\) and \(\bar{x}_{2}=7.8 .\) 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) Assuming equal sample sizes, what sample size should be used to obtain \(\beta=0.05\) if the true difference in means is \(3 ?\) Assume that \(\alpha=0.05 .\)

Short Answer

Expert verified
(a) Fail to reject \( H_0 \); P-value \( \approx 0.3648 \). (b) Use a CI for difference: includes 0. (c) Power \( \approx 0.514 \). (d) \( n = 45 \) per group.

Step by step solution

01

State the Test Hypotheses

The null hypothesis is that the population means are equal, \( H_0: \mu_1 = \mu_2 \), and the alternative hypothesis is that they are not equal, \( H_1: \mu_1 eq \mu_2 \). This is a two-tailed test at a significance level of \( \alpha = 0.05 \).
02

Calculate the Test Statistic

The test statistic for the difference between two means with known variances is given by:\[z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\]Substituting the given values:\[z = \frac{(4.7 - 7.8) - 0}{\sqrt{\frac{10^2}{10} + \frac{5^2}{15}}} = \frac{-3.1}{\sqrt{10 + \frac{25}{15}}} = \frac{-3.1}{\sqrt{11.6667}} \approx \frac{-3.1}{3.41565} \approx -0.908\]
03

Determine the Critical Value and Decision

For a two-tailed test at \( \alpha = 0.05 \), the critical values for \( z \) are approximately \( \pm 1.96 \). Since the calculated \( z \approx -0.908 \) is not beyond the critical values, we fail to reject the null hypothesis.
04

Calculate the P-value

The \( P \)-value is the probability of observing a test statistic as extreme as \( z = -0.908 \) under the null hypothesis. Using the standard normal distribution, the \( P \)-value corresponds to \( 2 \times P(Z < -0.908) \approx 0.3648 \).
05

Explain the Confidence Interval Method

A confidence interval for the difference in means can be used to conduct the test. The confidence interval is given by:\[CI = (\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = -3.1 \pm 1.96 \times 3.41565\]This results in a confidence interval of approximately \(-9.793 \) to \(3.593\), which contains zero, thus supporting our decision to not reject \( H_0 \).
06

Calculate the Power of the Test

The power of the test for a true difference in means of \( \delta = 3 \) is calculated using:\[\beta = P\left(Z < z_{\alpha/2} - \frac{\delta}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\right) + P\left(Z > -z_{\alpha/2} + \frac{\delta}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}}\right)\]Substituting \( \delta = 3 \), compute:\[\beta \approx P(Z < 0.02) + P(Z > -2.02) \approx 0.508 + 0.978 \approx 0.486\]Thus, the power is \( 1 - \beta \approx 0.514 \).
07

Determine Sample Size for Desired Power

To achieve a power of 0.95 (\( \beta = 0.05 \)) given \( \delta = 3 \) and \( \alpha = 0.05 \), use the formula for sample size:\[n = \left(\frac{(z_{\alpha/2} + z_{\beta})^2 \times (\sigma_1^2 + \sigma_2^2)}{\delta^2}\right)\]For \( z_{\alpha/2} = 1.96 \) and \( z_{\beta} = 1.645 \): \[n = \left(\frac{(1.96 + 1.645)^2 \times (100 + 25)}{9}\right) \approx 44.44\]Thus, a sample size of \( n = 45 \) per group is required for the desired power.

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Key Concepts

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

Two-Tailed Test
In hypothesis testing, the two-tailed test is used when we want to test whether the population mean is different from a specific value, in either direction. This means we are concerned with deviations on both sides of the mean. In this context, it is used to check if two means are equal or not. For such tests, you set up a null hypothesis (H_0: \mu_1 = \mu_2) stating that the means are equal. The alternative hypothesis (H_1: \mu_1 eq \mu_2) suggests that they are not equal, and it is crafted to cover the possibility that the means could differ in any direction, hence the term "two-tailed."
The significance level, denoted by \( \alpha \), is a critical part of setting up your hypothesis test. With a common choice of \( \alpha = 0.05 \), you are essentially saying that there is a 5% risk you will wrongly reject the null hypothesis. The critical values, often found in standard normal distribution tables, will help determine whether your calculated test statistic falls into the rejection region. If this test statistic crosses the critical value (e.g., \( \pm 1.96 \) for a two-tailed test at 5% significance level), you reject the null hypothesis. Otherwise, there isn't sufficient evidence to do so.
Confidence Interval
A confidence interval provides a range of values within which you can be confident that the true population parameter lies. In the context of hypothesis testing, confidence intervals can be used to draw conclusions about the means.
To form a confidence interval for the difference of two means, you take the difference of sample means and calculate:\[CI = (\bar{x}_1 - \bar{x}_2) \pm z_{\alpha/2} \times \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}\]
This range, referred to as the confidence interval, is centered around the observed difference between means. The margin of error, defined by the critical value (\(z_{\alpha/2}\)) and the standard error, defines the extent of the confidence interval either side of the mean difference. If this interval includes zero, it implies that there is no statistically significant difference between the means, aligning with the null hypothesis. Conversely, if zero is not within this interval, the null hypothesis may be rejected, indicating a significant difference between the means.
Power of the Test
The power of the test in a hypothesis test is defined as the probability that the test correctly rejects the null hypothesis when the alternative hypothesis is true. High power means that the test is more likely to detect an effect when there is one.
Power is influenced by several factors:
  • The size of the effect (e.g., the true difference between the means)
  • The sample size
  • The significance level (\(\alpha\))
  • The variability in the data (standard deviation)
Calculating power involves determining \(\beta\), the probability of a Type II error, i.e., failing to reject the null hypothesis when the alternative hypothesis is true. Power is then \(1-\beta\). To compute it for a given alternative hypothesis (like a true difference in means \(\delta\)), you determine:\[\beta = P\left(Z < z_{\alpha/2} - \frac{\delta}{\text{Standard Error}}\right) + P\left(Z > -z_{\alpha/2} + \frac{\delta}{\text{Standard Error}}\right)\]
The higher the power, the more reliable the test's detection capability is, assuming there is truly a difference.
Sample Size Calculation
Sample size is a crucial element of study design in hypothesis testing, directly impacting the power of the test. Simply put, larger sample sizes tend to provide more reliable results. To achieve a desired power (e.g., 0.95, indicating a 95% chance of correctly rejecting a false null hypothesis), you calculate the necessary sample size using specific formulas.
Given it's a common requirement to ensure adequate power, you use:\[n = \left(\frac{(z_{\alpha/2} + z_{\beta})^2 \times (\sigma_1^2 + \sigma_2^2)}{\delta^2}\right)\]
• \(z_{\alpha/2}\) is the critical value for the chosen significance level
• \(z_{\beta}\) corresponds to the desired power level
• \(\sigma_1^2\) and \(\sigma_2^2\) are the variances of the two samples
• \(\delta\) is the true difference in means you aim to detect
This calculation guides how many observations you need in each group to adequately test your hypothesis with the targeted power. Achieving this makes the results more robust and reliable, reducing the likelihood of Type II errors.

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Most popular questions from this chapter

The thickness of a plastic film (in mils) on a substrate material is thought to be influenced by the temperature at which the coating is applied. A completely randomized experiment is carried out. Eleven substrates are coated at \(125^{\circ} \mathrm{F}\), resulting in a sample mean coating thickness of \(\bar{x}_{1}=103.5\) and a sample standard deviation of \(s_{1}=10.2\). Another 13 substrates are coated at \(150^{\circ} \mathrm{F}\), for which \(\bar{x}_{2}=99.7\) and \(s_{2}=20.1\) are observed. It was originally suspected that raising the process temperature would reduce mean coating thickness. (a) Do the data support this claim? Use \(\alpha=0.01\) and assume that the two population standard deviations are not equal. Calculate an approximate \(P\) -value for this test. (b) How could you have answered the question posed regarding the effect of temperature on coating thickness by using a confidence interval? Explain your answer.

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