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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 \(\alpha=0.05 .\)

Short Answer

Expert verified
Fail to reject the null hypothesis; there is not enough evidence to show \( \mu_1 \neq \mu_2 \). Power is approximately 0.783 and each sample size should be 10 for \( \beta=0.05 \).

Step by step solution

01

State the Hypotheses

We need to define our null hypothesis and alternative hypothesis. \( H_0: \mu_1 = \mu_2 \) and \( H_1: \mu_1 eq \mu_2 \). This is a two-tailed test because we are testing for equality, not specifically greater or less.
02

Calculate the Test Statistic

First, calculate the pooled standard deviation because \( \sigma_1^2 = \sigma_2^2 \).The pooled variance \( s_p^2 \) is \(\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2} = \frac{(10 - 1)4 + (10 - 1)9}{10 + 10 - 2} = \frac{36 + 81}{18} = 6.5\).The pooled standard deviation \( s_p \) is \( \sqrt{6.5} \approx 2.55 \).Then calculate the test statistic \( t \):\[t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{7.8 - 5.6}{2.55 \sqrt{\frac{1}{10} + \frac{1}{10}}} = \frac{2.2}{1.138} \approx 1.93.\]
03

Find the Critical Value and P-value

For a two-tailed test with \( \alpha = 0.05 \) and degrees of freedom \( n_1 + n_2 - 2 = 18 \), the critical t-value is approximately \( \pm 2.101 \). The calculated \( t \) statistic of 1.93 does not exceed the critical value.Use a t-distribution table or calculator to find the P-value. The two-tailed P-value for \( t = 1.93 \) and \( df = 18 \) is approximately 0.069.
04

Conclusion of Hypothesis Test

Since the P-value (0.069) is greater than \( \alpha \) (0.05), we fail to reject the null hypothesis. There is not enough evidence to suggest that \( \mu_1 eq \mu_2 \).
05

Confidence Interval Approach

To use a confidence interval, compute a 95% confidence interval for \( \mu_1 - \mu_2 \), using the result from Step 2:\[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2}(s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}) = 2.2 \pm 2.101 \times 1.138 = 2.2 \pm 2.388.\]The interval is \([-0.188, 4.788]\). Since 0 is within this interval, we do not reject \( H_0 \).
06

Calculate the Power of the Test

Assuming \( \mu_1 = \mu_2 + 3 \), we calculate the non-centrality parameter \( \delta = \frac{\Delta}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} = \frac{3}{1.138} \approx 2.63 \).The power is the probability of rejecting \( H_0 \) when \( H_1 \) is true. Using a table or software for non-central t-distributions, find the probability of \( t \) being more extreme than the critical value of \( 2.101 \) with `δ=2.63`. The power is found to be approximately 0.783.
07

Determine Required Sample Size for Power

Using the relationship for power, given \( \beta = 0.05 \) and \( \alpha = 0.05 \), solve for \( n \):\(z_{1 - \alpha/2} + z_{1 - \beta} = \delta \sqrt{n} \rightarrow n = \left( \frac{z_{1-\alpha/2} + z_{1-\beta}}{\frac{\mu_1 - \mu_2}{\sigma}}\right)^2\).Using standard Z values: \( z_{0.975} = 1.96 \) and \( z_{0.95} = 1.645 \).\(n = \left(\frac{1.96 + 1.645}{3/2.55}\right)^2 = \left(\frac{3.605}{1.176}\right)^2 \approx 9.4 \text{ (round up to 10)}.\)Thus, each sample should be size 10.

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Key Concepts

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

P-value
The P-value is a crucial concept in hypothesis testing. It helps determine the strength of the evidence against the null hypothesis. When conducting a test, we compare the P-value to a significance level, often denoted as \( \alpha \). This significance level is typically set at common thresholds like 0.05 or 0.01.

When the P-value is less than or equal to \( \alpha \), we reject the null hypothesis. This suggests that the observed data is unusual under the assumption that the null hypothesis is true. However, if the P-value is greater, we fail to reject the null hypothesis, implying insufficient evidence to support the alternative hypothesis. It's important to note that failing to reject the null is not the same as accepting it as true.

In the original exercise, a two-tailed test was used to determine the P-value of approximately 0.069. Since this is greater than \( \alpha = 0.05 \), the null hypothesis \( \mu_1 = \mu_2 \) cannot be rejected. This indicates no strong evidence of a difference between \( \mu_1 \) and \( \mu_2 \).
Confidence Interval
A confidence interval provides a range of values for an estimate and gives an idea of the uncertainty surrounding that estimate. For a 95% confidence interval, there's a 95% probability that this interval includes the true population parameter. Confidence intervals are useful because they provide more information than a hypothesis test, offering insight into the possible range of the parameter difference.

In hypothesis testing, if a confidence interval for a parameter difference does not include zero, it suggests significant evidence against the null hypothesis. For our test involving \( \mu_1 - \mu_2 \), the calculated interval was \([-0.188, 4.788]\).

Since zero is within this interval, it supports the decision to not reject the null hypothesis. This means you're likely to find \( \mu_1 \) equal to \( \mu_2 \) based on the data collected. A larger interval suggests greater variability and uncertainty in the estimate.
Power of a Test
The power of a test measures its ability to correctly reject a false null hypothesis. In other words, it indicates the probability of avoiding a Type II error (failing to reject a false null hypothesis). A high test power, generally above 0.8, implies a strong test capable of detecting an actual effect when it exists.

Power depends on various factors, including the chosen significance level \( \alpha \), sample size, and effect size. Larger samples or more pronounced differences increase power. In the exercise, the test power was calculated with given conditions: \( \mu_1 \) being 3 units greater than \( \mu_2 \). The result showed a power of approximately 0.783. Although somewhat less than the preferred threshold of 0.8, it still implies a reasonably good likelihood of correctly detecting a real difference.

Increasing the sample size or raising the significance level could further boost this power, leading to more reliable results.
Sample Size Determination
Calculating the appropriate sample size is crucial in hypothesis testing to ensure adequate power. The aim is to have a sufficient number of observations to detect a true effect without being influenced by random chance. The bigger the sample size, the more trustworthy the results; however, excessively large samples can lead to unnecessary resource use.

In sample size determination, the desired power level (commonly 0.8 or 80%), significance level \( \alpha \), and effect size need consideration. Rounding decisions often accompany sample size calculations, ensuring practical application.

In this exercise, the task involved determining the sample size to achieve a power of 0.95, assuming \( \mu_1 \) is 3 units greater than \( \mu_2 \). Calculations suggested each group require 10 samples. It's key to evaluate these elements carefully since sample size impacts the study's capability to draw valid conclusions.

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Most popular questions from this chapter

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