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Suppose a level \(.05\) test of \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{2}: \mu_{1}-\mu_{2}>0\) is to be performed, assuming \(\sigma_{1}=\sigma_{2}=10\) and normality of both distributions, using equal sample sizes \((m=n)\). Evaluate the probability of a type II error when \(\mu_{1}-\mu_{2}=1\) and \(n=25,100,2500\), and 10,000 . Can you think of real problems in which the difference \(\mu_{1}-\mu_{2}=1\) has little practical significance? Would sample sizes of \(n=10,000\) be desirable in such problems?

Short Answer

Expert verified
Type II error decreases with larger sample sizes, showing no practical significance in large samples for small differences.

Step by step solution

01

Define Key Terms and Parameters

A Type II error occurs when we fail to reject the null hypothesis when the alternative hypothesis is true. Here we want to find the probability of Type II error when the true difference between means \( \mu_1 - \mu_2 = 1 \) (alternative hypothesis) while testing \( H_0: \mu_1 - \mu_2 = 0 \) at a significance level of 0.05.
02

Determine the Test Statistic

The test statistic for comparing two means with known variances under normality is given by \( Z = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\sigma^2/m + \sigma^2/n}} \). Since \( m = n \), the formula simplifies to \( Z = \frac{(\bar{X}_1 - \bar{X}_2)}{\sigma \sqrt{2/n}} \). We know \( \sigma = 10 \), so \( Z = \frac{(\bar{X}_1 - \bar{X}_2)}{10 \sqrt{2/n}} \).
03

Determine the Critical Value

For a one-sided test with \( \alpha = 0.05 \), the critical value for \( Z \) from the standard normal distribution is approximately 1.645. We want the probability that the test statistic falls less than this value when \( \mu_1 - \mu_2 = 1 \), or equivalently, \( Z < 1.645 \).
04

Calculate the Power Function

When \( \mu_1 - \mu_2 = 1 \), the distribution of \( Z \) becomes \( Z = \frac{(\bar{X}_1 - \bar{X}_2 - 1)}{10 \sqrt{2/n}} \). Adjusting this for the non-centrality parameter, \( \lambda = \frac{1}{10/\sqrt{n/2}} = \frac{\sqrt{n/2}}{10} \). So, \( Z \) follows a normal distribution with a mean of \( \lambda \) and a standard deviation of 1.
05

Calculate Type II Error Probability for Different Sample Sizes

We calculate \( P(Z < 1.645 - \lambda) \) for each sample size, which gives the probability of a Type II error. \- For \( n = 25 \), \( \lambda = \frac{\sqrt{12.5}}{10} \approx 0.3536 \). The probability is \( \Phi(1.645 - 0.3536) \approx 0.6793 \).- For \( n = 100 \), \( \lambda = 0.7071 \). The probability is \( \Phi(1.645 - 0.7071) \approx 0.3821 \).- For \( n = 2500 \), \( \lambda = 3.5355 \). The probability is \( \Phi(1.645 - 3.5355) \approx 0.0002 \).- For \( n = 10000 \), \( \lambda = 7.0711 \). The probability is \( \Phi(1.645 - 7.0711) \approx 0 \).
06

Interpret the Results

As the sample size increases, the probability of a Type II error decreases significantly, implying a higher power of the test. Small differences such as \( \mu_1 - \mu_2 = 1 \) might be statistically significant with large samples like \( n = 10000 \), but they could lack practical significance. In real-world problems, where such small differences are not impactful, large sample sizes might be unnecessary and economically undesirable.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method that helps us make decisions about the population based on sample data. It involves formulating two competing hypotheses: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)). The null hypothesis represents the baseline or a default assumption, which often posits no effect or no difference.
The alternative hypothesis is what we aim to support, indicating that there is an effect or a difference. For example, in our exercise, \( H_0: \mu_1 - \mu_2 = 0 \) suggests no difference, while \( H_a: \mu_1 - \mu_2 > 0 \) suggests a positive difference.
To assess these hypotheses, we utilize a significance level, usually denoted by \( \alpha \), which reflects the threshold for rejecting \( H_0 \). A common choice is \( \alpha = 0.05 \), meaning there's a 5% risk of incorrectly rejecting the null hypothesis.
Variance
Variance measures the spread of a set of numbers. It is a crucial concept in hypothesis testing because it affects the calculations of test statistics. Higher variance means that data points are more spread out from the mean, making precise predictions harder.
In the given scenario, both groups have the same variance of \( \sigma^2 = 100 \) (since \( \sigma = 10 \)). If variances were different, we would have to account for this discrepancy in our calculations, which might influence our test's power and the likelihood of Type II error.
  • Variance determines how data is distributed.
  • It impacts the reliability of statistical tests.
  • Consistent variance between groups simplifies comparative analysis.
Normal Distribution
The normal distribution is a continuous probability distribution characterized by its symmetrical bell shape. Most occurrences (68%) distribute within one standard deviation of the mean, hence why it's foundational in statistics. This property is essential when conducting hypothesis tests because test statistics often rely on assumptions of normality.
For our exercise, the distributions of the sample means are assumed to be normal, which allows us to use the \( Z \)-statistic in testing. The \( Z \)-test uses the standard normal distribution to calculate probabilities, making it easier to make inferences about population parameters based on samples.
  • Normal distribution's symmetry simplifies calculations.
  • It underpins many statistical tests and confidence interval estimations.
  • Assumption of normality can hold via the Central Limit Theorem, even if data aren't normally distributed.
Sample Size
Sample size significantly influences the power of a hypothesis test, impacting the probability of Type I and Type II errors. Larger sample sizes generally offer more reliable results, reducing the chance of Type II error (failing to reject a false \( H_0 \)). In the exercise, different sample sizes (\( n = 25, 100, 2500, 10000 \)) highlight how increasing sample size decreases the probability of a Type II error.
When the sample size is small, it's easier for random variation to mimic or conceal effects, increasing the probability of Type II errors. However, exceedingly large sample sizes, as noted in real-world problems, may detect statistically significant differences that aren't necessarily practical or meaningful, such as when \( \mu_1 - \mu_2 = 1 \) with \( n = 10000 \). Thus, choosing an appropriate sample size is essential for balancing statistical power and practical significance.
  • Small sample sizes may not adequately represent the population.
  • Large samples provide more accurate estimates of population parameters.
  • It's important to consider both statistical and practical significance when deciding on sample size.

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Most popular questions from this chapter

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