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Chapter 18: Problem 52

Find the Error Probabilities. You have an SRS of size \(n=25\) from a Normal distribution with \(\sigma=2.0\). You wish to test $$ \begin{aligned} &H_{0}: \mu=0 \\ &H_{a}: \mu>0 \end{aligned} $$ You decide to reject \(H_{0}\) if \(x>0\) and not reject \(H_{0}\) otherwise. a. Find the probability of a Type I error. That is, find the probability that the test rejects \(H_{0}\) when in fact \(\mu=0\). b. Find the probability of a Type II error when \(\mu=0.5\). This is the probability that the test fails to reject \(H_{0}\) when in fact \(\mu=0.5\). c. Find the probability of a Type II error when \(\mu=1.0\).

Short Answer

Expert verified
a. 0.5; b. \( \approx 0.1056 \); c. \( \approx 0.0062 \).

Step by step solution

01

Understanding the Problem

We need to calculate the probabilities of Type I and Type II errors for a hypothesis test involving a normal distribution. The null hypothesis is \( H_0: \mu=0 \) and the alternative hypothesis is \( H_a: \mu>0 \). The sample size \( n=25 \) and population standard deviation \( \sigma=2.0 \). The rejection rule is to reject \( H_0 \) if the sample mean \( \bar{x}>0 \).
02

Calculate Standard Error

Calculate the standard error of the mean using the formula \( SE = \frac{\sigma}{\sqrt{n}} = \frac{2.0}{\sqrt{25}} = 0.4 \). This will be used to determine the critical values for the hypothesis test.
03

Find Probability of Type I Error (\( \alpha \))

The Type I error occurs when we reject \( H_0 \) when it is true. Since we reject \( H_0 \) if \( \bar{x} > 0 \), determine \( P(\bar{x} > 0 \mid \mu = 0) \). Under \( H_0 \), \( \bar{x} \sim N(0, 0.4) \). The probability is \( P(Z > 0) = 0.5 \), where \( Z \) is the standard normal variable. Thus, the Type I error probability is 0.5.
04

Find Probability of Type II Error when \( \mu = 0.5 \)

Type II error occurs when \( H_0 \) fails to be rejected while in fact \( \mu=0.5 \). We find \( P(\bar{x} \leq 0 \mid \mu = 0.5) \). Under this condition, \( \bar{x} \sim N(0.5, 0.4) \). Calculate the Z-score: \( Z = \frac{0 - 0.5}{0.4} = -1.25 \). Thus, \( P(\bar{x} \leq 0\mid \mu=0.5) = P(Z \leq -1.25) \approx 0.1056 \).
05

Find Probability of Type II Error when \( \mu = 1.0 \)

Type II error occurs when \( H_0 \) fails to be rejected while in fact \( \mu=1.0 \). We find \( P(\bar{x} \leq 0 \mid \mu = 1.0) \). Under this condition, \( \bar{x} \sim N(1.0, 0.4) \). Calculate the Z-score: \( Z = \frac{0 - 1.0}{0.4} = -2.5 \). Thus, \( P(\bar{x} \leq 0\mid \mu=1.0) = P(Z \leq -2.5) \approx 0.0062 \).

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Key Concepts

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

Type I Error
In the context of hypothesis testing, a **Type I Error** occurs when the null hypothesis \( H_0 \) is rejected when it is actually true. This error is often denoted by the Greek letter \( \alpha \), and represents the probability of making this mistake. In this exercise, the null hypothesis is \( \mu = 0 \), and the decision rule is to reject \( H_0 \) if the sample mean \( \bar{x} > 0 \).
  • Since \( \bar{x} \sim N(0, 0.4) \) under the null hypothesis, \( P(\bar{x} > 0) = 0.5 \).
  • This means there is a 50% chance of incorrectly rejecting \( H_0 \) when it is actually true, showcasing a high Type I error rate.
Thus, understanding and controlling Type I error is crucial because it helps to maintain the validity of the hypothesis test. A high Type I error rate can lead to false positives, where we conclude there is an effect when there isn't one.
Type II Error
A **Type II Error** occurs when the null hypothesis \( H_0 \) is not rejected when it is actually false. This type of error is represented by \( \beta \), and it reflects the probability of missing a true effect.
  • In this scenario, failing to reject \( H_0 \) when \( \mu = 0.5 \) results in a Z-score of \(-1.25\).
  • The probability of this happening is \( P(Z \leq -1.25) \approx 0.1056 \), indicating about a 10.56% chance of making this error.
Similarly,
  • For \( \mu = 1.0 \), the Z-score becomes \(-2.5\).
  • The probability \( P(Z \leq -2.5) \approx 0.0062 \), showing a reduced Type II error probability of about 0.62% when the true mean is further away from the hypothesized value.
Understanding how to mitigate Type II errors is essential since they lead to false negatives, potentially overlooking real effects in studies.
Normal Distribution
The **Normal Distribution** is a fundamental concept in statistics, often referred to as the Gaussian distribution. It is a symmetric, bell-shaped distribution that describes how the values of a variable are distributed around the mean.
  • In this exercise, the sample mean \( \bar{x} \) is assumed to follow a normal distribution because the population from which it is drawn is normal.
  • This means \( \bar{x} \sim N(\mu, \sigma^2/n) \) with known values \( \mu \) depending on the hypothesis scenario and \( \sigma = 2.0 \).
The properties of the normal distribution allow us to calculate probabilistic measures, such as the Z-scores, which help assess Type I and Type II errors. The centrality, spread, and shape of the normal distribution make it a useful tool for hypothesis testing, as it helps in determining critical regions and calculating error rates.
Standard Error
The **Standard Error (SE)** measures the variability or dispersion of a sample statistic, typically the sample mean, due to sampling error. It is calculated from the population's standard deviation \( \sigma \) and the sample size \( n \).
  • In this scenario, the standard error of the mean is computed as \( SE = \frac{\sigma}{\sqrt{n}} \), simplifying to \( SE = \frac{2.0}{5} = 0.4 \).
  • Knowing the standard error enables us to relate the sample mean's distribution to the standard normal distribution, which is crucial for determining the probability of observing certain results under the null hypothesis assumption.
The standard error provides insight into how much the sample mean \( \bar{x} \) would vary from the population mean \( \mu \) if we were to repeatedly draw samples of the same size. Smaller standard errors suggest less variability and more reliable sample estimates.

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Most popular questions from this chapter

Divorces. Divorce rates vary from city to city in the United States. We have lots of data on many U.S. cities. Statistical software makes it easy to perform dozens of significance tests on dozens of variables to see which ones best predict divorce. One interesting finding is that those cities with Major League ballparks tend to have significantly more divorces than other cities. To improve your chances of a lasting marriage, should you use this "significant" variable to decide where to live? Explain your answer.

Sampling at the Gour met Food Store. A market researcher chooses at random from men entering a gourmet food store. One outcome of the study is a \(95 \%\) confidence interval for the mean of "the highest price you would pay for a bottle of wine." a. Explain why this confidence interval does not give useful information about the population of all men. b. Explain why it may give useful information about the population of men who shop at gourmet food stores.

Reducing the Gender Gap. In many science disciplines, women are outperformed by men on test scores. Will "values affirmation training" improve self- confidence and hence performance of women relative to men in science courses? A study conducted at a large university compares the scores of men and women at the end of a large introductory physics course on a nationally normed standardized test of conceptual physics, the Force and Motion Conceptual Evaluation (FMCE). Half the women in the course were randomly assigned to values affirmation training during the course; the other half received no training. The study reports that there was a significant difference \((P<0.01)\) in the gap between men's and women's scores, although the gap for women who received the values affirmation training was much smaller than that for women who did not receive training. As evidence that this gap was reduced for woman who received the training, the study also reports that a \(95 \%\) confidence interval for the difference in mean scores on the FMCE exam between women who received the training and those who didn't is \(13 \pm 8\) points. You are a faculty member in the physics department, and the provost, who is interested in women in science, asks you about the study. a. Explain in simple language what "a significant difference \((P<0.01)\) " means. b. Explain clearly and briefly what "95\% confidence" means. c. Is this study good evidence that requiring values affirmation training of all female students would greatly reduce the gender gap in scores on science tests in college courses?

Two Types of Error. Your company markets a computerized medical diagnostic program used to evaluate thousands of people. The program scans the results of routine medical tests (pulse rate, blood tests, etc.) and refers the case to a doctor if there is evidence of a medical problem. The program makes a decision about each person. a. What are the two hypotheses and the two types of error that the program can make? Describe the two types of error in terms of false-positive and false- negative test results. b. The program can be adjusted to decrease one error probability, at the cost of an increase in the other error probability. Which error probability would you choose to make smaller, and why? (This is a matter of judgment. There is no single correct answer.)

The First Child Has Higher IQ. Does the birth order of a family's children influence their IQ scores? A careful study of 241,310 Norwegian 18- and 19 -year-olds found that firstborn children scored \(2.3\) points higher on the average than second children in the same family. This difference was highly significant \((P<0.001)\). A commentator said, "One puzzle highlighted by these latest findings is why certain ot her within-family studies have failed to show equally consistent results. Some of these previous null findings, which have all been obtained in much smaller samples, may be explained by inadequate stat istical power." \(\underline{12}\) a. Explain in simple language why tests having low power often fail to give evidence against a null hypothesis even when the hypothesis is really false. b. Do you think a difference of \(2.3\) points in IQ scores is an important difference?
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