/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 Suppose you wish to detect a dif... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you wish to detect a difference between \(\mu_{1}\) and \(\mu_{2}\) (either \(\mu_{1}>\mu_{2}\) or \(\mu_{1}<\mu_{2}\) ) and, instead of running a two-tailed test using \(\alpha=.05,\) you use the following test procedure. You wait until you have collected the sample data and have calculated \(\bar{x}_{1}\) and \(\bar{x}_{2}\). If \(\bar{x}_{1}\) is larger than \(\bar{x}_{2},\) you choose the alternative hypothesis \(H_{\mathrm{a}}: \mu_{1}>\mu_{2}\) and run a one-tailed test placing \(\alpha_{1}=.05\) in the upper tail of the \(z\) distribution. If, on the other hand, \(\bar{x}_{2}\) is larger than \(\bar{x}_{1},\) you reverse the procedure and run a one-tailed test, placing \(\alpha_{2}=.05\) in the lower tail of the \(z\) distribution. If you use this procedure and if \(\mu_{1}\) actually equals \(\mu_{2},\) what is the probability \(\alpha\) that you will conclude that \(\mu_{1}\) is not equal to \(\mu_{2}\) (i.e., what is the probability \(\alpha\) that you will incorrectly reject \(H_{0}\) when \(H_{0}\) is true)? This exercise demonstrates why statistical tests should be formulated prior to observing the data.

Short Answer

Expert verified
Answer: The overall probability of incorrectly rejecting the null hypothesis when it's true is α = 0.05, using the two-step procedure. However, this procedure is not recommended due to potential biases and issues arising from choosing the alternative hypothesis based on the data itself.

Step by step solution

01

Recap the problem and notation

The null hypothesis (\(H_{0}\)) states that the population means \(\mu_{1}\) and \(\mu_{2}\) are equal, i.e., \(\mu_{1} = \mu_{2}\). The alternative hypothesis (\(H_{a}\)) depends on the observed difference in sample means (\(\bar{x}_1\) and \(\bar{x}_2\)). Instead of running a two-tailed test with significance level \(\alpha = 0.05\), we use a two-step procedure: - If \(\bar{x}_1 > \bar{x}_2\), we choose \(H_{a}: \mu_1 > \mu_2\) and run a one-tailed test with \(\alpha_1=0.05\) in the upper tail of the \(z\) distribution. - If \(\bar{x}_2 > \bar{x}_1\), we choose \(H_{a}: \mu_1 < \mu_2\) and run a one-tailed test with \(\alpha_2=0.05\) in the lower tail of the \(z\) distribution. Our task is to find the overall probability \(\alpha\) of incorrectly rejecting \(H_{0}\) when it's true.
02

Identify the type I error for each one-tailed test

When \(H_0\) is true and we run a one-tailed test with significance level \(\alpha_1=0.05\) in the upper tail, the probability of incorrectly rejecting \(H_0\) is \(\alpha_1=0.05\). Similarly, for the one-tailed test with significance level \(\alpha_2=0.05\) in the lower tail, the probability of incorrectly rejecting \(H_0\) is \(\alpha_2=0.05\).
03

Calculate the overall probability of type I error

Since we choose one of the tests at random based on the sample data, the overall probability of type I error, denoted by \(\alpha\), is the average of \(\alpha_1\) and \(\alpha_2\): \(\alpha = \dfrac{\alpha_1+\alpha_2}{2}\) Plugging the values, we get: \(\alpha = \dfrac{0.05+0.05}{2} = \dfrac{0.1}{2} = 0.05\)
04

Interpret the result

The overall probability of incorrectly rejecting the null hypothesis when it's true is \(\alpha = 0.05\), which is the same as the significance level in a two-tailed test. However, this procedure is problematic, because it involves choosing the alternative hypothesis based on the data itself, which can lead to biased results and increase the likelihood of making incorrect conclusions. This exercise demonstrates the importance of formulating the statistical test prior to observing the data, as it helps to avoid potential issues arising from post-hoc adjustments and data-driven decisions.

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Key Concepts

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

Type I Error
A Type I error occurs when we mistakenly reject a true null hypothesis. This is like sounding the alarm when everything is actually fine. In the context of hypothesis testing, it's important to understand that this error represents the probability of falsely concluding that there is a significant effect or difference when there is none.

Let's consider real-life implications: imagine we're testing whether a new drug is more effective than a placebo. A Type I error would occur if we conclude that the drug works when it doesn't, leading to wasted resources and potential harm. In statistical terms, this error probability is denoted by \( \alpha \), the significance level, meaning it reflects our tolerance for incorrectly rejecting the null hypothesis.

In the provided exercise, the significance level for each one-tailed test is set at 0.05. This means there's a 5% chance of committing a Type I error in each individual test. When we add the probability of making these errors in both tail tests as shown in the example, we finally reach the overall significance level used in a two-tailed test, maintaining a balance in hypothesis testing.
Two-Tailed Test
A two-tailed test in statistics is used when we want to detect an effect or a difference in either direction. This means we're interested in any major deviation from the null hypothesis, whether it's an increase or a decrease.

Think of a two-tailed test like checking if a string tied between two walls is too loose or too tight, without caring which one it might be. It's this openness to both possibilities that double our vigilance in monitoring unusual events at both ends.

In the scenario described, a two-tailed test was initially recommended to detect a difference between two means. However, instead of sticking with it, a one-tailed alternative was selected, based on the sample data. This affects the test's foundation because, unlike two-tailed tests, one-tailed versions only consider if one side of the distribution is significant. Two-tailed tests distribute the significance level \( \alpha \) equally across both tails, ensuring an equal chance of detecting an effect in either direction. This is an important concept in hypothesis testing as it ensures that the conclusions drawn are robust and less likely to be biased by random sample fluctuations.
Significance Level
The significance level, symbolized as \( \alpha \), is the threshold at which we determine whether an observed effect is statistically significant. In simpler terms, it's the risk we're willing to take to make a Type I error. Conventionally, \( \alpha \) is set at 0.05, meaning we accept a 5% probability of incorrectly rejecting a true null hypothesis.

In hypothesis testing, \( \alpha \) plays a crucial role by setting the criteria for decision making. A lower significance level means stricter criteria for rejecting \( H_0 \), hence reducing the risk of a Type I error, though it might increase the risk of missing a true effect, which is a Type II error.

In the original problem's context, the significance level was set at 0.05 for each of the one-tailed tests. By flipping between two one-tailed tests based on sample outcomes, we explore the implications of \( \alpha \). However, this reactive approach is unconventional and potentially detriment throws the rigor of the test into question. It's vital to set \( \alpha \) before the data is examined, sticking with predefined plans to prevent bias and adjust for significance only based on theoretical premises rather than empirical results. By keeping the significance level constant (in this case, at 0.05 for a two-tailed test), it ensures a consistent framework within which evidence against the null hypothesis can be evaluated.

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Most popular questions from this chapter

Independent random samples of \(n_{1}=140\) and \(n_{2}=140\) observations were randomly selected from binomial populations 1 and \(2,\) respectively. Sample 1 had 74 successes, and sample 2 had 81 successes. a. Suppose you have no preconceived idea as to whic parameter, \(p_{1}\) or \(p_{2}\), is the larger, but you want onl to detect a difference between the two parameters one exists. What should you choose as the alternative hypothesis for a statistical test? The null hypothesis? b. Calculate the standard error of the difference in the two sample proportions, \(\left(\hat{p}_{1}-\hat{p}_{2}\right) .\) Make sure to use the pooled estimate for the common value of \(p\) c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population proportions are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population proportions at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population proportions?

Cheaper Airfares Looking for a great airfare? Consumer Reports \(^{8}\) has several hints about how to minimize your costs, which include being flexible about travel dates and times, checking multiple websites, and knowing when to book your flight. One suggestion involved checking fares at "secondary" airports - airports that might be slightly farther from your home, but where fares are lower. For example, the average of all domestic ticket prices at Los Angeles International Airport (LAX) was quoted as \(\$ 349\) compared to an average price of \(\$ 287\) at nearby Ontario International Airport (ONT). Suppose that these estimates were based on random samples of 1000 domestic tickets at each airport and that the standard deviation of the prices at both airports was \(\$ 200 .\) a. Is there sufficient evidence to indicate that the mean ticket prices differ for these two airports at the \(\alpha=\) .05 level of significance? Use the large-sample ztest. What is the \(p\) -value of this test? b. Construct a \(95 \%\) confidence interval for \(\left(\mu_{1}-\mu_{2}\right) .\) Does this interval confirm your conclusions in part a?

Clopidogrel and Aspirin A large study was conducted to test the effectiveness of clopidogrel in combination with aspirin in warding off heart attacks and strokes. \({ }^{15}\) The trial involved more than 15.500 people 45 years of age or older from 32 countries, including the United States, who had been diagnosed with cardiovascular disease or had multiple risk factors. The subjects were randomly assigned to one of two groups. After 2 years, there was no difference in the risk of heart attack, stroke, or dying from heart disease between those who took clopidogrel and low-dose aspirin daily and those who took low-dose aspirin plus a dummy pill. The 2 -drug combination actually increased the risk of dying \((5.4 \%\) versus \(3.8 \%)\) or dying specifically from cardiovascular disease \((3.9 \%\) versus \(2.2 \%\) ). a. The subjects were randomly assigned to one of the two groups. Explain how you could use the random number table to make these assignments. b. No sample sizes were given in the article: however, let us assume that the sample sizes for each group were \(n_{1}=7720\) and \(n_{2}=7780 .\) Determine whether the risk of dying was significantly different for the two groups. c. What do the results of the study mean in terms of practical significance?

Find the appropriate rejection regions for the large-sample test statistic \(z\) in these cases: a. A left-tailed test at the \(1 \%\) significance level. b. A two-tailed test with \(\alpha=.01\). c. Suppose that the observed value of the test statistic was \(z=-2.41 .\) For the rejection regions constructed in parts a and b, draw the appropriate conclusion for the tests. If appropriate, give a measure of the reliability of your conclusion.

Flextime A company was contemplating the installation of a flextime schedule in which a worker schedules his or her work hours or compresses work weeks. The company estimates that it needs a minimum mean of 7 hours per day per assembly worker in order to operate effectively. Each of a random sample of 80 of the company's assemblers was asked to submit a tentative flextime schedule. If the mean number of hours per day for Monday was 6.7 hours and the standard deviation was 2.7 hours, do the data provide sufficient evidence to indicate that the mean number of hours worked per day on Mondays, for all of the company's assemblers, will be less than 7 hours? Test using \(\alpha=.05 .\)

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