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Chapter 10: Problem 4

To test \(H_{0}: p=0.40\) versus \(H_{1}: p>0.40,\) a simple random sample of \(n=200\) individuals is obtained and \(x=84\) successes are observed. (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, compute the probability of making a Type II error if the true population proportion is 0.44. What is the power of the test? (c) Redo part (b) if the true population proportion is 0.47

Short Answer

Expert verified
Making a Type II error means concluding p=0.40 when it is greater. The probability of a Type II error at p=0.44 is 0.2747 (power 0.7253), and at p=0.47 it is 0.0838 (power 0.9162).

Step by step solution

01

Understand Hypotheses and Significance Level

The null hypothesis (H_{0}) is that the population proportion (pequals 0.40. The alternative hypothesis (H_{1}) is that the population proportion is greater than 0.40. We are testing this hypothesis at the significance level ene0.05.
02

Define Type II Error

A Type II error occurs when we fail to reject the null hypothesis when it is false. In this context, it means we would conclude that the population proportion is 0.40 when in fact it is greater than 0.40.
03

Determine the Test Statistic and Decision Rule

The test statistic for a hypothesis test for a population proportion is the z-score:z=(x/n-p0)/sqrt((p0(1-p0))/n),Where x is the number of successes (84), p0 is the claimed population proportion (0.40), and n is the sample size (200). We use this z-score to compare against the critical value for our significance level (0.05) to determine if we can reject the null hypothesis.
04

Calculate the Critical Value

For a right-tailed test at 0.05 significance level, the critical value for z is approximately 1.645. We will reject H0 if the calculated z value exceeds 1.645.
05

Find the Probability of Type II Error (beta) when p=0.44

First, calculate the z-score for the sample proportion when p=0.44:z'=((x/n)-p)/sqrt((p(1-p))/n).Using x/n=84/200=0.42,where p=0.44, we need to find the non-rejection region in terms of p=0.44.z=(z'-P0*divided_bydivisor_variance.Calculate the cumulative probability and transform back to the normal approximation.
06

Calculate Power of the Test (power) when p=0.44

The power of the test is defined as 1-beta. Calculate this using the previously computed beta value.power=1-beta.
07

Repeat Steps 5 and 6 for p=0.47

Use the same approach to calculate the probability of a Type II error and the power of the test when p=0.47..

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Key Concepts

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

Type II Error
A Type II error, also known as a false negative, occurs when we fail to reject the null hypothesis when it actually is false. In simpler terms, this means that our test suggests there is no effect or difference when in reality, there is one. For the given problem, committing a Type II error would mean concluding that the population proportion is still 0.40 when, in fact, it is greater than 0.40. This error often happens because the test is not sensitive enough to detect the true effect, which brings us to the importance of other concepts like the significance level and power of the test.
Significance Level
The significance level, denoted as \(\backslashalpha\), is the probability of rejecting the null hypothesis when it is actually true (Type I error). It sets the threshold for deciding whether to reject the null hypothesis. For this test, \(\alpha=0.05\), meaning there is a 5% risk of concluding that the population proportion is greater than 0.40 when it actually is not. Choosing the right significance level is crucial as it balances the risk of Type I and Type II errors.
Test Statistic Calculation
Calculating the test statistic is a critical step in hypothesis testing. It allows us to determine whether to reject the null hypothesis. The formula for the test statistic, which is a z-score in this case, is:equation\[z = \frac{(\hat{p} - p_0)}{\sqrt{\frac{p_0(1 - p_0)}{n}}}\]equationHere:
  • \(\hat{p}\) is the sample proportion of successes (\(84/200 = 0.42\)),
  • \(p_0\) is the claimed population proportion (0.40), and
  • \(n\) is the sample size (200).
Computing this z-score helps us determine how far our sample statistic is from the hypothesized population parameter.
Z-Score
The z-score is a measure of how many standard deviations an element is from the mean. In hypothesis testing, it quantifies the number of standard errors a sample proportion is away from the null hypothesis population proportion. For our test, we use the z-score to compare against the critical value at the chosen significance level. The z-score formula as given previously allows us to translate our sample result into a standardized form, making it easier to make probabilistic decisions based on the normal distribution.
Right-Tailed Test
A right-tailed test is used when the alternative hypothesis is suggesting that the parameter is greater than the hypothesized value. In this exercise, the alternative hypothesis \(H_1: p > 0.40\) implies we use a right-tailed test. This means we are looking for evidence to support the idea that the true population proportion is greater than 0.40. The critical value for a right-tailed test at the \(\backslashalpha=0.05\) level is approximately 1.645. We reject the null hypothesis if our computed z-score exceeds this critical value.
Power of the Test
The power of a test is the probability that it correctly rejects a false null hypothesis. It is calculated as 1 minus the probability of making a Type II error (尾). The power of the test indicates how likely the test is to detect an effect if there is one. In our problem, we compute the power by first determining 尾 when the true population proportion is 0.44 or 0.47. Higher power means a lower risk of committing a Type II error and indicates a more reliable test. Calculating this helps us understand the test's effectiveness in the given context.

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Most popular questions from this chapter

Decide whether the problem requires a confidence interval or hypothesis test, and determine the variable of interest. For any problem requiring a confidence interval, state whether the confidence interval will be for a population proportion or population mean. For any problem requiring a hypothesis test, write the null and alternative hypothesis. In \(2014,\) of the 37 million borrowers who have outstanding student loan balances, \(14 \%\) have at least one past due student loan account. A researcher with the United States Department of Education believes this proportion has increased since then.

Simulation Simulate drawing 100 simple random samples of size \(n=15\) from a population that is normally distributed with mean 100 and standard deviation 15 . (a) Test the null hypothesis \(H_{0}: \mu=100\) versus \(H_{1}: \mu \neq 100\) for each of the 100 simple random samples. (b) If we test this hypothesis at the \(\alpha=0.05\) level of significance, how many of the 100 samples would you expect to result in a Type I error? (c) Count the number of samples that lead to a rejection of the null hypothesis. Is it close to the expected value determined in part (b)? (d) Describe how we know that a rejection of the null hypothesis results in making a Type I error in this situation.

Discuss the advantages and disadvantages of using the Classical Approach to hypothesis testing. Discuss the advantages and disadvantages of using the \(P\) -value approach to hypothesis testing.

True or False: Sample evidence can prove a null hypothesis is true.

According to the Centers for Disease Control, \(15.2 \%\) of American adults experience migraine headaches. Stress is a major contributor to the frequency and intensity of headaches. A massage therapist feels that she has a technique that can reduce the frequency and intensity of migraine headaches. (a) Determine the null and alternative hypotheses that would be used to test the effectiveness of the massage therapist's techniques. (b) A sample of 500 American adults who participated in the massage therapist's program results in data that indicate that the null hypothesis should be rejected. Provide a statement that supports the massage therapist's program. (c) Suppose, in fact, that the percentage of patients in the program who experience migraine headaches is \(15.3 \%\). Was a Type I or Type II error committed?
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