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Chapter 10: Problem 3

To test \(H_{0}: p=0.30\) versus \(H_{1}: p<0.30,\) a simple random sample of \(n=300\) individuals is obtained and \(x=86\) 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 \(0.28 .\) What is the power of the test? (c) Redo part (b) if the true population proportion is 0.25 .

Short Answer

Expert verified
Type II error means failing to reject \( H_0 \) when \( H_1 \) is true. For \( p = 0.28 \), \( \beta = 0.0075 \) and power is 99.25%. For \( p = 0.25 \), \( \beta = 0.00013 \) and power is 99.987%.

Step by step solution

01

Title - Understanding Type II Error

A Type II error occurs when the null hypothesis ( \(H_{0}: p=0.30 \)) is not rejected, even though the alternative hypothesis ( \(H_{1}: p<0.30 \)) is true. In this context, it means concluding that the population proportion is 0.30 or more when in reality it is less than 0.30.
02

Title - Calculation of the Standard Error

Compute the standard error for the sample proportion: \( SE = \sqrt{\frac{p \cdot (1 - p)}{n}} \), where \( p = 0.30 \) and \( n = 300 \). This becomes \( SE = \sqrt{\frac{0.30 \cdot 0.70}{300}} = \sqrt{\frac{0.21}{300}} \approx 0.0265 \).
03

Title - Determine the Test Statistic

Find the test statistic using the sample proportion \( \hat{p} = \frac{x}{n} \). Here, \( \hat{p} = \frac{86}{300} \approx 0.287 \). The test statistic is calculated as \( z = \frac{\hat{p} - p}{SE} \). Using \( p = 0.30 \), we get \( z = \frac{0.287 - 0.30}{0.0265} \approx -0.49 \).
04

Title - Critical Value for the Significance Level

For \( \alpha = 0.05 \), find the critical value \( z_{\alpha} = -1.645 \) from the standard normal distribution, since this is a one-tailed test.
05

Title - Calculation of Beta (Type II Error) for \( p = 0.28 \)

Compute the new test statistic under \( p = 0.28 \). Recalculate the standard error: \( SE_{new} = \sqrt{\frac{0.28 \cdot 0.72}{300}} \approx 0.0254 \). The new z-score: \( z_{new} = \frac{0.30 - 0.28}{0.0254} \approx 0.787 \). Find the probability \( P(Z < -1.645 - 0.787) \) using the standard normal distribution, leading to \( P(Z < -2.432) \approx 0.0075 \). Thus, \( \beta = 0.0075 \) or 0.75%.
06

Title - Calculate the Power of the Test for \( p = 0.28 \)

The power of the test is \( 1 - \beta \). Hence, the power is \( 1 - 0.0075 \approx 0.9925 \) or 99.25%.
07

Title - Repeat Beta Calculation for \( p = 0.25 \)

Now, compute \( \beta \) for \( p = 0.25 \). Calculate \( SE_{new} = \sqrt{\frac{0.25 \cdot 0.75}{300}} \approx 0.0250 \). The new z-score: \( z_{new} = \frac{0.30 - 0.25}{0.0250} = 2 \). Find the probability \( P(Z < -1.645 - 2) = P(Z < -3.645) \approx 0.00013 \). Thus, \( \beta = 0.00013 \) or 0.013%.
08

Title - Calculate the Power of the Test for \( p = 0.25 \)

The power of the test is \( 1 - \beta \). Hence, the power is \( 1 - 0.00013 \approx 0.99987 \) or 99.987%.

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Key Concepts

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

Probability of Type II Error
A Type II error happens when we don't reject the null hypothesis even though the alternative hypothesis is true. In terms of this exercise, a Type II error means that we conclude the population proportion is 0.30 or greater when it is actually less than 0.30.
The probability of making this error is called beta (\( \beta \)).
For example, if the true population proportion is 0.28 and our study doesn't detect this difference, we have committed a Type II error. Calculating this probability involves using the test statistic, standard error, and critical values.
Hypothesis Testing
Hypothesis testing is a method used to make decisions about a population parameter based on sample data.
It starts with forming two hypotheses:
  • The null hypothesis (\( H_{0}: p = 0.30 \))
  • The alternative hypothesis (\( H_{1}: p < 0.30 \))
Next, we select a significance level (\( \alpha \)), calculate the test statistic, and compare it with critical values to reach a decision on whether to reject the null hypothesis. If the test statistic shows enough evidence against the null hypothesis, it is rejected in favor of the alternative hypothesis.
Power of a Test
The power of a test is the probability of correctly rejecting the null hypothesis when the alternative hypothesis is true. It can be calculated as \( 1 - \beta \), where \( \beta \) is the probability of making a Type II error.
Higher power means the test is more likely to detect a true effect. For our example, if the true proportion is 0.28, the power of the test is \( 1 - 0.0075 = 0.9925 \) or 99.25%. This high power indicates a strong ability to detect the actual population proportion.
Significance Level
The significance level (\( \alpha \)) represents the probability of making a Type I error, which occurs when we wrongly reject the null hypothesis.
In this exercise, the chosen significance level is 0.05. This means we accept a 5% chance of mistakenly concluding that the proportion is less than 0.30 when it is not.
The corresponding critical value for \( \alpha = 0.05 \) in a one-tailed test is -1.645. Any test statistic below this threshold leads to rejecting the null hypothesis.
Standard Error Calculation
The standard error (SE) measures the variability of the sample proportion and helps determine how much it might differ from the true population proportion.
It is calculated using the formula:
\( SE = \sqrt{ \frac{p \cdot (1 - p)}{n} } \)

For this exercise, where \( p = 0.30 \) and \( n = 300 \), the SE becomes:

\( SE = \sqrt{ \frac{0.30 \cdot 0.70}{300} } \approx 0.0265 \)
This value is used to calculate the test statistic and the probability of Type II error.
Test Statistic Calculation
The test statistic shows how far the sample proportion is from the null hypothesis value, measured in terms of standard errors.
It is calculated as:
\( z = \frac{ \hat{p} - p }{ SE } \)

In our exercise, with a sample proportion \( \hat{p} = \frac{86}{300} \approx 0.287 \) and \( p = 0.30 \), the test statistic is:
\( z = \frac{ 0.287 - 0.30 }{ 0.0265 } \approx -0.49 \)
This result is then compared to the critical value to decide whether to reject the null hypothesis.

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Most popular questions from this chapter

In Problems \(9-14,\) the null and alternative hypotheses are given. Determine whether the hypothesis test is lefi-tailed, right-tailed, or two-tailed. What parameter is being tested? \(H_{0}: \sigma=4.2\) \(H_{1}: \sigma \neq 4.2\)

The website pundittracker.com keeps track of predictions made by individuals in finance, politics, sports, and entertainment. Jim Cramer is a famous TV financial personality and author. Pundittracker monitored 678 of his stock predictions (such as a recommendation to buy the stock) and found that 320 were correct predictions. Treat these 678 predictions as a random sample of all of Cramer's predictions. (a) Determine the sample proportion of predictions Cramer got correct. (b) Suppose that we want to know whether the evidence suggests Cramer is correct less than half the time. State the null and alternative hypotheses. (c) Verify the normal model may be used to determine the \(P\) -value for this hypothesis test. (d) Draw a normal model with area representing the \(P\) -value shaded for this hypothesis test. (e) Determine the \(P\) -value based on the model from part (d). (f) Interpret the \(P\) -value. (g) Based on the \(P\) -value, what does the sample evidence suggest? That is, what is the conclusion of the hypothesis test? Assume an \(\alpha=0.05\) level of significance.

Throwing darts at the stock pages to decide which companies to invest in could be a successful stock-picking strategy. Suppose a researcher decides to test this theory and randomly chooses 100 companies to invest in. After 1 year, 53 of the companies were considered winners; that is, they outperformed other companies in the same investment class. To assess whether the dart-picking strategy resulted in a majority of winners, the researcher tested \(H_{0}: p=0.5\) versus \(H_{1}: p>0.5\) and obtained a \(P\) -value of \(0.2743 .\) Explain what this \(P\) -value means and write a conclusion for the researcher.

Suppose you wish to determine if the mean IQ of students on your campus is different from the mean IQ in the general population, \(100 .\) To conduct this study, you obtain a simple random sample of 50 students on your campus, administer an IQ test, and record the results. The mean IQ of the sample of 50 students is found to be 107.3 with a standard deviation of \(13.6 .\) (a) Conduct a hypothesis test (preferably using technology) \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0}\) for \(\mu_{0}=103,104,105,106,107,108,109,110,111,112\) at the \(\alpha=0.05\) level of significance. For which values of \(\mu_{0}\) do you not reject the null hypothesis? (b) Construct a \(95 \%\) confidence interval for the mean IQ of students on your campus. What might you conclude about how the lower and upper bounds of a confidence interval relate to the values for which the null hypothesis is rejected? (c) Suppose you changed the level of significance in conducting the hypothesis test to \(\alpha=0.01\). What would happen to the range of values of \(\mu_{0}\) for which the null hypothesis is not rejected? Why does this make sense?

The headline reporting the results of a poll conducted by the Gallup organization stated "Majority of Americans at Personal Best in the Morning." The results indicated that a survey of 1100 Americans resulted in \(55 \%\) stating they were at their personal best in the morning. The poll's results were reported with a margin of error of \(3 \% .\) Explain why the Gallup organization's headline is accurate.
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