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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

To test \(H_{0}: \mu=45\) versus \(H_{1}: \mu \neq 45,\) a simple random sample of size \(n=40\) is obtained. (a) Does the population have to be normally distributed to test this hypothesis by using the methods presented in this section? Why? (b) If \(\bar{x}=48.3\) and \(s=8.5,\) compute the test statistic. (c) Draw a \(t\) -distribution with the area that represents the P-value shaded. (d) Approximate and interpret the \(P\) -value. (e) If the researcher decides to test this hypothesis at the \(\alpha=0.01\) level of significance, will the researcher reject the null hypothesis? Why? (f) Construct a \(99 \%\) confidence interval to test the hypothesis.

A simple random sample of size \(n=320\) adults was asked their favorite ice cream flavor. Of the 320 individuals surveyed, 58 responded that they preferred mint chocolate chip. Do less than \(25 \%\) of adults prefer mint chocolate chip ice cream? Use the \(\alpha=0.01\) level of significance.

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

From Super Bowl I (1967) through Super Bowl XXXI (1997), the stock market increased if an NFL team won the Super Bowl and decreased if an AFL team won. This condition held 28 out of 31 years. (a) Suppose the likelihood of predicting the direction of the stock market (increasing or decreasing) in any given year is \(0.50 .\) Decide on the appropriate null and alternative hypotheses to test whether the outcome of the Super Bowl can be used to predict the direction of the stock market. (b) Use the binomial probability distribution to determine the \(P\) -value for the hypothesis test from part (a). (c) Comment on the dangers of using the outcome of the hypothesis test to judge investments. Be sure your comment includes a discussion of circumstances in which associations have a causal relationship.

The piston diameter of a certain hand pump is 0.5 inch. The quality-control manager determines that the diameters are normally distributed, with a mean of 0.5 inch and a standard deviation of 0.004 inch. The machine that controls the piston diameter is recalibrated in an attempt to lower the standard deviation. After recalibration, the quality-control manager randomly selects 25 pistons from the production line and determines that the standard deviation is 0.0025 inch. Was the recalibration effective? Use the \(\alpha=0.01\) level of significance.
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