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Simulate drawing 100 simple random samples of size \(n=40\) from a population whose proportion is 0.3 (a) Test the null hypothesis \(H_{0}: p=0.3\) versus \(H_{1}: p \neq 0.3\) for each simulated sample. (b) If we test the hypothesis at the \(\alpha=0.1\) 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) How do we know that a rejection of the null hypothesis results in making a Type I error in this situation?

Short Answer

Expert verified
For a test at \( \alpha=0.1 \), we expect about 10 Type I errors in 100 samples. Count the number of rejections and compare with this expectation.

Step by step solution

01

Understanding the Problem

We need to draw 100 simple random samples of size \(n=40\) from a population where the proportion is 0.3. We will then test the hypothesis for each sample and analyze the results.
02

Simulate Drawing Samples

Use a computer program or a statistical software like R or Python to simulate drawing 100 random samples. Each sample should have 40 elements, and the probability of success (for a proportion of 0.3) is 0.3.
03

State the Hypotheses

For each sample, state the null hypothesis \(H_0: p=0.3\) and the alternative hypothesis \(H_1: p eq 0.3\).
04

Calculate Test Statistics

For each sample, calculate the sample proportion \( \hat{p} \) and the test statistic, which is the Z-score: \[Z = \frac{ \hat{p} - p_0 }{ \sqrt{ \frac{p_0(1 - p_0)}{n} } }\]where \( p_0 = 0.3 \) and \( n = 40 \).
05

Determine the Critical Value

At the \( \alpha = 0.1 \) level of significance, find the critical Z values corresponding to this \( \alpha \) for a two-tailed test. These values are approximately \( \pm 1.645 \).
06

Test Each Sample

Compare the Z-score obtained from each sample to the critical values \( \pm 1.645 \). If the absolute value of the Z-score is greater than 1.645, reject the null hypothesis \( H_0 \). Otherwise, do not reject \( H_0 \).
07

Count Rejections

Count the number of samples out of the 100 that lead to a rejection of the null hypothesis \( H_0 \).
08

Expected Type I Errors

Given that the significance level \( \alpha = 0.1 \), we expect 10% of the samples to result in a Type I error. Therefore, in 100 samples, we would expect about 10 rejections of \( H_0 \).
09

Compare Results

Compare the number of rejections found in Step 6 to the expected value from Step 7. Analyze if the actual number is close to the expected 10 rejections.
10

Validating Type I Errors

In this scenario, if we reject the null hypothesis \( H_0 \) when \( H_0 \) is true (since the population proportion is actually 0.3), we are making a Type I error.

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Key Concepts

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

Type I error
Type I error occurs when we reject a true null hypothesis. In the context of our exercise, we are testing the null hypothesis that the population proportion is 0.3. If this hypothesis is actually true and we incorrectly reject it based on our sample data, then we have made a Type I error. It's like a false alarm—saying something has changed when it hasn't.

When we perform multiple tests, like in our simulation of drawing and testing 100 samples, we expect a certain number of Type I errors to occur purely by chance. Since our significance level \(\alpha\) is set to 0.1, we'd expect 10 out of 100 samples to result in a Type I error.
significance level
The significance level, denoted by \(\alpha\), is the probability of rejecting the null hypothesis when it is true. In our exercise, we use a significance level of 0.1. This means we have a 10% risk of committing a Type I error with each test.

Significance levels are pre-determined thresholds that help us decide whether our sample data provide strong enough evidence against the null hypothesis. Common significance levels are 0.05, 0.01, and 0.1, with 0.05 being most widely used. In our case, by choosing \(\alpha = 0.1\), we are allowing a relatively higher margin for error, which is useful when we want to be more inclusive of potential findings.
Z-score calculation
The Z-score helps us determine how far our sample proportion is from the hypothesized population proportion in terms of standard errors. In this exercise, we calculate the Z-score for each of our 100 samples to test the null hypothesis \(H_0: p = 0.3\).

The formula for the Z-score is:
\[ Z = \frac{ \hat{p} - p_0 }{ \sqrt{ \frac{p_0(1 - p_0)}{n} } } \] Here, \(\hat{p}\) is the sample proportion, \(p_0\) is the hypothesized proportion (0.3), and \(n\) is the sample size (40).

We then compare the Z-score to the critical values for our chosen significance level. For a two-tailed test at \(\alpha = 0.1\), the critical Z-values are approximately \(\pm 1.645\). If the absolute value of our calculated Z-score exceeds 1.645, we reject the null hypothesis.
simple random sample
A simple random sample is a subset of individuals chosen from a larger set, where each individual is selected randomly and entirely by chance. In our exercise, we simulate drawing 100 simple random samples, each of size \(n = 40\), from a population with a proportion of 0.3.

Using statistical software or programming languages like R or Python can help ensure these samples are truly random. Simple random sampling is essential as it ensures that each sample is representative of the population, eliminating bias and ensuring the validity of our hypothesis test results. This randomness is crucial in making inferences about the entire population based on sample data.

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Most popular questions from this chapter

According to menstuff.org, \(22 \%\) of married men have "strayed" at least once during their married lives. (a) Describe how you might go about administering a survey to assess the accuracy of this statement. (b) A survey of 500 married men indicated that 122 have "strayed" at least once during their married life. Construct a \(95 \%\) confidence interval for the population proportion of married men who have strayed. Use this interval to assess the accuracy of the statement made by menstuff.org.

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?

A simple random sample of size \(n=200\) individuals with a valid driver's license is asked if they drive an American-made automobile. Of the 200 individuals surveyed, 115 responded that they drive an American-made automobile. Determine if a majority of those with a valid driver's license drive an American made automobile at the \(\alpha=0.05\) level of significance.

Simulation The exponential probability distribution can be used to model waiting time in line or the lifetime of electronic components. Its density function is skewed right. Suppose the wait time in a line can be modeled by the exponential distribution with \(\mu=\sigma=5\) minutes. (a) Simulate obtaining 100 simple random samples of size \(n=10\) from the population described. That is, simulate obtaining a simple random sample of 10 individuals waiting in a line where the wait time is expected to be 5 minutes. (b) Test the null hypothesis \(H_{0}: \mu=5\) versus the alternative \(H_{1}: \mu \neq 5\) for each of the 100 simulated simple random samples. (c) 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? (d) Count the number of samples that lead to a rejection of the null hypothesis. Is it close to the expected value determined in part (c)? What might account for any discrepancies?

If we do not reject the null hypothesis when the statement in the alternative hypothesis is true, we have made a Type ____ error.

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