91Ó°ÊÓ

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.

Step by step solution

01

Simulate Random Samples

Generate 100 simple random samples, each of size 15, from a population that is normally distributed with mean 100 and standard deviation 15.

02

State the Hypotheses

For each sample, we need to test the null hypothesis: \(H_0: \mu = 100\) against the alternative hypothesis: \(H_1: \mu eq 100\).

03

Conduct t-tests

Conduct a t-test for each of the 100 random samples to determine whether to reject \( H_0 \) or not at the \( \alpha = 0.05 \) significance level. Use the formula for the t-statistic: \[ t = \frac{\overline{X} - 100}{\left( \frac{S}{\sqrt{15}} \right)} \]where \( \overline{X} \) is the sample mean and \( S \) is the sample standard deviation.

04

Critical t-value

Determine the critical t-values for a two-tailed test at \( \alpha = 0.05 \) significance level. This value will be approximately \( t_{critical} = \pm2.145 \) given \( df = 14 \).

05

Decision Rule

Reject the null hypothesis \( H_0 \) if the computed t-statistic falls outside the range \( -2.145 \) to \( +2.145 \).

06

Calculate Expected Type I Errors

Since \( \alpha = 0.05 \), expect about 5% of the 100 samples to result in Type I errors. This means we should expect around 5 samples to falsely reject \( H_0 \).

07

Count Rejections

Count the number of samples for which \( H_0 \) is rejected based on the t-tests performed in Step 3. This gives the actual number of rejections.

08

Compare Observed and Expected Results

Compare the number of rejections observed in Step 7 with the expected value of approximately 5 rejections calculated in Step 6. This will verify whether the observed rejections align with our expectations.

09

Explain Type I Error

A Type I error occurs when \( H_0 \) is true, but we erroneously reject it. Since the null hypothesis assumes \( \mu = 100 \) and our samples come from a population with \( \mu = 100 \), rejecting \( H_0 \) when it is actually true, directly constitutes a Type I error.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Type I Error

A Type I error occurs in hypothesis testing when we incorrectly reject a true null hypothesis. Think of it as a 'false positive'. For example, if we test the null hypothesis that the average is 100 (\( H_0: \mu \ = 100 \)), and it actually is 100, but our test suggests otherwise, we’ve made a Type I error.

• In practice, this means we believe there is an effect or difference when there isn’t one.
• Type I errors are directly connected to the significance level, often denoted by alpha (\( \alpha \)).

For a significance level of \( \alpha = 0.05 \), this error rate implies that about 5% of the time, we will reject the null hypothesis, even when it is true.
So, in the context of the exercise, if 100 simple random samples are tested, we'd expect about 5 samples to wrongly reject the null hypothesis, leading to Type I errors.

t-test

A t-test is used to determine whether there is a significant difference between the means of two groups. It’s particularly useful when we have small sample sizes and the population standard deviation is unknown.
In this exercise, we use a one-sample t-test to compare the sample mean to the population mean. Here’s what you need to know:

• Calculate the t-statistic using the formula:
  \[ t = \frac{\barn\X-100}{\left(\frac{S}{\sqrt{15}}\right)} \]
  
• Where \( \overline{X} \) is the sample mean, and \( S \) is the sample standard deviation.
• Compare the t-statistic to the critical t-value to make a decision.

This calculated t-value is then compared against the critical t-value (based on the chosen significance level and degrees of freedom) to decide whether to reject the null hypothesis. In this context, the critical t-value at \( \alpha = 0.05 \, \) and 14 df is \( \pm2.145 \).

Simple Random Sample

A simple random sample (SRS) is a subset of a statistical population in which each member of the subset has an equal probability of being chosen. This is crucial for ensuring the sample represents the population accurately.
Here’s why SRS is important:

• Ensures every sample has an equal chance of selection, removing bias.
• Allows for the application of standard statistical methods, like the t-test.
• Helps to guarantee the sample's results are representative of the population.

In the exercise, we generated 100 simple random samples of size 15 from a normally distributed population. This approach helps ensure that each sample's mean and standard deviation are unbiased estimates of the population parameters.

Null Hypothesis

The null hypothesis (\( H_0 \)) is a statement that there is no effect or no difference, and it serves as the starting point for statistical testing. It’s what we assume to be true until we have enough evidence to suggest otherwise.
In hypothesis testing:

• We start with \( H_0: \mu \ = 100 \), which means the population mean is 100.
• The alternative hypothesis (\( H_1 \)) challenges this assumption, suggesting \( \mu \eq 100 \).
• We gather sample data and conduct a test to determine whether to reject \( H_0 \).

In our case, if the computed t-statistic falls outside of the critical range (\( -2.145 \) to \(+2.145 \)), we reject \( H_0 \). If it does not, we do not have enough evidence to reject \( H_0 \), meaning we assume the population mean is still 100.

Significance Level

The significance level (\( \alpha \)) is the probability of making a Type I error. It is chosen by the researcher before conducting the test and signifies the threshold for rejecting the null hypothesis.
Common significance levels include 0.05, 0.01, and 0.10. In this exercise:

• The chosen significance level is \( \alpha = 0.05 \).
• This means we accept a 5% risk of rejecting a true null hypothesis.
• At \( \alpha = 0.05 \), the critical t-value for a two-tailed test with 14 degrees of freedom is \( \pm2.145 \).

If our test statistic falls outside the range (\( -2.145 \) to \(+2.145 \)), we reject the null hypothesis, suggesting the sample mean significantly differs from 100. Therefore, only results beyond these critical values provide strong enough evidence to reject the null hypothesis.