91Ó°ÊÓ

To test \(H_{0}: \mu=50\) versus \(H_{1}: \mu<50,\) a simple random sample of size \(n=24\) is obtained from a population that is known to be normally distributed, and the sample standard deviation is found to be 6 . (a) A researcher decides to test the hypothesis at the \(\alpha=0.05\) level of significance. Determine the sample mean that separates the rejection region from the nonrejection region. [Hint: Follow the same approach as that laid out on page \(514,\) but use Student's \(t\) -distribution to find the critical value.] (b) Suppose the true population mean is \(\mu=48.9 .\) Use technology to find the area under the \(t\) -distribution to the right of the sample mean found in part (a) assuming \(\mu=48.9\). [Hint: This can be accomplished by performing a one-sample \(t\) -test.] This represents the probability of making a Type II error, \(\beta\). What is the power of the test?

Step by step solution

01

- State the null and alternative hypotheses

The null hypothesis is stated as $$H_0: \mu = 50$$and the alternative hypothesis is $$H_1: \mu < 50$$

02

- Identify the significance level

The significance level is given as \( \alpha = 0.05 \).

03

- Determine the degrees of freedom

Use the formula for degrees of freedom:$$df = n - 1 = 24 - 1 = 23$$

04

- Find the critical value using the Student's \( t \)-distribution

For a one-tailed test with \( \alpha = 0.05 \) and \( df = 23 \), look up the critical value in the \( t \)-distribution table or use technology. The critical value \( t_{\alpha, df} \) is approximately $$ t_{0.05, 23} \approx -1.714$$

05

- Calculate the sample mean that separates the rejection region

Use the formula for the test statistic in the \( t \)-distribution: $$ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} $$Rearrange to solve for \( \bar{x} \):\( -1.714 = \frac{\bar{x} - 50}{6/\sqrt{24}} \)Solving this gives:\( \bar{x} = 50 - 1.714 * \frac{6}{\sqrt{24}} \)\( \bar{x} \approx 48.9768 \)

06

- Find the probability of making a Type II error (\( \beta \))

Given that the true population mean is \( \mu = 48.9 \):Use technology to find the probability of observing a sample mean less than 48.9768 when \( \mu = 48.9 \) and \( s = 6 \)Use a one-sample \( t \)-test with\( H_0 : \mu = 48.9768 \) and \( H_1 : \mu = 48.9 \)Find \( P(t > -0.514) \approx 1 - P(t < 0.514) \approx 0.3038 \)

07

- Determine the power of the test

Calculate the power of the test as: \( 1 - \beta \approx 1 - 0.3038 \approx 0.6962 \)

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.

null hypothesis

In hypothesis testing, the **null hypothesis** (H_0) is a statement we aim to test. It suggests there is no effect or no difference. In our example, the null hypothesis is stated as H_0: iu=50. This means we assume the population mean is 50 until we have enough evidence to suggest otherwise.
We gather data and use statistical methods to determine if this assumption can be rejected.

alternative hypothesis

The **alternative hypothesis** (H_1) contrasts the null hypothesis. It represents what we are trying to find evidence for. In this case, the alternative hypothesis is H_1: iu<50. This means that we are looking to see if the population mean is less than 50.
The goal of hypothesis testing is often to determine which hypothesis is more likely given the sample data.

t-distribution

The **t-distribution** is a type of probability distribution that is symmetric and bell-shaped, like the normal distribution, but with heavier tails. It is used when the sample size is small and/or the population standard deviation is unknown.
For our problem, we use the t-distribution because we have a small sample size (=24) and the sample standard deviation is provided (6) rather than the population standard deviation.

significance level

The **significance level** (alpha) is the probability of rejecting the null hypothesis when it is actually true. It represents the risk we are willing to take of making a Type I error. For this problem, the significance level is 0.05, which implies a 5% risk of making an incorrect rejection.

critical value

The **critical value** is the threshold that defines the boundary for rejecting the null hypothesis. For our one-tailed test with alpha=0.05 and df=23, the critical value from the t-distribution table is approximately −1.714. If our test statistic exceeds this value, we reject the null hypothesis.

degrees of freedom

In hypothesis testing, **degrees of freedom** (df) represent the number of independent values that can vary in our data sample. The formula is generally df=−1, where is the sample size. Here, our degrees of freedom are 24−1=23.

Type II error

A **Type II error** (ierror β) occurs when we fail to reject the null hypothesis even though it is false. In the problem, we calculate β using technology, finding that the probability of making a Type II error is approximately 0.3038.

power of the test

The **power of the test** is the probability of correctly rejecting the null hypothesis when it is false. It is calculated as 1−β. In our case, we find the power of the test to be approximately 0.6962, indicating a roughly 69.62% chance of detecting an effect if there is one.