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A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation of \(s=.29 .\) Your research objective is to show that the population mean \(\mu\) exceeds 2.3. Use this information to answer the questions. Calculate \(\beta=P\left(\right.\) accept \(H_{0}\) when \(\left.\mu=2.4\right)\)

Short Answer

Expert verified
Answer: The power of the test is approximately 0.946, which means there's a 94.6% chance of accepting the null hypothesis when the population mean is actually 2.4.

Step by step solution

01

Set up the hypotheses

The null hypothesis \(H_{0}\) states that the population mean \(\mu\) is less than or equal to 2.3: \(H_{0}: \mu \leq 2.3\) The alternative hypothesis \(H_{a}\) states that the population mean \(\mu\) is greater than 2.3: \(H_{a}: \mu > 2.3\)
02

Calculate the test statistic

We will use the sample mean \(\bar{x}\) to calculate the test statistic. The formula is: \(T = \frac{\bar{x} - \mu_{0}}{\frac{s}{\sqrt{n}}}\) where \(T\) is the test statistic, \(\bar{x}\) is the sample mean, \(\mu_{0}\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Using the given values, we have: \(T = \frac{2.4 - 2.3}{\frac{0.29}{\sqrt{35}}}\) Now, calculate the test statistic: \(T \approx 2.147\)
03

Find the critical value

To find the critical value, we need to determine the significance level, \(\alpha\). Assuming a significance level of 0.05 (5%), we will find the critical value corresponding to a one-tailed test in the t-distribution (because \(\sigma\) is unknown). For a one-tailed test with \(\alpha = 0.05\) and \(n-1 = 34\) degrees of freedom, the critical value (\(t\)) is approximately equal to 1.69.
04

Compare test statistic to critical value

Since the test statistic (\(T \approx 2.147\)) is greater than the critical value (\(t = 1.69\)), we reject the null hypothesis, \(H_{0}\).
05

Calculate the power of the test

To calculate the power of the test, we need to find the probability of making a Type II error when \(\mu = 2.4\). We have: 1. \(\alpha = 0.05\) (significance level) 2. \(T = \frac{2.4 - 2.4}{\frac{0.29}{\sqrt{35}}}\) (test statistic) Now, calculate the new test statistic when \(\mu = 2.4\): \(T = 0\)
06

Calculate \(\beta\)

We must find the probability of accepting \(H_{0}\) when the test statistic \(T = 0\). To do so, we must find the probability of observing a test statistic less than the critical value. This is the probability of observing a value less than 1.69 in the t-distribution with 34 degrees of freedom: \(\beta = P(T < 1.69)\) Using a t-distribution table or calculator, we have: \(\beta \approx 0.946\) So, the power of the test is approximately 0.946, which means there's a 94.6% chance of accepting the null hypothesis when the population mean is actually 2.4.

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Key Concepts

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

Type II Error
When we're talking about hypothesis testing in statistics, a Type II error represents a scenario where the test fails to reject the null hypothesis (\(H_0\)) even though the alternative hypothesis (\(H_a\)) is true. In other words, it's the error of not detecting an effect or difference when there actually is one.

This is often symbolized as \beta, which is the probability of a Type II error occurring. It's important to distinguish between this and \balpha, which is the probability of a Type I error—the incorrect rejection of a true null hypothesis. The power of a test, which is 1 - \beta, measures a test's ability to correctly reject the null hypothesis when the alternative hypothesis is true.

In the given exercise, calculating \beta serves to estimate the likelihood that the statistical test will accept the null hypothesis that the population mean is less than or equal to 2.3 when, in fact, it is 2.4. This type of analysis is crucial because it helps researchers understand the potential effectiveness of their tests and whether they have a high probability of missing an actual significant result.
Test Statistic Calculation
The calculation of a test statistic is at the heart of hypothesis testing procedures. It's a standardized value that arises from sample data and is used to make decisions about the null hypothesis. The test statistic helps determine how far, in standard error measures, a sample statistic lies from the hypothesized parameter under the null hypothesis.

For instance, when we're assessing a sample mean against a population mean, the formula \(T = \frac{\bar{x} - \mu_{0}}{\frac{s}{\sqrt{n}}}\) is used, where \(\bar{x}\) is the sample mean, \(s\) the sample standard deviation, \(n\) the sample size, and \(\mu_{0}\) the hypothesized population mean. A higher absolute value of the test statistic indicates a greater difference between the sample statistic and the hypothesized parameter.

In the exercise, the test statistic was found to be approximately 2.147, suggesting the sample mean is farther from the null hypothesis value than what random chance would likely offer, leading us to consider rejecting the null hypothesis.
t-Distribution
The t-distribution, also known as the Student's t-distribution, is used in hypothesis testing when the sample size is small and the population standard deviation is unknown. It's a probability distribution that arises from estimating the mean of a normally-distributed population in situations where the sample size is small and population standard deviation is not known.

It's similar to the standard normal distribution but has heavier tails, meaning it is more prone to producing values that fall far from its mean. This properties of the t-distribution are particularly important when working with small sample sizes, as these heavy tails increase the chances of observing extreme values, which helps to compensate for the uncertainty and variability that arise from the smaller samples.

In hypothesis testing, we compare the test statistic to a critical value from the t-distribution to decide whether to reject the null hypothesis. For example, in the exercise provided, with 34 degrees of freedom and a significance level of 0.05, the critical value from the t-distribution was approximately 1.69. The computed test statistic, which is greater than this critical value, leads to the rejection of the null hypothesis.

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Most popular questions from this chapter

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