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True or False: To test \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0}\) using a \(5 \%\) level of significance, we could construct a \(95 \%\) confidence interval.

Short Answer

Expert verified
True, a 95% confidence interval can be used to test \(H_{0}\) at a 5% significance level.

Step by step solution

01

Understand the hypothesis

The null hypothesis is given as \(H_{0}: \mu=\mu_{0}\), and the alternative hypothesis is \(H_{1}: \mu \eq \mu_{0}\). We are tasked with determining the validity of constructing a 95% confidence interval to test this hypothesis at a 5% significance level.
02

Level of significance and confidence interval relationship

A key point to understand is that a 95% confidence interval corresponds to a 5% level of significance. This is because the level of significance is the probability of rejecting the null hypothesis when it is true, while the confidence interval gives the range of values that should contain the true parameter value 95% of the time.
03

Confidence interval construction

Constructing a 95% confidence interval around the sample mean estimates the range of values in which the population mean \(\mu\) is likely to fall with 95% confidence. If \(\mu_{0}\) lies outside this interval, we reject the null hypothesis.
04

Making a decision

If \(\mu_{0}\) falls outside the 95% confidence interval, it suggests that there is enough evidence to reject the null hypothesis \(H_{0}\) at a 5% level of significance. If \(\mu_{0}\) is inside the interval, we fail to reject \(H_{0}\).

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Key Concepts

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

Level of Significance
The level of significance, often denoted as \(\alpha\), is a key concept in hypothesis testing. It represents the probability of rejecting the null hypothesis (\(H_{0}\)) when it is, in fact, true. In simpler terms, it's the threshold we set for determining whether the observed data is significantly different from what we would expect if \(H_{0}\) were true.
For example, in this exercise, a 5% level of significance (\(\alpha = 0.05\)) is used. This means there is a 5% risk of concluding that there is an effect or a difference when there is none. It implies that we are 95% confident in our decision when we do not reject the null hypothesis.
In practice, a lower \alpha\ value (like 1% or 0.01) means stricter criteria to reject \(H_{0}\), reducing the risk of Type I error (false positive). Conversely, a higher \alpha\ value makes it easier to reject \(H_{0}\) but increases the risk of a Type I error.
Confidence Interval
A confidence interval provides a range of values in which the true population parameter is expected to fall. For instance, a 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each sample, we would expect the true parameter to fall within these intervals in about 95 of those samples.
The relationship between confidence intervals and the level of significance is inverse. If we use a 5% level of significance, this corresponds to a 95% confidence interval. This is because the confidence level (\(1 - \alpha\)) defines how confident we are that the interval contains the true population parameter.
In the context of this exercise, constructing a 95% confidence interval helps us to check if \(\mu_{0}\) lies within this interval. If it does not, we reject the null hypothesis. If it does, we do not have enough evidence to reject \(H_{0}\).
Null Hypothesis
The null hypothesis (\(H_{0}\)) is a fundamental concept in statistical testing. It represents a statement of no effect or no difference and is the default assumption that there is no relationship between two measured phenomena.
In our exercise, the null hypothesis is stated as \(H_{0}: \mu = \mu_{0}\), meaning we assume the population mean \(\mu\) is equal to a specific value \(\mu_{0}\). The alternative hypothesis (\(H_{1}\)) suggests that the population mean is not equal to \(\mu_{0}\).
Hypothesis testing involves comparing the null hypothesis with the observed data. If the data provides sufficient evidence against \(H_{0}\), we reject it in favor of \(H_{1}\). Using the 95% confidence interval, we check whether \(\mu_{0}\) falls within this interval. If it does not, we conclude that there is significant evidence to reject \(H_{0}\) at the 5% level of significance. If \(\mu_{0}\) is within the interval, we do not reject \(H_{0}\).
In summary, the null hypothesis acts as the foundation of hypothesis testing, guiding us in making statistical decisions based on our data.

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Most popular questions from this chapter

To test \(H_{0}: \mu=45\) versus \(H_{1}: \mu \neq 45,\) a random sample of size \(n=40\) is obtained from a population whose standard deviation is known to be \(\sigma=8\) (a) Does the population need to be normally distributed to compute the \(P\) -value? (b) If the sample mean is determined to be \(\bar{x}=48.3\) compute and interpret the \(P\) -value. (c) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, will the researcher reject the null hypothesis? Why?

Test the hypothesis, using (a) the classical approach and then (b) the P-value approach. Be sure to verify the requirements of the test. $$\begin{aligned}&H_{0}: p=0.55 \text { versus } H_{1}: p<0.55\\\&n=150 ; x=78 ; \alpha=0.1\end{aligned}$$

(a) determine the null and alternative hypotheses, (b) explain what it would mean to make a Type I error, and (c) explain what it would mean to make a Type II error. Federal law requires that a jar of peanut butter that is labeled as containing 32 ounces must contain at least 32 ounces. A consumer advocate feels that a certain peanut butter manufacturer is shorting customers by under filling the jars so that the mean content is less than the 32 ounces stated on the label.

To test \(H_{0}: \mu=50\) versus \(H_{1}: \mu<50,\) a random sample of size \(n=24\) is obtained from a population that is known to be normally distributed with \(\sigma=12\) (a) If the sample mean is determined to be \(\bar{x}=47.1\) compute the test statistic. (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, determine the critical value. (c) Draw a normal curve that depicts the critical region. (d) Will the researcher reject the null hypothesis? Why?

Carl Reinhold August Wunderlich said that the mean temperature of humans is \(98.6^{\circ} \mathrm{F}\). Researchers Philip Mackowiak, Steven Wasserman, \(\begin{array}{lllllll}\text { and } & \text { Myron } & \text { Levine } & {[\mathrm{JAMA},} & \text { Sept. } & 23-30 & 1992\end{array}\) \(268(12): 1578-80]\) thought that the mean temperature of humans is less than \(98.6^{\circ}\) F. They measured the temperature of 148 healthy adults 1 to 4 times daily for 3 days, obtaining 700 measurements. The sample data resulted in a sample mean of \(98.2^{\circ} \mathrm{F}\) and a sample standard deviation of \(0.7^{\circ} \mathrm{F}\) (a) Test whether the mean temperature of humans is less than \(98.6^{\circ} \mathrm{F}\) at the \(\alpha=0.01\) level of significance using the classical approach. (b) Determine and interpret the \(P\) -value.

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