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Chapter 10: Problem 14

To test \(H_{0}: \mu=80\) versus \(H_{1}: \mu<80,\) a random sample of size \(n=22\) is obtained from a population that is known to be normally distributed with \(\sigma=11\) (a) If the sample mean is determined to be \(\bar{x}=76.9\) compute the test statistic. (b) If the researcher decides to test this hypothesis at the \(\alpha=0.02\) 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?

Short Answer

Expert verified
Test statistic is -1.396. Critical value is -2.054. Fail to reject the null hypothesis.

Step by step solution

01

- State the given information

Given: \[n = 22, \ \bar{x} = 76.9, \ \mu = 80, \ \sigma = 11, \ \alpha = 0.02\] The population is known to be normally distributed.
02

- Formulate the hypotheses

The null hypothesis is: \(H_{0}: \mu=80\) and the alternative hypothesis is: \(H_{1}: \mu<80\).
03

- Calculate the test statistic

The formula for the test statistic is given by: \[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\] Substituting the given values, we get: \[z = \frac{76.9 - 80}{\frac{11}{\sqrt{22}}} \approx -1.396\]
04

- Determine the critical value

For a one-tailed test at \(\alpha = 0.02\), the critical value (\(z_{\alpha}\)) can be found using the standard normal distribution table. \[z_{\alpha} = -2.054\]
05

- Draw the normal curve

Draw a standard normal distribution curve. The critical region is shaded to the left of the critical value \(z_{\alpha} = -2.054\).
06

- Make a decision and conclusion

Compare the test statistic to the critical value: \(-1.396 > -2.054\). Since the test statistic is not in the critical region, we fail to reject the null hypothesis. Therefore, there is not enough evidence to support the claim that \(\mu < 80\) at the \(\alpha = 0.02\) level of significance.

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Key Concepts

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

Null Hypothesis
A null hypothesis, symbolized by \(H_0\), is a statement that there is no effect or no difference. It is a common assumption held true unless proven otherwise with statistical evidence. In our example, the null hypothesis is \(H_0: \mu = 80\). Here, it states that the population mean, \(\mu\), is equal to 80. The objective of hypothesis testing is to determine if there is sufficient evidence to reject this assumption.
Alternative Hypothesis
The alternative hypothesis, denoted by \(H_1\) or \(H_a\), is the statement we aim to prove. It stands in contrast to the null hypothesis. In this case, our alternative hypothesis is \(H_1: \mu < 80\). This implies we are testing if the population mean is less than 80. When performing hypothesis testing, if the data significantly contradicts the null hypothesis, the alternative hypothesis is considered more likely.
Test Statistic
A test statistic is a standardized value calculated from sample data during a hypothesis test. It is used to determine whether to reject the null hypothesis. The formula for the test statistic in our scenario is given by: \[ z = \frac{ \bar{x} - \mu }{ \frac{ \sigma }{ \sqrt{n} } } \] Substituting the given values, we compute: \[ z = \frac{ 76.9 - 80 }{ \frac{ 11 }{ \sqrt{ 22 } } } \approx -1.396 \] This result is the calculated z-score which will be compared against the critical value.
Critical Value
A critical value defines the boundary or cut-off point of the region where we would reject the null hypothesis. For a one-tailed test at a 0.02 significance level, the critical value can be found using the standard normal (Z) distribution table, yielding \(z_{\alpha} = -2.054\). This suggests that if our test statistic is less than -2.054, we will reject the null hypothesis. Otherwise, we do not have enough evidence to do so.
Level of Significance
The level of significance, represented by \(\alpha\), is the threshold probability for determining whether to reject the null hypothesis. It signifies the risk of rejecting \(H_0\) when it is actually true (Type I error). In our test, the level of significance is set to 0.02, meaning there is a 2% risk of making a Type I error. By comparing the test statistic to the critical value, we ensure our decision aligns with this risk threshold.

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Most popular questions from this chapter

(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.

A null and alternative hypothesis is given. Determine whether the hypothesis test is left-tailed, right-tailed, or two tailed. What parameter is being tested? $$\begin{aligned}&H_{0}: p=0.2\\\&H_{1}: p<0.2\end{aligned}$$

Load the hypothesis tests for a mean applet. (a) Set the shape to right skewed, the mean to \(50,\) and the standard deviation to \(10 .\) Obtain 1000 simple random samples of size \(n=8\) from this population, and test whether the mean is different from \(50 .\) How many of the samples led to a rejection of the null hypothesis if \(\alpha=0.05 ?\) How many would we expect to lead to a rejection of the null hypothesis if \(\alpha=0.05 ?\) What might account for any discrepancies? (b) Set the shape to right skewed, the mean to \(50,\) and the standard deviation to \(10 .\) Obtain 1000 simple random samples of size \(n=40\) from this population, and test whether the mean is different from \(50 .\) How many of the samples led to a rejection of the null hypothesis if \(\alpha=0.05 ?\) How many would we expect to lead to a rejection of the null hypothesis if \(\alpha=0.05 ?\)

To test \(H_{0}: \mu=20\) versus \(H_{1}: \mu<20,\) a random sample of size \(n=18\) is obtained from a population that is known to be normally distributed with \(\sigma=3\) (a) If the sample mean is determined to be \(\bar{x}=18.3\) compute and interpret the \(P\) -value. (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, 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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