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In 2001 , the U.S. Department of Labor reported the average hourly earnings for U.S. production workers to be \(\$ 14.32\) per hour (The World Almanac, 2003 ). A sample of 75 production workers during 2003 showed a sample mean of \(\$ 14.68\) per hour. Assuming the population standard deviation \(\sigma=\$ 1.45,\) can we conclude that an increase occurred in the mean hourly earnings since \(2001 ?\) Use \(\alpha=.05\).

Short Answer

Expert verified
Yes, the mean hourly earnings increased.

Step by step solution

01

State the Hypotheses

We start by defining the null hypothesis and the alternative hypothesis for this test. The null hypothesis \( H_0 \) states that there is no increase in the mean hourly earnings, i.e., \( \mu = 14.32 \). The alternative hypothesis \( H_a \) states that there is an increase in the mean hourly earnings, i.e., \( \mu > 14.32 \).
02

Determine the Significance Level

The significance level \( \alpha \) is given as 0.05. This means we have a 5% risk of rejecting the null hypothesis when it is true.
03

Compute the Test Statistic

To compare the sample mean with the population mean under the assumption of known population standard deviation, we use the z-test. The test statistic \( z \) is calculated using the formula:\[z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}\]Where:- \( \bar{x} = 14.68 \) is the sample mean,- \( \mu = 14.32 \) is the population mean under the null hypothesis,- \( \sigma = 1.45 \) is the population standard deviation,- \( n = 75 \) is the sample size.Substituting in the values:\[z = \frac{14.68 - 14.32}{1.45/\sqrt{75}} = \frac{0.36}{0.1674} \approx 2.15\]
04

Determine the Critical Value

For a significance level \( \alpha = 0.05 \) in a right-tailed test, we look up the critical z-value from the standard normal distribution table. The critical value corresponding to \( \alpha = 0.05 \) is approximately 1.645.
05

Make a Decision

Compare the computed z-test statistic to the critical value. Since \( z = 2.15 \) is greater than the critical value of 1.645, we reject the null hypothesis \( H_0 \). This indicates that there is statistically significant evidence to conclude that the mean hourly earnings have increased since 2001.

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Key Concepts

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

Statistical Significance
Statistical significance is a vital concept in hypothesis testing, used to determine whether the results of an experiment or study are likely to be true or if they may have occurred by random chance. When conducting a hypothesis test, researchers decide in advance how much risk of error they are willing to accept, which is called the significance level, denoted as \( \alpha \).
For example, a significance level of \( \alpha = 0.05 \) means the researcher is willing to accept a 5% probability of rejecting the null hypothesis when it is actually true. This is a common threshold used in many fields.
When a test's result is statistically significant, it means the evidence is strong enough to reject the null hypothesis in favor of the alternative hypothesis. This does not imply absolute certainty, but rather that the observed data is unlikely to have occurred under the assumption that the null hypothesis is correct.
A critical part of this decision is comparing the test statistic to a critical value, which helps to measure how extreme the observed data are. If the test statistic exceeds the critical value, the result is deemed statistically significant.
Z-Test
The Z-Test is a statistical method used to determine if there is a significant difference between sample data points and the population from which the sample was drawn. It is particularly useful when the population standard deviation is known, and the sample size is large (typically \( n \geq 30 \)).
The formula for calculating the Z-Test statistic is:
  • \( z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \)
where:
  • \( \bar{x} \) is the sample mean,
  • \( \mu \) is the population mean under the null hypothesis,
  • \( \sigma \) is the population standard deviation,
  • \( n \) is the sample size.
The resulting \( z \)-value is then compared against a critical value, which is derived from the significance level \( \alpha \) and the direction of the test (left, right, or two-tailed).
The Z-Test is quite powerful for hypotheses testing because it allows for a more straightforward interpretation when standardized against the standard normal distribution, making the decision-making process clear: if the \( z \)-value transcends the critical threshold, it provides evidence against the null hypothesis.
Right-Tailed Test
A right-tailed test, sometimes called a one-tailed test, is a type of hypothesis test where the area of interest is in the right tail of the distribution curve. This test is appropriate when the researcher is testing for the possibility of an effect in only one direction. In the context of hypothesis testing, this means the alternative hypothesis posits that the parameter of interest is greater than a certain value.
In our scenario of testing whether the mean hourly earnings have increased, the null hypothesis \( H_0 \) states that the mean is \( \mu = 14.32 \) dollars, whereas the alternative hypothesis \( H_a \) suggests \( \mu > 14.32 \) dollars. If the test statistic falls into the critical region (extreme right of the curve), we reject \( H_0 \).
The critical region is determined using the significance level \( \alpha \). For instance, with \( \alpha = 0.05 \), the critical value is the \( z \)-score where only 5% of the standard normal distribution lies to its right, typically \( z \approx 1.645 \). If the computed test statistic \( z \) is greater than 1.645, the test results are statistically significant toward the right tail, indicating the mean hourly earnings have indeed increased.

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Most popular questions from this chapter

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu=22 \\ H_{\mathrm{a}}: \mu \neq 22 \end{array} \\] A sample of 75 is used and the population standard deviation is \(10 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\bar{x}=23\) b. \(\bar{x}=25.1\) c. \(\bar{x}=20\)

For the United States, the mean monthly Internet bill is \(\$ 32.79\) per household (CNBC, January 18,2006 ). A sample of 50 households in a southern state showed a sample mean of \(\$ 30.63 .\) Use a population standard deviation of \(\sigma=\$ 5.60\) a. Formulate hypotheses for a test to determine whether the sample data support the conclusion that the mean monthly Internet bill in the southern state is less than the national mean of \(\$ 32.79\) b. What is the value of the test statistic? c. What is the \(p\) -value? d. \(\quad\) At \(\alpha=.01,\) what is your conclusion?

CCN and ActMedia provided a television channel targeted to individuals waiting in supermarket checkout lines. The channel showed news, short features, and advertisements. The length of the program was based on the assumption that the population mean time a shopper stands in a supermarket checkout line is 8 minutes. A sample of actual waiting times will be used to test this assumption and determine whether actual mean waiting time differs from this standard. a. Formulate the hypotheses for this application. b. \(\quad\) A sample of 120 shoppers showed a sample mean waiting time of 8.5 minutes. Assume a population standard deviation \(\sigma=3.2\) minutes. What is the \(p\) -value? c. \(\quad\) At \(\alpha=.05,\) what is your conclusion? d. Compute a \(95 \%\) confidence interval for the population mean. Does it support your conclusion?

Speaking to a group of analysts in January \(2006,\) a brokerage firm executive claimed that at least \(70 \%\) of investors are currently confident of meeting their investment objectives. A UBS Investor Optimism Survey, conducted over the period January 2 to January \(15,\) found that \(67 \%\) of investors were confident of meeting their investment objectives (CNBC, January \(20,2006)\) a. Formulate the hypotheses that can be used to test the validity of the brokerage firm executive's claim. b. Assume the UBS Investor Optimism Survey collected information from 300 investors. What is the \(p\) -value for the hypothesis test? c. At \(\alpha=.05,\) should the executive's claim be rejected?

Consider the following hypothesis test: \\[ \begin{array}{l} H_{0}: \mu \geq 80 \\ H_{\mathrm{a}}: \mu<80 \end{array} \\] A sample of 100 is used and the population standard deviation is \(12 .\) Compute the \(p\) -value and state your conclusion for each of the following sample results. Use \(\alpha=.01\) a. \(\quad \bar{x}=78.5\) b. \(\bar{x}=77\) c. \(\quad \bar{x}=75.5\) d. \(\bar{x}=81\)

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