/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 A researcher estimates that the ... [FREE SOLUTION] | 91Ó°ÊÓ

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A researcher estimates that the average revenue of the largest businesses in the United States is greater than \(\$ 24\) billion. A random sample of 50 companies is selected, and the revenues (in billions of dollars) are shown. At \(\alpha=0.05,\) is there enough evidence to support the researcher's claim? Assume \(\sigma=28.7\). $$ \begin{array}{rrrrr} 178 & 122 & 91 & 44 & 35 \\ 61 & 56 & 46 & 20 & 32 \\ 30 & 28 & 28 & 20 & 27 \\ 29 & 16 & 16 & 19 & 15 \\ 41 & 38 & 36 & 15 & 25 \\ 31 & 30 & 19 & 19 & 19 \\ 24 & 16 & 15 & 15 & 19 \\ 25 & 25 & 18 & 14 & 15 \\ 24 & 23 & 17 & 17 & 22 \\ 22 & 21 & 20 & 17 & 20 \end{array} $$

Short Answer

Expert verified
There is not enough evidence to support the claim at the 0.05 significance level.

Step by step solution

01

Define the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) states that the average revenue of the largest businesses in the United States is equal to \(24\) billion. The alternative hypothesis (\(H_1\)) claims that the average revenue is greater than \(24\) billion. Thus, \(H_0: \mu = 24\) and \(H_1: \mu > 24\), where \(\mu\) is the population mean revenue.
02

Calculate the Sample Mean

First, find the sum of all sample revenues: \( \sum x = 178+122+91+\ldots+20 = 1395 \). The number of observations, \(n\), is \(50\). The sample mean \(\bar{x}\) is \(\bar{x} = \frac{1395}{50} = 27.9\).
03

Determine the Test Statistic

Use the formula for the z-test: \( z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \). Plug in the known values: \( z = \frac{27.9 - 24}{\frac{28.7}{\sqrt{50}}} \). Calculate the standard error: \( \frac{28.7}{\sqrt{50}} \approx 4.06 \). Then, calculate z: \( z = \frac{3.9}{4.06} \approx 0.96 \).
04

Determine the Critical Value

Since this is a one-tailed test with \(\alpha = 0.05\), look for the z-value in the standard normal distribution table for \(0.95\). The critical z-value is approximately \(1.645\).
05

Compare Test Statistic and Critical Value

Compare the calculated z-value ( $0.96$ ) to the critical value ( $1.645$ ). Since $0.96 < 1.645$ , we fail to reject the null hypothesis.
06

Conclusion

With the calculated z-value being less than the critical value, there is not enough statistical evidence at the \(\alpha = 0.05\) level to support the researcher's claim that the average revenue of the largest businesses in the United States is greater than \(24\) billion.

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Key Concepts

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

Z-Test
A Z-test is a statistical method used to determine whether there is a significant difference between a sample mean and a population mean. It is particularly useful when the sample size is large (usually n > 30) and the population standard deviation is known. In our exercise, we utilized the Z-test to assess whether the average revenue of the largest businesses exceeds \(24 billion. By knowing the sample mean (\)27.9 billion), the population mean hypothesized (\(24 billion), and the population standard deviation (\)28.7 billion), we formed our equation. The formula for the Z-test is:
\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. By applying this formula, we calculated a z-value of approximately 0.96, providing a key metric for further hypothesis testing.
Significance Level
The significance level, denoted as \(\alpha\), is the threshold used to determine when to reject the null hypothesis. Essentially, it represents the probability of making an incorrect decision—specifically, rejecting the null hypothesis when it is true. In most cases, a common choice for \(\alpha\) is 0.05. This means that if the probability of observing the test results under the null hypothesis is less than 5%, the null hypothesis is rejected.
The chosen significance level dictates the critical value threshold from which we compare our test statistic. In our exercise, we have selected \(\alpha = 0.05\), which implies a 5% risk of concluding that the average revenue is greater than $24 billion when it might not be. This choice directly affects the criteria for decision-making in the Z-test, aligning with standard practices for hypothesis testing, providing a balanced approach to both types of errors in statistical hypothesis testing.
Critical Value
The critical value is a point on the scale of the test statistic beyond which you reject the null hypothesis. It is determined based on the chosen significance level and the type of test conducted (one-tailed or two-tailed). In our specific case (a one-tailed test with \(\alpha = 0.05\)), the critical value can be found in a Z-table by looking at the z-score corresponding to a cumulative probability of \(1 - \alpha\), or 0.95.
This critical z-value acts as a cutoff point: if our calculated z-value is beyond this critical value, we will reject the null hypothesis. For our exercise, the critical z-value is approximately 1.645. Because the calculated z-value (0.96) is less than 1.645, we do not have enough statistical evidence to reject the null hypothesis. Understanding the role of the critical value is crucial as it dictates the boundary for decision-making, thereby concluding whether the sample data offers sufficient evidence to support the researcher's claim.

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Most popular questions from this chapter

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At \(\alpha=0.10\), is he correct?

Workers with a formal arrangement with their employer to be paid for time worked at home worked an average of 19 hours per week. A random sample of 15 mortgage brokers indicated that they worked a mean of 21.3 hours per week at home with a standard deviation of 6.5 hours. At \(\alpha=0.05,\) is there sufficient evidence to conclude a difference? Construct a \(95 \%\) confidence interval for the true mean number of paid working hours at home. Compare the results of your confidence interval to the conclusion of your hypothesis test and discuss the implications.

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A manufacturer of cigarettes wishes to test the claim that the variance of the nicotine content of the cigarettes the company manufactures is equal to 0.638 milligram. The variance of a random sample of 25 cigarettes is 0.930 milligram. At \(\alpha=0.05,\) test the claim.

A study found that the average stopping distance of a school bus traveling 50 miles per hour was 264 feet. A group of automotive engineers decided to conduct a study of its school buses and found that for 20 randomly selected buses, the average stopping distance of buses traveling 50 miles per hour was 262.3 feet. The standard deviation of the population was 3 feet. Test the claim that the average stopping distance of the company's buses is actually less than 264 feet. Find the \(P\) -value. On the basis of the \(P\) -value, should the null hypothesis be rejected at \(\alpha=0.01 ?\) Assume that the variable is normally distributed.

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A machine fills 12 -ounce bottles with soda. For the machine to function properly, the standard deviation of the population must be less than or equal to 0.03 ounce. A random sample of 8 bottles is selected, and the number of ounces of soda in each bottle is given. At \(\alpha=0.05,\) can we reject the claim that the machine is functioning properly? Use the \(P\) -value method. \(\begin{array}{llll}12.03 & 12.10 & 12.02 & 11.98 \\ 12.00 & 12.05 & 11.97 & 11.99\end{array}\)

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