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Raftelis Financial Consulting reported that the mean quarterly water bill in the United States is \(\$ 47.50(U . S . \text { News \& World Report, August } 12,2002\) ). Some water systems are operated by public utilities, whereas other water systems are operated by private companies. An economist pointed out that privatization does not equal competition and that monopoly powers provided to public utilities are now being transferred to private companies. The concern is that consumers end up paying higher-than-average rates for water provided by private companies. The water system for Atlanta, Georgia, is provided by a private company. A sample of 64 Atlanta consumers showed a mean quarterly water bill of \(\$ 51\) with a sample standard deviation of \(\$ 12 .\) At \(\alpha=.05,\) does the Atlanta sample support the conclusion that above- average rates exist for this private water system? What is your conclusion?

Short Answer

Expert verified
The Atlanta sample supports that water rates are above average increasing consumer expenses.

Step by step solution

01

Identify the hypothesis statements

We must identify the null hypothesis \( H_0 \) and the alternative hypothesis \( H_1 \). Here, \( H_0: \mu = 47.50 \) implies that the mean quarterly water bill in Atlanta is the same as the national average. \( H_1: \mu > 47.50 \) implies that the mean quarterly water bill in Atlanta is higher than the national average.
02

Gather population and sample data

Mean of the national water bill, \( \mu = 47.50 \). Sample mean, \( \bar{x} = 51 \). Sample standard deviation, \( s = 12 \). Sample size, \( n = 64 \). Significance level, \( \alpha = 0.05 \).
03

Calculate the Standard Error (SE)

The standard error of the sample mean is calculated using the formula \( SE = \frac{s}{\sqrt{n}} \). Plugging in the values, \( SE = \frac{12}{\sqrt{64}} = 1.5 \).
04

Calculate the test statistic (t)

The test statistic for the sample mean is calculated using the formula \( t = \frac{\bar{x} - \mu}{SE} \). Substituting the values, we get \( t = \frac{51 - 47.5}{1.5} \approx 2.33 \).
05

Determine the critical t-value and decision rule

For a significance level of \( \alpha = 0.05 \) and degrees of freedom \( df = n - 1 = 63 \), use the t-distribution table to find the critical t-value. Since it's a one-tailed test, the critical value is approximately 1.67. The decision rule is to reject \( H_0 \) if \( t \) is greater than 1.67.
06

Compare test statistic to critical value

Our calculated test statistic is 2.33, which is greater than the critical value of 1.67. Thus, we reject the null hypothesis \( H_0 \).
07

Conclusion

Since the null hypothesis is rejected, there is significant statistical support at \( \alpha = 0.05 \) that the mean quarterly water bill in Atlanta is higher than the national average of \( \$47.50 \).

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Key Concepts

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

Sample Mean
The concept of a sample mean is fundamental in statistics. When we want to understand a population's characteristics, it's often impractical to examine every single member of that population.
Instead, we analyze a smaller subset or sample of the population. The average from this sample is what we call the sample mean.
For example, in the case of the Atlanta water bills, we have a sample of 64 consumers. By calculating their average quarterly water bill, which turned out to be $51, we have our sample mean. Why is the sample mean important? It provides an estimate of the population mean, which is the average value for the entire population.
  • The larger the sample size, the more reliable the sample mean is as an estimate of the population mean.
  • The sample mean helps us make inferences about the whole population using only a fraction of it.
Standard Error
The standard error (SE) is a crucial concept that helps gauge the reliability of the sample mean. It represents the standard deviation of the sample mean's distribution if we were to take an infinite number of samples from the population.
The formula for calculating SE is: \[ SE = \frac{s}{\sqrt{n}} \] where \( s \) stands for the sample standard deviation and \( n \) is the sample size. In our example, the standard error is calculated as follows:\[ SE = \frac{12}{\sqrt{64}} = 1.5 \] This value provides us with a measure of how much we expect the sample mean to vary from the true population mean. Key takeaways about standard error:
  • A smaller SE indicates a more accurate estimate of the population mean.
  • As the sample size increases, the standard error decreases, implying more precision in our estimate.
t-distribution
The t-distribution is a probability distribution used when estimating population parameters when the sample size is small and/or the population standard deviation is unknown. It is similar to a normal distribution but has heavier tails, meaning there is a greater chance of values farther from the mean.
The t-distribution becomes especially relevant in hypothesis testing to determine how far the sample mean is from the population mean.In our exercise, the calculated test statistic was 2.33, derived using the formula:\[ t = \frac{\bar{x} - \mu}{SE} \] For hypothesis tests, the degree of freedom (df) is crucial, calculated as \( n - 1 \) where \( n \) is the sample size. Key concepts about the t-distribution:
  • With larger sample sizes, the t-distribution approaches the standard normal distribution.
  • It provides a critical t-value by which our test statistic is compared to make decisions on the hypothesis.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold used to determine when to reject the null hypothesis. It reflects the risk of concluding a difference exists when there is none.
Typically, a common significance level is \( \alpha = 0.05 \), as seen in our example.It means there is a 5% risk of rejecting the true null hypothesis.
The significance level helps set the critical value, which we use to compare against our test statistic. In our case, the critical value was approximately 1.67. Key points about significance levels:
  • A lower \( \alpha \) means a stricter criterion for rejecting \( H_0 \), reducing the risk of false positives.
  • The choice of \( \alpha \) should reflect the context and potential consequences of making an incorrect decision.
In hypothesis testing, a proper understanding of the significance level ensures robust and coherent conclusions.

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Most popular questions from this chapter

During the 2004 election year, new polling results were reported daily. In an IBD/TIPP poll of 910 adults, 503 respondents reported that they were optimistic about the national outlook, and President Bush's leadership index jumped 4.7 points to 55.3 (Investor's Business Daily, January 14,2004 ). a. What is the sample proportion of respondents who are optimistic about the national outlook? b. A campaign manager wants to claim that this poll indicates that the majority of adults are optimistic about the national outlook. Construct a hypothesis test so that rejection of the null hypothesis will permit the conclusion that the proportion optimistic is greater than \(50 \%\) c. Use the polling data to compute the \(p\) -value for the hypothesis test in part (b). Explain to the manager what this \(p\) -value means about the level of significance of the results.

Carpetland salespersons average \(\$ 8000\) per week in sales. Steve Contois, the firm's vice president, proposes a compensation plan with new selling incentives. Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per salesperson. a. Develop the appropriate null and alternative hypotheses. b. What is the Type I error in this situation? What are the consequences of making this error? c. What is the Type II error in this situation? What are the consequences of making this error?

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. \(\bar{x}=81\)

An airline promotion to business travelers is based on the assumption that two-thirds of business travelers use a laptop computer on overnight business trips. a. State the hypotheses that can be used to test the assumption. b. What is the sample proportion from an American Express sponsored survey that found 355 of 546 business travelers use a laptop computer on overnight business trips? c. What is the \(p\) -value? d. Use \(\alpha=.05 .\) What is your conclusion?

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. \(\bar{x}=20\)

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