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The American Water Works Association reports that the per capita water use in a single-family home is 69 gallons per day. Legacy Ranch is a relatively new housing development. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Thirty-six owners responded, and the sample mean water use per day was 64 gallons with a standard deviation of 8.8 gallons per day. At the .10 level of significance, is that enough evidence to conclude that residents of Leqacy Ranch use less water on average?

Short Answer

Expert verified
Yes, there's enough evidence to conclude residents use less water on average.

Step by step solution

01

Define the Hypotheses

First, we need to set up our null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the mean water usage at Legacy Ranch is equal to the national average, 69 gallons per day. The alternative hypothesis (\(H_1\)) is that the mean water usage at Legacy Ranch is less than the national average.\[ H_0: \mu = 69 \] \[ H_1: \mu < 69 \]
02

Determine the Significance Level and Test Type

The significance level is given as 0.10. Since we are testing if the water usage is "less than" the national average, this will be a one-tailed test.
03

Calculate the Test Statistic

To test the hypothesis, we'll use the t-statistic formula for a sample:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where \(\bar{x} = 64\) is the sample mean, \(\mu = 69\) is the population mean, \(s = 8.8\) is the sample standard deviation, and \(n = 36\) is the sample size.Plugging in these values:\[ t = \frac{64 - 69}{8.8/\sqrt{36}} = \frac{-5}{1.467} \approx -3.41 \]
04

Determine the Critical Value

Using a t-table or calculator for a t-distribution with \(n-1 = 35\) degrees of freedom at the 0.10 level of significance, the critical t-value for a one-tailed test is approximately -1.306.
05

Compare and Make a Decision

Now, compare the calculated t-statistic (-3.41) with the critical t-value (-1.306). Since -3.41 is less than -1.306, we reject the null hypothesis.

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Key Concepts

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

t-statistic
The t-statistic is a key component of hypothesis testing, especially when dealing with small sample sizes. It helps us determine how much the sample mean deviates from the population mean, in terms of standard errors. This is crucial when the population standard deviation is unknown, as is often the case in real-world scenarios like the water usage test.

To calculate the t-statistic, we use the formula:\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}}\]Here, \(\bar{x}\) represents the sample mean, \(\mu\) is the population mean, \(s\) is the sample standard deviation, and \(n\) is the number of samples.

The calculated t-statistic allows us to compare the sample statistic to what is expected under the null hypothesis. A significant t-statistic indicates that the observed sample mean is likely not just due to random chance.
significance level
The significance level, often denoted as \(\alpha\), reflects the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it's the threshold for determining when the evidence is strong enough to support rejecting the null hypothesis.

Common significance levels include 0.05, 0.01, and, as in our water usage problem, 0.10. This chosen level informs us of how often we are willing to make a Type I error—wrongly rejecting a true null hypothesis.

Choosing a significance level is crucial as it influences the critical value used to decide the fate of the null hypothesis. Lower significance levels require stronger evidence to reject the null hypothesis.
one-tailed test
A one-tailed test is used in hypothesis testing when we are interested in a specific direction of the effect. For instance, it tests if a sample statistic is significantly greater than or less than a population parameter.

In the context of the water usage test, we set up a one-tailed test to determine if Legacy Ranch uses **less** water on average than the national norm. We are focused only on one specific direction—reduction in water use—hence the name "one-tailed." This orientation affects the placement of the critical region on the distribution chart.

One-tailed tests can be more powerful than two-tailed tests as they require less evidence to detect an effect in one direction, but riskier as they ignore any possibility of an effect in the other direction.
null hypothesis
The null hypothesis, denoted as \(H_0\), is the default assumption that there is no effect or no difference. It serves as a baseline for statistical testing. In the water usage context, the null hypothesis is that Legacy Ranch's average water usage is the same as the national average, specifically 69 gallons per day.

Formally, it is expressed as:

\[ H_0: \mu = 69 \]

Testing the null hypothesis allows us to determine if there is enough statistical evidence to declare a significant difference. Typically, hypothesis tests focus on rejecting the null hypothesis, thus supporting an alternative hypothesis.
alternative hypothesis
The alternative hypothesis, denoted as \(H_1\), is complementary to the null hypothesis. It represents the assumption that there is indeed a difference or effect. For our water usage example, the alternative hypothesis posits that Legacy Ranch uses less water on average than the national average of 69 gallons per day.

It is expressed as:

\[ H_1: \mu < 69 \]

This hypothesis aligns with our main question about the water efficiency of the new housing development. Statistical tests aim to see if there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis. A significant result supports the notion that Legacy Ranch consumes less water, verifying the effectiveness of the efficient fixtures.

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Most popular questions from this chapter

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