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A statewide real estate sales agency, Farm Associates, specializes in selling farm property in the state of Nebraska. Their records indicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, they believe that the mean selling time is now greater than 90 days. A statewide survey of 100 farms sold recently revealed that the mean selling time was 94 days, with a standard deviation of 22 days. At the .10 significance level, has there been an increase in selling time?

Short Answer

Expert verified
There is an increase in selling time.

Step by step solution

01

Define Hypotheses

We first define the null and alternative hypotheses. The null hypothesis (H0) is that there has been no increase in the mean selling time, i.e., \( \mu = 90 \). The alternative hypothesis (H1) is that the mean selling time has increased, i.e., \( \mu > 90 \).
02

Determine Test Type and Significance Level

We will use a one-sample z-test since the sample size is large (n=100) and the population standard deviation is unknown. The significance level is given as \( \alpha = 0.10 \).
03

Calculate the Test Statistic

The test statistic for a one-sample z-test is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where \( \bar{x} = 94 \) (sample mean), \( \mu = 90 \) (population mean), \( \sigma = 22 \) (sample standard deviation), and \( n = 100 \). Calculating this gives: \[ z = \frac{94 - 90}{\frac{22}{\sqrt{100}}} = \frac{4}{2.2} \approx 1.818 \]
04

Find the Critical Z-value

For a right-tailed test at \( \alpha = 0.10 \), we find the critical z-value from the z-table, which is approximately 1.28.
05

Make a Decision

Compare the calculated z-value with the critical z-value. If \( z \) is greater than the critical value, we reject the null hypothesis. In this case, \( 1.818 > 1.28 \), hence we reject the null hypothesis.
06

State the Conclusion

Since the calculated z-value is greater than the critical z-value, there is sufficient evidence at the 0.10 significance level to conclude that the mean selling time has increased.

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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 test used to determine whether there is a significant difference between the means of two populations or between the mean of a sample and the mean of a population. It is particularly useful when the sample size is large and the population standard deviation is not known.

In the scenario where Farm Associates wants to test if the mean selling time has increased due to drought conditions, a one-sample z-test is the appropriate choice.

Here's why a z-test is chosen in this scenario:
  • The sample size is large (n=100), which satisfies one of the conditions for using a z-test.
  • The test compares the sample mean (selling time of 94 days) with a known population mean (90 days).
By calculating the z-value, you can assess how far the sample mean has deviated from the population mean in units of the standard deviation. This helps to make an informed decision about whether the difference is statistically significant.
Significance Level
The significance level, often denoted as \( \alpha \), is a threshold used in hypothesis testing to determine whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true.

In our exercise with Farm Associates, the significance level is set at 0.10, or 10%. This means there's a 10% risk of concluding that selling times have increased when they haven't.

Choosing a significance level is crucial because it influences the confidence in the results of the test.

Some points to consider about significance levels:
  • A lower significance level (like 0.05 or 0.01) means stricter criteria for rejecting the null hypothesis, reducing the risk of a Type I error.
  • A higher significance level, like 0.10, is less strict, which is often acceptable in preliminary research or when the consequences of a Type I error are not severe.
Understanding the significance level helps interpret the results of the hypothesis test more accurately.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement of no effect or no difference. It provides a baseline that the data will be tested against in statistical hypothesis testing.

For the Farm Associates case, the null hypothesis is that the mean selling time has not changed from the known mean of 90 days, or \( \mu = 90 \).

Here are some characteristics of a null hypothesis:
  • It is assumed true until evidence suggests otherwise.
  • Failing to reject the null hypothesis simply suggests insufficient evidence against it, not proving it true.
  • It is usually expressed with an "=" sign, indicating no effect or change.
Rejecting or failing to reject the null hypothesis depends on comparing the test statistic to a critical value derived from the chosen significance level.
Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \) or sometimes \( H_a \), is a statement that contradicts the null hypothesis. It indicates a new effect or difference that the test aims to support.

In the context of the Farm Associates problem, the alternative hypothesis suggests that the mean selling time is greater than 90 days, or \( \mu > 90 \).

Key points about the alternative hypothesis:
  • It represents the research question or what you aim to prove.
  • Unlike the null hypothesis, it is often expressed with inequalities (such as \( > \), \( < \), or \( eq \)).
  • The choice of a one-tailed or two-tailed test affects the structure of the alternative hypothesis. In this case, a one-tailed test is used, as the interest is in testing if the selling time is greater, not just different.
Evaluating the alternative hypothesis involves determining if the sample data provides enough evidence to reject the null hypothesis in favor of this new claim.

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Most popular questions from this chapter

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The following information is available. $$\begin{array}{l}H_{0}: \mu=50 \\\H_{1}: \mu \neq 50\end{array}$$ The sample mean is \(49,\) and the sample size is \(36 .\) The population follows the normal distribution and the standard deviation is \(5 .\) Use the .05 significance level.

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