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A statewide real estate sales agency, Farm Associates, specializes in selling farm property in the state of Nebraska. Its records indicate that the mean selling time of farm property is 90 days. Because of recent drought conditions, the agency believes that the mean selling time is now greater than 90 days. A statewide survey of 100 recently sold farms revealed a mean selling time of 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 sufficient evidence to conclude that selling time has increased.

Step by step solution

01

State the Hypotheses

First, let's establish our null and alternative hypotheses. The null hypothesis (H_0) states that the mean selling time is equal to 90 days: \( H_0: \mu = 90 \). The alternative hypothesis (H_1) claims that the mean selling time has increased, so: \( H_1: \mu > 90 \).
02

Identify the Significance Level and Standardized Test

The significance level given is 0.10. Because the sample size is 100, we use a z-test. We will calculate the z-score to determine if the increase is statistically significant.
03

Calculate the Standard Error

The standard error of the mean is computed using the formula: \[ SE = \frac{\sigma}{\sqrt{n}} \] where \( \sigma = 22 \) and \( n = 100 \). Substituting these values, we get: \[ SE = \frac{22}{\sqrt{100}} = 2.2 \].
04

Find the Z-Score

Compute the z-score to compare the sample mean to the population mean using the formula: \[ z = \frac{\bar{x} - \mu_0}{SE} \] where \( \bar{x} = 94 \), \( \mu_0 = 90 \), and \( SE = 2.2 \). Plugging in the values gives: \[ z = \frac{94 - 90}{2.2} \approx 1.82 \].
05

Determine the Critical Value

For a right-tailed test at the 0.10 significance level, we look up the critical z-value in the standard normal distribution table, which is approximately 1.28.
06

Compare the Z-Score with the Critical Value

Compare the calculated z-score (1.82) against the critical z-value (1.28). Since 1.82 > 1.28, we have enough evidence to reject the null hypothesis.
07

Draw a Conclusion

Because the calculated z-score is greater than the critical value, this means there is sufficient statistical evidence at the 0.10 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.

Null Hypothesis
The null hypothesis ( H鈧 ) is a fundamental concept in hypothesis testing. It represents a statement of no effect or no difference. In the context of this real estate example, the null hypothesis is that the mean selling time of farm properties is still 90 days. This is the default position we start with, which assumes there has been no change despite the drought conditions. When formulating a null hypothesis, it typically includes an equal sign. This is what we test against any observed changes in the data. If the null hypothesis is true, any observed differences would be due to random sampling variability. In mathematical terms, we state this as: - H鈧: 渭 = 90 Remember, the goal of testing is often to find evidence against the null hypothesis. If the data provides enough evidence to prove that conditions have changed, we may reject the null hypothesis.
Alternative Hypothesis
The alternative hypothesis ( H鈧 or H鈧 ) is what we consider if we find sufficient evidence against the null hypothesis. It proposes that there is a genuine effect or difference. In this scenario, because the firm expects selling times to have increased, the alternative hypothesis is that the mean selling time is now greater than 90 days. The alternative hypothesis is based on what we wish to prove or suspect. It often includes signs like "greater than," "less than," or "not equal to." In this case, the agency strongly suspects longer selling times due to drought conditions, which is mathematically stated as: - H鈧: 渭 > 90 Testing the alternative hypothesis tells us whether we have enough statistical evidence to claim a change or effect has occurred, thus influencing decisions based on significant findings.
Significance Level
The significance level, often denoted by 伪 (alpha), is a critical threshold in hypothesis testing. It defines the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. In this example, the significance level is set at 0.10. This means we are willing to accept a 10% chance of mistakenly concluding that the mean selling time has increased when it actually hasn't. Choosing a significance level also impacts the critical value used to assess our findings. For instance, at a 0.10 significance level, the corresponding critical z-value for a right-tailed test is approximately 1.28. This z-value helps determine whether the observed changes in sample data (such as the mean selling time) are statistically significant when compared against the null hypothesis. Operations or studies where precision is paramount might require a smaller significance level like 0.05 or 0.01, but in broader applications, a 0.10 level offers a balance between risk of error and effort to demonstrate notable differences.
Z-Test
The z-test is a statistical method used to determine if there is a significant difference between the sample mean and the population mean, assuming the population standard deviation is known and the sample size is sufficient (usually 30 or more). In our farm property scenario, a z-test was chosen because the sample size is 100, and the standard deviation is known to be 22 days.The z-test involves calculating a z-score, which quantifies the deviation of the sample mean from the population mean in units of standard error. The formula used in this context is:- \[ z = \frac{\bar{x} - \mu_0}{SE} \]Where \(\bar{x}\) is the sample mean, \(\mu_0\) is the population mean, and \(SE\) is the standard error of the mean.By comparing our calculated z-score (1.82) to the critical z-value from the standard normal distribution table (1.28 for 伪 = 0.10), we found the z-score exceeds the critical value. This result allows us to reject the null hypothesis and conclude with statistical significance that the mean selling time of farm properties has increased.

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Most popular questions from this chapter

During recent seasons, Major League Baseball has been criticized for the length of the games. A report indicated that the average game lasts 3 hours and 30 minutes. A sample of 17 games revealed the following times to completion. (Note that the minutes have been changed to fractions of hours, so that a game that lasted 2 hours and 24 minutes is reported at 2.40 hours.) \(\begin{array}{lllllllll}2.98 & 2.40 & 2.70 & 2.25 & 3.23 & 3.17 & 2.93 & 3.18 & 2.80 \\ 2.38 & 3.75 & 3.20 & 3.27 & 2.52 & 2.58 & 4.45 & 2.45 & \end{array}\) Can we conclude that the mean time for a game is less than 3.50 hours? Use the .05 significance level.

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A recent study revealed the typical American coffee drinker consumes an average of 3.1 cups per day. A sample of 12 senior citizens revealed they consumed the following amounts of coffee, reported in cups, yesterday. \(\begin{array}{lllllllllll}3.1 & 3.3 & 3.5 & 2.6 & 2.6 & 4.3 & 4.4 & 3.8 & 3.1 & 4.1 & 3.1 & 3.2\end{array}\) At the .05 significance level, do these sample data suggest there is a difference between the national average and the sample mean from senior citizens?

A recent survey by nerdwallet.com indicated Americans paid a mean of \(\$ 6,658\) interest on credit card debt in 2017 . A sample of 12 households with children revealed the following amounts. At the .05 significance level, is it reasonable to conclude that these households paid more interest? \(\begin{array}{lllllllll}7,077 & 5,744 & 6,753 & 7,381 & 7,625 & 6,636 & 7,164 & 7,348 & 8,060 & 5,848 & 9,275 & 7,052\end{array}\)

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