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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 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
Yes, the mean selling time has significantly increased to more than 90 days at the 0.10 significance level.

Step by step solution

01

Define Hypotheses

The first step is to define the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the mean selling time is 90 days, i.e., \( \mu = 90 \). The alternative hypothesis \( H_a \) is that the mean selling time is greater than 90 days, i.e., \( \mu > 90 \).
02

Identify Test Type and Significance Level

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

Calculate Test Statistic

The formula for the t-test statistic is \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), where \( \bar{x} = 94 \) is the sample mean, \( \mu_0 = 90 \) is the hypothesized population mean, \( s = 22 \) is the sample standard deviation, and \( n = 100 \) is the sample size. Substitute the values to get: \[ t = \frac{94 - 90}{22/\sqrt{100}} = \frac{4}{2.2} \approx 1.818 \]
04

Determine the Critical Value

Using a t-distribution table or calculator for \( \alpha = 0.10 \) and \( df = n-1 = 99 \), the critical value for a one-tailed test is found to be approximately 1.290.
05

Compare Test Statistic to Critical Value

Compare the calculated test statistic \( t = 1.818 \) to the critical value 1.290. Since 1.818 > 1.290, we reject the null hypothesis.
06

Conclusion

Based on the comparison, there is enough evidence at the 0.10 significance level to conclude that the mean selling time of farm properties has increased to more than 90 days due to the drought conditions.

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Key Concepts

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

Understanding the Null Hypothesis
In hypothesis testing, the null hypothesis is your starting assumption. It's like saying, "There is no change." For the real estate agency in Nebraska, the null hypothesis (\( H_0 \)) assumes that the average selling time of farm properties remains at 90 days. This is a baseline or default position. It asserts that the recent drought has not affected the selling time significantly.

Writing the null hypothesis mathematically helps clarify this point: \( H_0: \mu = 90 \), where \( \mu \) is the mean selling time. Notice how straightforward this is? You're essentially trying to see if there's enough evidence to move away from this assumption, to show that the selling time has actually changed due to conditions like drought.

Keeping it simple, the null hypothesis acts as the claim you're testing against. You're looking for evidence strong enough to reject this claim.
Exploring the Alternative Hypothesis
The alternative hypothesis steps onto the stage when you have reasonable evidence that the initial assumption (null hypothesis) is outdated. For the situation with Farm Associates, the alternative hypothesis (\( H_a \)) posits that the average selling time is more than 90 days. This is directly derived from the agency's suspicion about increased selling times.

In mathematical terms, it's expressed as: \( H_a: \mu > 90 \). The alternative hypothesis provides a specific direction for change, suggesting that the drought conditions have indeed affected the selling time.

Think of the alternative hypothesis as where the action happens - it's the new theory you're testing for. You'd accept it only if you have enough statistical evidence to back it up. Here, we're interested in seeing if the selling time is indeed longer than before, which would have practical implications for the real estate agency in question.
Deciphering the T-Test
The t-test is a common statistical method used to determine if there's a significant difference between a sample mean and a known population mean. It's particularly useful when the population standard deviation is unknown, and the sample size isn't too small, making it ideal for the Farm Associates exercise with 100 samples.

Here's a simple breakdown of how the t-test works for this scenario:
  • Calculate the t-test statistic using the given formula: \( t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \), where \( \bar{x} = 94 \), \( \mu_0 = 90 \), \( s = 22 \), and \( n = 100 \).
  • Plug in the numbers: \( t = \frac{94 - 90}{22/\sqrt{100}} \approx 1.818 \).
  • Compare this computed t-value to a critical value from the t-distribution table at a \(0.10d\) significance level. If the t-value is greater, you reject the null hypothesis.
The t-test essentially helps in making a decision about the null hypothesis. It provides the evidence necessary to either stick with or reject the initial assumption. In this exercise, since the computed t-value is greater than the critical value, there is enough evidence to conclude that the drought indeed increased the average selling time.

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Most popular questions from this chapter

(a) Is this a one- or two-tailed test? (b) What is the decision rule? (c) What is the value of the test statistic? (d) What is your decision regarding \(H_{0} ?\) (e) What is the \(p\) -value? Interpret it. The following information is available. \(H_{0}: \mu=50\) \(H_{i}: \mu \neq 50\) The sample mean is \(49,\) and the sample size is \(36 .\) The population standard deviation is 5\. Use the .05 significance level.

The MacBurger restaurant chain claims that the mean waiting time of customers is 3 minutes with a population standard deviation of 1 minute. The quality- assurance department found in a sample of 50 customers at the Warren Road MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, "You can average more than \(\$ 80\) a day in tips." Assume the standard deviation of the population distribution is 3.24 . Over the first 35 days she was employed at the restaurant, the mean daily amount of her tips was 84.85 . At the .01 significance level, can Ms. Brigden conclude that she is earning an average of more than 80 in tips?

NBC TV news, in a segment on the price of gasoline, reported last evening that the mean price nationwide is 2.50 per gallon for self-serve regular unleaded. A random sample of 35 stations in the Milwaukee, Wisconsin, area revealed that the mean price was \(\$ 2.52\) per gallon and that the standard deviation was 0.05 per gallon. At the .05 significance level, can we conclude that the price of gasoline is higher in the Milwaukee area? Determine the p -value.

The cost of weddings in the United States has skyrocketed in recent years. As a result many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than \(\$ 10,000 .\) Listed below is a total cost in thousands of dollars for a sample of 8 Caribbean weddings $$\begin{array}{|llllllll|}\hline 9.7 & 9.4 & 11.7 & 9.0 & 9.1 & 10.5 & 9.1 & 9.8 \\\\\hline\end{array}$$ At the .05 significance level is it reasonable to conclude the mean wedding cost is less than \(\$ 10,000\) as advertised?

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