/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Nationally, about \(11 \%\) of t... [FREE SOLUTION] | 91Ó°ÊÓ

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Nationally, about \(11 \%\) of the total U.S. wheat crop is destroyed each year by hail (Reference: Agricultural Statistics, U.S. Department of Agriculture). An insurance company is studying wheat hail damage claims in Weld County, Colorado. A random sample of 16 claims in Weld County gave the following data (\% wheat crop lost to hail). $$\begin{array}{cccccccc} 15 & 8 & 9 & 11 & 12 & 20 & 14 & 11 \\ 7 & 10 & 24 & 20 & 13 & 9 & 12 & 5 \end{array}$$ The sample mean is \(\bar{x}=12.5 \% .\) Let \(x\) be a random variable that represents the percentage of wheat crop in Weld County lost to hail. Assume that \(x\) has a normal distribution and \(\sigma=5.0 \% .\) Do these data indicate that the percentage of wheat crop lost to hail in Weld County is different (either way) from the national mean of \(11 \% ?\) Use \(\alpha=0.01\).

Short Answer

Expert verified
The data do not indicate a difference from the national mean.

Step by step solution

01

Define the Hypotheses

We will use a hypothesis test to determine if the percentage of wheat crop lost to hail in Weld County is different from the national average. Let the null hypothesis (H0) be that the mean percentage loss in Weld County is equal to the national mean: \[ H_0: \mu = 11\% \]The alternative hypothesis (Ha) is that the mean percentage loss in Weld County is different from the national mean:\[ H_a: \mu eq 11\% \]
02

Identify the Test Statistic

Since we have a sample mean and known population standard deviation, we can use the z-test for the hypothesis. The test statistic is calculated using the formula:\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where:- \(\bar{x} = 12.5\%\) is the sample mean,- \(\mu = 11\%\) is the population mean,- \(\sigma = 5.0\%\) is the population standard deviation,- \(n = 16\) is the sample size.
03

Calculate the Test Statistic

Substitute the known values into the z-test formula:\[ z = \frac{12.5 - 11}{\frac{5.0}{\sqrt{16}}} = \frac{1.5}{1.25} = 1.2 \]So, the calculated z-statistic is 1.2.
04

Determine the Critical Value and Decision Rule

For a two-tailed test at \(\alpha = 0.01\), we look up the critical z-values. The critical z-values for \(\alpha = 0.01\) are approximately \(-2.576\) and \(2.576\). The decision rule is: - Reject the null hypothesis if \(z\) is less than \(-2.576\) or greater than \(2.576\). - Do not reject the null hypothesis if \(-2.576 \leq z \leq 2.576\).
05

Make a Decision

The calculated z-value of 1.2 does not fall in the critical region, as it is between \(-2.576\) and \(2.576\). Therefore, we do not reject the null hypothesis.
06

Conclusion

At a significance level of \(\alpha = 0.01\), there is not enough evidence to conclude that the percentage of wheat crop lost to hail in Weld County is different from the national mean of 11%.

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Key Concepts

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

Z-test
The Z-test is a statistical method used to determine if there's a significant difference between the sample mean of a population and a known population mean. It is particularly useful when you have a known population standard deviation. In the context of the wheat crop hail damage, the Z-test helps us decide if the mean percentage of crop loss in Weld County deviates from the national mean of 11%.

To perform a Z-test, we calculate a test statistic, which is the ratio of the difference between the sample mean and the population mean to the standard error of the mean. This ratio tells us how many standard deviations the sample mean is from the population mean. The formula is given by:
  • z = \( \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \)
  • \( \bar{x} \) is the sample mean
  • \( \mu \) is the population mean
  • \( \sigma \) is the population standard deviation
  • \( n \) is the sample size
The result is then compared against critical values in a Z-table to determine if we should accept or reject the null hypothesis.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. A low standard deviation means that the data points tend to be close to the mean of the dataset, whereas a high standard deviation indicates that the data points are spread out over a wider range.

In hypothesis testing scenarios like the wheat crop example, the standard deviation ( 5.0% ) represents the expected variation in the percentage of wheat crop loss due to hail across the whole country. It is a crucial component in calculating the standard error and thereby the Z-test statistic.

Understanding standard deviation is essential because it helps us evaluate how much observed data deviates from the expected norm, which in turn allows us to test claims about population parameters accurately.
Normal Distribution
The normal distribution is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. It is often referred to as a "bell curve" because of its shape.

In the context of the exercise, we assume that the percentage of wheat crop loss in Weld County is normally distributed. This assumption is critical as it allows us to apply the Z-test, which relies on normal distribution properties to determine statistical significance.

Key characteristics of a normal distribution include:
  • Symmetry: The left and right sides of the distribution are mirror images.
  • Mean, median, and mode are all located at the center of the distribution.
  • The total area under the curve is equal to 1.
Knowing these properties is important for correctly interpreting the results of statistical tests like the Z-test.
Critical Value
The critical value is the threshold or boundary of the acceptance region for a test statistic in hypothesis testing. When the calculated test statistic exceeds the critical value, we reject the null hypothesis.

In a two-tailed Z-test at a significance level of \( \alpha = 0.01 \), the critical values are approximately \(-2.576\) and \(2.576\). These values tell us the points beyond which the test statistic would indicate a statistically significant deviation from the null hypothesis mean. If the test statistic falls within these boundaries, as with 1.2 in the wheat crop example, the null hypothesis is not rejected.

  • Critical values depend on the chosen significance level (\( \alpha \)).
  • Lower \( \alpha \) leads to more stringent critical values, implying a smaller probability of rejecting the null hypothesis falsely.
  • It is important to understand the role of critical values for making informed decisions based on statistical tests.
Accurately determining and interpreting critical values helps define the reliability and outcome of the statistical inference.

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Most popular questions from this chapter

Myers-Briggs: Extroverts Are most student government leaders extroverts? According to Myers-Briggs estimates, about \(82 \%\) of college student government leaders are extroverts (Source: Myers -Briggs Type Indicator Atlas of Type Tables). Suppose that a Myers-Briggs personality preference test was given to a random sample of 73 student government leaders attending a large national leadership conference and that 56 were found to be extroverts. Does this indicate that the population proportion of extroverts among college student government leaders is different (either way) from \(82 \% ?\) Use \(\alpha=0.01.\)

Consider a test for \(\mu\). If the \(P\) -value is such that you can reject \(H_{0}\) at the \(5 \%\) level of significance, can you always reject \(H_{0}\) at the \(1 \%\) level of significance? Explain.

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Tree-ring dating from archaeological excavation sites is used in conjunction with other chronologic evidence to estimate occupation dates of prehistoric Indian ruins in the southwestern United States. It is thought that Burnt Mesa Pueblo was occupied around 1300 A.D. (based on evidence from potsherds and stone tools). The following data give tree-ring dates (A.D.) from adjacent archaeological sites (Bandelier Archaeological Excavation Project: Summer 1990 Excavations at Burnt Mesa Pueblo, edited by T. Kohler, Washington State University Department of Anthropology): $$\begin{aligned} &\begin{array}{ccccc} 1189 & 1267 & 1268 & 1275 & 1275 \end{array}\\\ &1271 \quad 1272 \quad 1316 \quad 1317 \quad1230 \end{aligned}$$ i. Use a calculator with mean and standard deviation keys to verify that \(\bar{x}=1268\) and \(s \approx 37.29\) years. ii. Assuming the tree-ring dates in this excavation area follow a distribution that is approximately normal, does this information indicate that the population mean of tree-ring dates in the area is different from (either higher or lower than) that in 1300 A.D.? Use a \(1 \%\) level of significance.

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Management: Lost Time In her book Red Ink Behaviors, Jean Hollands reports on the assessment of leading Silicon Valley companies regarding a manager's lost time due to inappropriate behavior of employees. Consider the following independent random variables. The first variable \(x_{1}\) measures a manager's hours per week lost due to hot tempers, flaming e-mails, and general unproductive tensions: The variable \(x_{2}\) measures a manager's hours per week lost due to disputes regarding technical workers' superior attitudes that their colleagues are "dumb and dispensable": i. Use a calculator with sample mean and standard deviation keys to verify that \(\bar{x}_{1} \approx 4.86, s_{1} \approx 3.18, \bar{x}_{2}=6.5,\) and \(s_{2} \approx 2.88\) ii. Does the information indicate that the population mean time lost due to hot tempers is different (either way) from population mean time lost due to disputes arising from technical workers' superior attitudes? Use \(\alpha=0.05 .\) Assume that the two lost-time population distributions are mound-shaped and symmetric.

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