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Invasive Species In a study of the pernicious giant hogweed, Jan Pergl \(^{1}\) and associates compared the density of these plants in two different sites within the Caucasus region of Russia. In its native area, the average density was found to be 5 plants/m \(^{2}\). In an invaded area in the Czech Republic, a sample of \(n=50\) plants produced an average density of 11.17 plants/m \(^{2}\) with a standard deviation of 3.9 plants \(/ \mathrm{m}^{2}\) a. Does the invaded area in the Czech Republic have an average density of giant hogweed that is different from \(\mu=5\) at the \(\alpha=.05\) level of significance? b. What is the \(p\) -value associated with the test in part a? Can you reject \(H_{0}\) at the \(5 \%\) level of significance using the \(p\) -value?

Short Answer

Expert verified
Answer: Yes, we can conclude that the average density of giant hogweed in the invaded area is significantly different from 5 plants/m² at the 5% level of significance.

Step by step solution

01

Set up the null and alternative hypotheses

We're asked to test if the average density of the invaded area is different from 5 plants/m². So the null and alternative hypotheses are as follows: H0 (null hypothesis): μ = 5 (the average density in the invaded area is the same as the native area) H1 (alternative hypothesis): μ ≠ 5 (the average density in the invaded area is different from the native area)
02

Calculate the test statistic

The test statistic can be calculated using the formula: t = (x̄ - μ) / (s / √n) Where x̄ is the sample mean, μ is the population mean, s is the sample standard deviation, and n is the sample size. We have: x̄ = 11.17 plants/m² μ = 5 plants/m² s = 3.9 plants/m² n = 50 t = (11.17 - 5) / (3.9 / √50) = 6.17 / (3.9 / √50) ≈ 15.77
03

Determine the critical value(s)

Since we are conducting a two-tailed hypothesis test at the α = 0.05 level of significance, we can find the critical values by determining the t-values with 49 degrees of freedom (n-1). Using a t-table or a calculator, the critical values for a two-tailed test with 49 degrees of freedom are approximately ±2.01.
04

Compare the test statistic with the critical value(s)

Now we compare the calculated test statistic (15.77) with the critical values (±2.01). Since 15.77 > 2.01, the test statistic falls in the rejection region.
05

Calculate the p-value

The p-value is the probability of obtaining a test statistic as extreme or more so, given that the null hypothesis is true. To find the p-value for a two-tailed test with a t-value of 15.77 and 49 degrees of freedom, we can use a calculator or a software package. The p-value is approximately 0, which is much less than the significance level of 0.05.
06

Make a decision based on the p-value

Because the p-value (approx 0) is less than the significance level (0.05), we reject the null hypothesis in favor of the alternative hypothesis. Therefore, we can conclude that the average density of giant hogweed in the invaded area is significantly different from 5 plants/m² at the 5% level of significance.

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Key Concepts

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

Invasive Species
Invasive species, like the giant hogweed, are plants, animals, or other organisms that are introduced to an area where they are not native. They often spread rapidly and outcompete native species for resources. This can lead to significant ecological changes and impacts on local biodiversity. The giant hogweed, notable for its massive size, is originally from the Caucasus region of Russia. However, when it spreads to other areas, such as the Czech Republic, it can cause environmental harm. Understanding the density and spread of such species is crucial for managing their impacts. Researchers often study these species by comparing their density and abundance in native versus invaded environments using hypothesis testing.
Sample Mean
The sample mean, denoted as \( \bar{x} \), is an important statistical measure that represents the average value of a sample. It acts as an estimate of the population mean, which is the average of the entire population. In the context of our study, the sample mean for the invaded area is 11.17 plants per square meter. This was calculated from a sample of 50 plants. By comparing this sample mean to known values (like the 5 plants per square meter in the native area), researchers can infer differences or changes in the population's distribution. The sample mean is pivotal in hypothesis testing as it helps us draw conclusions about the population based on the sample.
Standard Deviation
Standard deviation, represented by \( s \), is a measure of the amount of variation or spread in a set of values. A low standard deviation indicates that the data points tend to be close to the sample mean, while a high standard deviation indicates they are more spread out. In this study, the standard deviation is 3.9 plants per square meter. This provides insight into how much the individual densities of plants vary from the average density in the sample. Understanding standard deviation is key in hypothesis testing because it helps in calculating the test statistic, which determines if the observed differences are statistically significant. It's part of the formula used to calculate the test statistic: \( t = \frac{(\bar{x} - \mu)}{(s / \sqrt{n})} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the hypothesis population mean, \( s \) is the sample standard deviation, and \( n \) is the sample size.
Two-tailed Test
A two-tailed test is a hypothesis test where the region of rejection is on both sides of the sampling distribution. This test is used when we need to determine whether there is evidence to reject the null hypothesis in both directions – meaning the sample mean could be higher or lower than the population mean. In the current exercise, we're testing whether the average plant density in the invaded area is significantly different from the native density of 5 plants per square meter. To conduct this test, we compare our calculated test statistic to critical values from the t-distribution. If the test statistic falls beyond these critical values in either direction, we reject the null hypothesis. In our example, the critical values were approximately ±2.01, and our test statistic of 15.77 is well beyond this range, leading us to reject the null hypothesis. This supports the conclusion that the average plant density is indeed different in the invaded area.

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Most popular questions from this chapter

Does College Pay Off? An article in Time describing various aspects of American life indicated that higher educational achievement paid off! College grads work 7.4 hours per day, fewer than those with less than a college education. \({ }^{2}\) Suppose that the average work day for a random sample of \(n=100\) indivic uals who had less than a 4 -year college education was calculated to be \(\bar{x}=7.9\) hours with a standard deviation of \(s=1.9\) hours. a. Use the \(p\) -value approach to test the hypothesis tha the average number of hours worked by individual having less than a college degree is greater than individuals having a college degree. At what level can you reject \(H_{0} ?\) b. If you were a college graduate, how would you state your conclusion to put yourself in the best possible light? c. If you were not a college graduate, how might you state your conclusion?

A random sample of \(n=1000\) observations from a binomial population produced \(x=279 .\) a. If your research hypothesis is that \(p\) is less than .3 , what should you choose for your alternative hypothesis? Your null hypothesis? b. What is the critical value that determines the rejection region for your test with \(\alpha=.05 ?\) c. Do the data provide sufficient evidence to indicate that \(p\) is less than \(.3 ?\) Use a \(5 \%\) significance level.

Early Detection of Breast Cancer Of those women who are diagnosed to have early-stage breast cancer, one-third eventually die of the disease. Suppose a community public health department instituted a screening program to provide for the early detection of breast cancer and to increase the survival rate \(p\) of those diagnosed to have the disease. A random sample of 200 women was selected from among those who were periodically screened by the program and who were diagnosed to have the disease. Let \(x\) represent the number of those in the sample who survive the disease. a. If you wish to determine whether the community screening program has been effective, state the alternative hypothesis that should be tested. b. State the null hypothesis. c. If 164 women in the sample of 200 survive the disease, can you conclude that the community screening program was effective? Test using \(\alpha=.05\) and explain the practical conclusions from your test.

What's Normal? What is normal, when it comes to people's body temperatures? A random sample of 130 human body temperatures, provided by Allen Shoemaker \(^{3}\) in the Journal of Statistical Education, had a mean of \(98.25^{\circ}\) and a standard deviation of \(0.73^{\circ} .\) Does the data indicate that the average body temperature for healthy humans is different from \(98.6^{\circ},\) the usual average temperature cited by physicians and others? Test using both methods given in this section. a. Use the \(p\) -value approach with \(\alpha=.05\). b. Use the critical value approach with \(\alpha=.05\). c. Compare the conclusions from parts a and b. Are they the same? d. The 98.6 standard was derived by a German doctor in 1868 , who claimed to have recorded 1 million temperatures in the course of his research. \({ }^{4}\) What conclusions can you draw about his research in light of your conclusions in parts a and b?

Independent random samples of \(n_{1}=140\) and \(n_{2}=140\) observations were randomly selected from binomial populations 1 and \(2,\) respectively. Sample 1 had 74 successes, and sample 2 had 81 successes. a. Suppose you have no preconceived idea as to whic parameter, \(p_{1}\) or \(p_{2}\), is the larger, but you want onl to detect a difference between the two parameters one exists. What should you choose as the alternative hypothesis for a statistical test? The null hypothesis? b. Calculate the standard error of the difference in the two sample proportions, \(\left(\hat{p}_{1}-\hat{p}_{2}\right) .\) Make sure to use the pooled estimate for the common value of \(p\) c. Calculate the test statistic that you would use for the test in part a. Based on your knowledge of the standard normal distribution, is this a likely or unlikely observation, assuming that \(H_{0}\) is true and the two population proportions are the same? d. \(p\) -value approach: Find the \(p\) -value for the test. Test for a significant difference in the population proportions at the \(1 \%\) significance level. e. Critical value approach: Find the rejection region when \(\alpha=.01 .\) Do the data provide sufficient evidence to indicate a difference in the population proportions?

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