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Wildlife: Wolves The following is based on information from The Wolf in the Southwest: The Making of an Endangered Species by David E. Brown (University of Arizona Press). Before \(1918,\) the proportion of female wolves in the general population of all southwestern wolves was about \(50 \% .\) However, after 1918, southwestern cattle ranchers began a widespread effort to destroy wolves. In a recent sample of 34 wolves, there were only 10 females. One theory is that male wolves tend to return sooner than females to their old territories where their predecessors were exterminated. Do these data indicate that the population proportion of female wolves is now less than \(50 \%\) in the region? Use \(\alpha=0.01.\)

Short Answer

Expert verified
The data indicates that the proportion of female wolves is now less than 50%.

Step by step solution

01

Establish the Hypotheses

We need to determine if the proportion of female wolves is less than 50%. So, our null hypothesis \(H_0\) is that the proportion of female wolves \(p = 0.5\). The alternative hypothesis \(H_1\) is that \(p < 0.5\).
02

Identify Sample Statistics

The sample size \(n\) is 34, and the number of females \(X\) is 10. Therefore, the sample proportion \(\hat{p}\) of female wolves is calculated as \(\hat{p} = \frac{X}{n} = \frac{10}{34} \approx 0.294\).
03

Calculate the Test Statistic

The test statistic for a proportion is given by \(z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\), where \(p_0 = 0.5\). Substitute the values to get \(z = \frac{0.294 - 0.5}{\sqrt{\frac{0.5\times 0.5}{34}}}\).
04

Compute the Test Statistic Value

Calculating further, we find \(z \approx \frac{-0.206}{0.086}\approx -2.395\).
05

Determine the Critical Value

For a one-tailed test with \(\alpha = 0.01\), the critical value from the Z-table is approximately -2.33.
06

Make a Decision

Since the calculated \(z\)-value of about -2.395 is less than the critical value of -2.33, we reject the null hypothesis \(H_0\).
07

Conclusion

There is sufficient evidence at the \(\alpha = 0.01\) significance level to conclude that the population proportion of female wolves is less than 50%.

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Key Concepts

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

Proportion Test
The proportion test is a type of hypothesis test used to determine if the proportion of a certain attribute in a population is equal to a specified value. In our wolf example, we are interested in whether the proportion of female wolves is less than 50%. We do this by comparing the proportion derived from our sample data with a hypothesized population proportion.

A proportion test is used when the outcome is binary, meaning each case either has the attribute or does not. It involves:
  • Selecting a sample from the population.
  • Calculating the sample proportion.
  • Conducting the test to compare the sample proportion to the population proportion.
This method helps us understand how our sample relates to the whole population.
Z-Score
The Z-score is a statistical measure that tells us how far away a data point is from the mean of the data set, expressed in terms of standard deviations. In hypothesis testing, the Z-score is used to determine the stance of the sample proportion compared to the hypothesized population proportion, taking into account sampling variability.

For our wolf population, after identifying the sample statistics, we calculate the Z-score to see how far the sample proportion of female wolves is from 0.5, the hypothesized proportion. The formula used is \[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]where:
  • \(\hat{p}\) is the sample proportion,
  • \(p_0\) is the hypothesized population proportion,
  • \(n\) is the sample size.
A higher absolute value of the Z-score suggests a stronger evidence against the null hypothesis.
Significance Level
The significance level, denoted by \(\alpha\), is the threshold for deciding whether to reject the null hypothesis. It defines the probability of making a Type I error, which is rejecting a true null hypothesis. In simpler terms, it's how willing we are to be wrong if we claim there is an effect when, in fact, there isn't.

In our exercise, the significance level is set at 0.01, implying we are 99% confident in the results.
This strict level means that we require strong evidence against the null hypothesis to reject it. Lower significance levels demand more conclusive evidence, reducing the risk of making a Type I error.
Null Hypothesis
The null hypothesis (often symbolized as \(H_0\)) represents the baseline or default statement about a population parameter that there's no effect or no difference. It is what we attempt to test against using statistical methods.

In the case of the wolves, the null hypothesis is that the proportion of female wolves is 50%, or \(H_0: p = 0.5\). We run a hypothesis test to determine if there is sufficient evidence to support the claim that the actual proportion is less than 50%.
The beauty of hypothesis testing is that it provides a structured way to make decisions based on data. By analyzing sample evidence in light of the null hypothesis, we can make more informed conclusions about the larger population.

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Most popular questions from this chapter

A random sample of 49 measurements from a population with population standard deviation 3 had a sample mean of \(10 .\) An independent random sample of 64 measurements from a second population with population standard deviation 4 had a sample mean of \(12 .\) Test the claim that the population means are different. Use level of significance 0.01. (a) Check Requirements What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute \(\bar{x}_{1}-\bar{x}_{2}\) and the corresponding sample distribution value. (d) Find the \(P\) -value of the sample test statistic. (e) Conclude the test. (f) Interpret the results.

Two populations have normal distributions. The first has population standard deviation 2 and the second has population standard deviation \(3 .\) A random sample of 16 measurements from the first population had a sample mean of \(20 .\) An independent random sample of 9 measurements from the second population had a sample mean of \(19 .\) Test the claim that the population mean of the first population exceeds that of the second. Use a \(5 \%\) level of significance. (a) Check Requirements What distribution does the sample test statistic follow? Explain. (b) State the hypotheses. (c) Compute \(\bar{x}_{1}-\bar{x}_{2}\) and the corresponding sample distribution value. (d) Find the \(P\) -value of the sample test statistic. (e) Conclude the test (f) Interpret the results.

To use the normal distribution to test a proportion \(p,\) the conditions \(n p>5\) and \(n q>5\) must be satisfied. Does the value of \(p\) come from \(H_{0}\) or is it estimated by using \(\hat{p}\) from the sample?

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. Generation Gap: Education Education influences attitude and lifestyle. Differences in education are a big factor in the "generation gap." Is the younger generation really better educated? Large surveys of people age 65 and older were taken in \(n_{1}=32\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{1}=15.2 \%\) of the older adults had attended college. Large surveys of young adults (ages \(25-34\) ) were taken in \(n_{2}=35\) U.S. cities. The sample mean for these cities showed that \(\bar{x}_{2}=19.7 \%\) of the young adults had attended college. From previous studies, it is known that \(\sigma_{1}=7.2 \%\) and \(\sigma_{2}=5.2 \%\) (Reference: American Generations by S. Mitchell). Does this information indicate that the population mean percentage of young adults who attended college is higher? Use \(\alpha=0.05\)

Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let \(c\) be the level of confidence used to construct a confidence interval from sample data. Let \(\alpha\) be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance \(\alpha\) and null hypothesis \(H_{0}\) : \(\mu=k,\) we reject \(H_{0}\) whenever \(k\) falls outside the \(c=1-\alpha\) confidence interval for \(\mu\) based on the sample data. When \(k\) falls within the \(c=1-\alpha\) confidence interval, we do not reject \(H_{0}\) (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as \(p, \mu_{1}-\mu_{2},\) and \(p_{1}-p_{2},\) which we will study in Sections 8.3 and \(8.5 .\) ) Whenever the value of \(k\) given in the null hypothesis falls outside the \(c=1-\alpha\) confidence interval for the parameter, we reject \(H_{0} .\) For example, consider a two-tailed hypothesis test with \(\alpha=0.01\) and $$H_{0}: \mu=20 \quad H_{1}: \mu \neq 20$$ A random sample of size 36 has a sample mean \(\bar{x}=22\) from a population with standard deviation \(\sigma=4.\) (a) What is the value of \(c=1-\alpha ?\) Using the methods of Chapter \(7,\) construct a \(1-\alpha\) confidence interval for \(\mu\) from the sample data. What is the value of \(\mu\) given in the null hypothesis (i.e., what is \(k\) )? Is this value in the confidence interval? Do we reject or fail to reject \(H_{0}\) based on this information? (b) Using methods of this chapter, find the \(P\) -value for the hypothesis test. Do we reject or fail to reject \(H_{0}\) ? Compare your result to that of part (a).

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