/*! 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 7 Is the national crime rate reall... [FREE SOLUTION] | 91Ó°ÊÓ

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Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the \(1980 \mathrm{~s}\) and \(1990 \mathrm{~s}\) is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, \(70 \%\) of all arrests are of males aged 15 to 34 years. (Source: True Odds, by J. Walsh, Merritt Publishing.) Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 32 arrests last month, 24 were of males aged 15 to 34 years. Use a \(1 \%\) level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from \(70 \%\).

Short Answer

Expert verified
The data does not show a significant difference from a 70% arrest rate at 1% significance level.

Step by step solution

01

Define the Hypotheses

First, we need to set up the null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the proportion of arrests for males aged 15 to 34 is 70%, or \(p = 0.70\). The alternative hypothesis (\(H_a\)) is that the proportion is different from 70%, or \(p eq 0.70\).
02

Collect and Define Sample Data

From the problem, we have a sample size of \(n = 32\) and \(x = 24\) males aged 15 to 34 were arrested. Therefore, the sample proportion \(\hat{p}\) is calculated as \(\hat{p} = \frac{24}{32} = 0.75\).
03

Calculate the Test Statistic

We'll use the test statistic for a proportion, which is\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]Substituting our values, we have\[ z = \frac{0.75 - 0.70}{\sqrt{\frac{0.70 \times 0.30}{32}}} = \frac{0.05}{0.08165} \approx 0.612 \]
04

Determine the Critical Value and Decision Rule

At a \(1\%\) significance level, the critical values for a two-tailed test are approximately \(\pm 2.576\) from the standard normal distribution table. If the test statistic falls beyond these critical values, we reject the null hypothesis.
05

Compare and Conclude

Since \(z = 0.612\) is within the range \(-2.576\) to \(2.576\), we do not reject the null hypothesis. There is not enough evidence at the \(1\%\) significance level to conclude that the proportion of arrests is different from 70%.

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis serves as a statement of no effect or no difference. It's a starting point for statistical testing. In the case of our exercise, the null hypothesis ( H_0 ) claims that the population proportion of arrests for males aged 15 to 34 years is exactly 70%. This means that in our study, we assume there is no deviation from the national statistic provided. Why is this important? By assuming the null hypothesis is true, it sets a benchmark for comparison. We can determine if the observed data significantly differs from this assumption. It's crucial to remember that the null hypothesis is not something researchers aim to prove. Instead, it's a statement to be potentially disproven through statistical evidence. When our results don't provide enough evidence against it, we "fail to reject" the null hypothesis instead of accepting it.
Alternative Hypothesis
The alternative hypothesis ( H_a ) offers a contrary assertion to the null hypothesis. In our particular study, it suggests that the population proportion of arrests is different from 70%. This hypothesis represents a challenge to the status quo provided by the null hypothesis and usually reflects the researcher’s original claim. Depending on the researcher's expectations, the alternative hypothesis can be:
  • Two-tailed: Claims that the parameter is different from the hypothesized value (used in our case ( p eq 0.70 )).
  • One-tailed: Suggests that the parameter is either greater than or less than the hypothesized value.
The type of hypothesis is chosen based on the research question and guides how conclusions are drawn from statistical tests. It's what researchers are attempting to find evidence for in their study. In our example, finding evidence to support the alternative hypothesis means showing that the arrest rates in Rock Springs deviate from the national proportion.
Significance Level
The significance level in hypothesis testing ( alpha ) sets the threshold for statistical significance in an experiment. It reflects the probability of rejecting the null hypothesis when it is actually true (Type I error). In our example, we used a 1% level of significance. This implies a stringent criterion for evidence: only 1 in 100 samples might incorrectly lead us to reject a true null hypothesis. This low significance level suggests strong confidence in the test results. Different scenarios may dictate different levels of significance:
  • 1% level: Used for very high stakes situations, requiring substantial proof.
  • 5% level: A more common choice in various research fields.
  • 10% level: Opted for when evidence is less rigorous.
Choosing an appropriate significance level depends heavily on the context and consequences of potential errors.
Population Proportion
The population proportion is a parameter that represents the fraction of the whole population exhibiting a particular characteristic, such as males aged 15 to 34 years in arrests. In our exercise, the national arrest proportion is 70%. This forms the comparative basis for our hypothesis test. Calculating the sample proportion ( hat{p} ) involves dividing the number of favorable outcomes by the total sample size. Population proportion plays a vital role in hypothesis testing, especially when comparing a sample proportion against it. Here's why it's important:
  • Provides a benchmark to assess whether observed sample behavior is typical or atypical for the population.
  • Serves as key input in the formula for the test statistic.
Understanding the broader population characteristics, like the arrest rates, helps in calculating deviation for the hypothesis test. By analyzing differences from the population proportion, researchers can gain insights into whether underlying assumptions hold true.

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Most popular questions from this chapter

Consider independent random samples from two populations that are normal or approximately normal, or the case in which both sample sizes are at least \(30 .\) Then, if \(\sigma_{1}\) and \(\sigma_{2}\) are unknown but we have reason to believe that \(\sigma_{1}=\sigma_{2}\), we can pool the standard deviations. Using sample sizes \(n_{1}\) and \(n_{2}\), the sample test statistic \(\bar{x}_{1}-\bar{x}_{2}\) has a Student's \(t\) distribution, where \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{s \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\) with degrees of freedom d.f. \(=n_{1}+n_{2}-2\) and where the pooled standard deviation \(s\) is $$s=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}$$ Note: With statistical software, select the pooled variance or equal variance options. (a) There are many situations in which we want to compare means from populations having standard deviations that are equal. This method applies even if the standard deviations are known to be only approximately equal (see Section \(11.4\) for methods to test that \(\sigma_{1}=\sigma_{2}\) ). Consider Problem 17 regarding average incidence of fox rabies in two regions. For region I, \(n_{1}=16\), \(\bar{x}_{1}=4.75\), and \(s_{1} \approx 2.82\) and for region II, \(n_{2}=15, \bar{x}_{2} \approx 3.93\), and \(s_{2} \approx\) 2.43. The two sample standard deviations are sufficiently close that we can assume \(\sigma_{1}=\sigma_{2}\). Use the method of pooled standard deviation to redo Problem 17 , where we tested if there was a difference in population mean average incidence of rabies at the \(5 \%\) level of significance. (b) Compare the \(t\) value calculated in part (a) using the pooled standard deviation with the \(t\) value calculated in Problem 17 using the unpooled standard deviation. Compare the degrees of freedom for the sample test statistic. Compare the conclusions.

Based on information from Harper's Index, \(r_{1}=\) 37 people out of a random sample of \(n_{1}=100\) adult Americans who did not attend college believe in extraterrestrials. However, out of a random sample of \(n_{2}=100\) adult Americans who did attend college, \(r_{2}=47\) claim that they believe in extraterrestrials. Does this indicate that the proportion of people who attended college and who believe in extraterrestrials is higher than the proportion who did not attend college? Use \(\alpha=0.01\).

Athabasca Fishing Lodge is located on Lake Athabasca in northern Canada. In one of its recent brochures, the lodge advertises that \(75 \%\) of its guests catch northern pike over 20 pounds. Suppose that last summer 64 out of a random sample of 83 guests did, in fact, catch northern pike weighing over 20 pounds. Does this indicate that the population proportion of guests who catch pike over 20 pounds is different from \(75 \%\) (either higher or lower)? Use \(\alpha=0.05\).

Tree Rings 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 \(\mathrm{T}\). Kohler, Washington State University Department of Anthropology, 1992): \(\begin{array}{lllll} 1189 & 1267 & 1268 & 1275 & 1275 \\ 1271 & 1272 & 1316 & 1317 & 1230 \end{array}\) 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 on lower than) that in 1300 A.D.? Use a \(1 \%\) level of significance.

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 (age \(25-34\) ) 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, S. Mitchell). Does this information indicate that the population mean percentage of young adults who attended college is higher? Use \(\alpha=0.05\).

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