The FBI computed the proportion of violent crimes in the United States falling into each of four categories. A simple random sample of 500 violent crimes committed in California were categorized in the same way. The following table presents the results. (18. ~ 20.) Category U.S. Proportion California Frequency Murder 0.013 5 Forcible Rape 0.051 23 Robbery 0.360 206 Aggravated Assault 0.576 266 18. Compute the value of ?² (A) 6.026 (B) 6.027 (C) 6.028 (D) 6.029 19. Find the area to the right of ?² under the chi-square distribution. (A) 0.1102 (B) 0.1103 (C) 0.1104 (D) 0.1105 20. Can you conclude that the proportions of crimes in the various categories in California differ from the United States as a whole? Use the 0.05 level of significance. (A) since the P-value (0.1102) is not less than ? = 0.05. There is enough evidence to conclude that the proportions of violent crimes in the various categories in California differ from those in the United States as a whole. (B) since the P-value (0.1103) is not less than ? = 0.05. There is not enough evidence to conclude that the proportions of violent crimes in the various categories in California differ from those in the United States as a whole. (C) since the P-value (0.1104) is not less than ? = 0.05. There is enough evidence to conclude that the proportions of violent crimes in the various categories in California differ from those in the United States as a whole. (D) since the P-value (0.1105) is not less than ? = 0.05. There is not enough evidence to conclude that the proportions of violent crimes in the various categories in California differ from those in the United States as a whole.
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S. proportions. We can do this by multiplying the U.S. proportion by the total number of crimes in the California sample (500). Expected frequencies for California: Murder: 0.013 * 500 = 6.5 Forcible Rape: 0.051 * 500 = 25.5 Robbery: 0.360 * 500 = 180 Aggravated Show more…
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Crime rates: The FBI computed the proportion of violent crimes in the United States falling into each of four categories. A simple random sample of 500 violent crimes committed in California were categorized in the same way. The following table presents the results. Category | U.S. Proportion | California Frequency Murder | 0.013 | 5 Forcible Rape | 0.051 | 23 Robbery | 0.360 | 206 Aggravated Assault | 0.576 | 266 Can you conclude that the proportions of crimes in the various categories in California differ from the United States as a whole? Use the 0.05 level of significance. Where do you live? The U.S. Census Bureau computed the proportion of U.S. residents who lived in each of four geographic regions in 2010. Then a simple random sample was drawn of 1000 people living in the United States in 2017. The following table presents the results: Region | 2010 Proportion | 2017 Frequency Northeast | 0.179 | 173 Midwest | 0.217 | 205 South | 0.371 | 384 West | 0.233 | 238 Can you conclude that the proportions of people living in the various regions changed between 2010 and 2017? Use the 0.05 level of significance.
Maitreya E.
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. Crime Rate: FBI A random sample of $n_{1}=10$ regions in New England gave the following violent crime rates (per million population). $x_{1}:$ New England Crime Rate $3.5 \quad 3.7 \quad 4.0 \quad 3.9 \quad 3.3 \quad 4.1 \quad 1.8 \quad 4.8 \quad 2.9 \quad 3.1$ Another random sample of $n_{2}=12$ regions in the Rocky Mountain states gave the following violent crime rates (per million population). $x_{2}:$ Rocky Mountain States $\begin{array}{cccccccccccc}3.7 & 4.3 & 4.5 & 5.3 & 3.3 & 4.8 & 3.5 & 2.4 & 3.1 & 3.5 & 5.2 & 2.8\end{array}$ (Reference: Crime in the United States, Federal Bureau of Investigation.) Assume that the crime rate distribution is approximately normal in both regions. i. Use a calculator to verify that $\bar{x}_{1} \approx 3.51, s_{1} \approx 0.81, \bar{x}_{2} \approx 3.87,$ and $s_{2} \approx 0.94$ ii. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than that in New England? Use $\alpha=0.01$
Hypothesis Testing
Testing $\mu_{1}-\mu_{2}$ and $p_{1}-p_{2}$ (Independent Samples)
Crime concerns in China. A 2013 poll found that 24.9% of Chinese adults see crime as a very big problem, and the standard error for this estimate, which can reasonably be modeled using a normal distribution, is SE = 1.6%. Suppose an issue will get special attention from the Chinese government if more than 1-in-5 Chinese adults (20%) express concern on an issue. 1. Choose words from the dropdown choices to construct hypotheses regarding whether or not crime should receive special attention by the Chinese government according to the 1-in-5 guideline. Before making your choices consider the appropriateness of using a one-sided or two-sided test for this exercise. That is, for this decision process, would we care about one or both directions? H0: The proportion of adults in China who see crime as a very big problem is not different from 20%. HA: The proportion of adults in China who see crime as a very big problem is more than 20%. 2. Calculate a z-score using the observed percentage and the two model parameters. Round to four decimal places. z = 3. Use the normal model to calculate a p-value. Round to four decimal places. p = 4. Based on the p-value, we have: A. strong evidence B. extremely strong evidence C. some evidence D. very strong evidence E. little evidence that the null model is not a good fit for our observed data.
Kari H.
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