A random sample of n1 - 10 regions in New England gave the following violent crime rates (per million population). X1: New England Crime Rate 3.3 3.9 4.2 4.1 3.3 4.1 1.8 4.8 2.9 3.1 Another random sample of 12 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). X2: Rocky Mountain Crime Rate 3.5 4.1 4.7 5.1 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 i USE SALT Assume that the crime rate distribution is approximately normal in both regions. Do the data indicate that the violent crime rate in the Rocky Mountain region is higher than in New England? USC a = 0.01. Solve the problem using both the traditional method and the p-value method. (Test the difference M1 - M2. Round the test statistic and critical value to three decimal places.) -1.231 critical value 2.520 Enter a number test statistic Find (or estimate the p-value P-value > 0.250 0.125 < P-value < 0.250 0.050 < P-value < 0.125 0.025 < P-value < 0.050 0.005 < P-value < 0.025 P-value < 0.005 Conclusion Fail to reject the null hypothesis, there is sufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Reject the null hypothesis, there is insufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Reject the null hypothesis, there is sufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Fail to reject the null hypothesis, there is insufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Compare your conclusion with the conclusion abtained by using the P-value method. Are they the same? The conclusions obtained by using both methods are the same. These two methods differ slightly. We reject the null hypothesis using the traditional method, but fail to reject using the P-value method. We reject the null hypothesis using the P-value method, but fail to reject using the traditional method.
