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 n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.7 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 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? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value 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 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. Fail to reject the null hypothesis, there is insufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Fail to reject the null hypothesis, there is sufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same? 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. The conclusions obtained by using both methods are the same. These two methods differ slightly.

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 n2 = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population). x2: Rocky Mountain Crime Rate 3.7 4.1 4.7 5.5 3.3 4.8 3.5 2.4 3.1 3.5 5.2 2.8 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? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?1 − ?2. Round the test statistic and critical value to three decimal places.) test statistic critical value 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 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. Fail to reject the null hypothesis, there is insufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Fail to reject the null hypothesis, there is sufficient evidence that violent crime in the Rocky Mountain region is higher than in New England. Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same? 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. The conclusions obtained by using both methods are the same. These two methods differ slightly.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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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 n₂ = 12 regions in the Rocky Mountain states gave the following violent crime rates (per million population).

x₂: Rocky Mountain Crime Rate

3.7

4.1

4.7

5.5

3.3

4.8

3.5

2.4

3.1

3.5

5.2

2.8

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? Use ? = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference ?₁ − ?₂. Round the test statistic and critical value to three decimal places.)

test statistic
critical value

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

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.

Fail to reject the null hypothesis, there is insufficient evidence that violent crime in the Rocky Mountain region is higher than in New England.

Fail to reject the null hypothesis, there is sufficient evidence that violent crime in the Rocky Mountain region is higher than in New England.

Compare your conclusion with the conclusion obtained by using the P-value method. Are they the same?

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.

The conclusions obtained by using both methods are the same.

These two methods differ slightly.

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