Question

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

3.7

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

3.9

4.1

4.5

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 α = 0.01. Solve the problem using both the traditional method and the P-value method. (Test the difference μ₁ − μ₂. Round the test statistic and critical value to three decimal places.)

test statistic ?
critical value. ?

Answer 1

	X1	X2
mean	3.530	3.883
std. dev.	0.845	0.965
n	10	12

To Test :-

H0 :- µ1 = µ2
H1 :- µ1 < µ2

Test Statistic :-
t = (X̅1 - X̅2) / SP √ ( ( 1 / n1) + (1 / n2))

SP = 0.9126

t = -0.904

Test Criteria :-
Reject null hypothesis if t < - t(α, n1 + n2 - 2)
Critical value t(α, n1 + n1 - 2) = t( 0.01 , 10 + 12 - 2) = 2.528
t > -t(α, n1 + n2 - 2) = -0.9042 > -2.528
Result :- Fail to Reject Null Hypothesis

Decision based on P value
P - value = P ( t > 0.9042 ) = 0.1883
Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0.1883 > 0.01 ,hence we fail to reject null hypothesis
Conclusion :- Fail to Reject Null Hypothesis

There is insufficient evidence to support the claim that the violent crime rate in the Rocky Mountain region is higher than in New England.