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

4.0

3.9

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

3.5

4.1

4.5

5.1

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

Find (or estimate) the P-value.

P-value > 0.2500.125 < P-value < 0.250    0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005

Conclusion

Fail to 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 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 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.    These two methods differ slightly.The conclusions obtained by using both methods are the same.

Answer 1

Given that,
mean(x)=3.8167
standard deviation , s.d1=0.9144
number(n1)=12
y(mean)=3.49
standard deviation, s.d2 =0.8157
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.01
from standard normal table,right tailed t α/2 =2.821
since our test is right-tailed
reject Ho, if to > 2.821
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3.8167-3.49/sqrt((0.83613/12)+(0.66537/10))
to =0.885
| to | =0.885
critical value
the value of |t α| with min (n1-1, n2-1) i.e 9 d.f is 2.821
we got |to| = 0.88519 & | t α | = 2.821
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:right tail - Ha : ( p > 0.8852 ) = 0.19954
hence value of p0.01 < 0.19954,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 0.885
critical value: 2.821
decision: do not reject Ho
p-value: 0.19954
0.125 < P-value < 0.250
we do not have enough evidence to support the claim that the violent crime rate in the Rocky Mountain region is higher than in New England.
These two methods differ slightly.The conclusions obtained by using both methods are the same.