Question

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

3.3 | 3.7 | 4.2 | 4.1 | 3.3 | 4.1 | 1.8 | 4.8 | 2.9 | 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 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
*μ*_{1} − *μ*_{2}. 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.

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 3.5 4.3 4.5 5.3 3.3 4.8 3.5 2.4
3.1 3.5 5.2 2.8
Assume that the crime rate distribution...

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
3.5
4.3
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...

A random sample of n1 = 10 regions in New
England gave the following violent crime rates (per million
population).
x1: New England Crime
Rate
3.5
3.9
4.0
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
5.3
3.3
4.8
3.5
2.4
3.1
3.5
5.2
2.8
Assume that the crime rate distribution...

A random sample of n1 = 10 regions in New
England gave the following violent crime rates (per million
population).
x1: New England Crime
Rate
3.5
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
5.1
3.3
4.8
3.5
2.4
3.1
3.5
5.2
2.8
Assume that the crime rate distribution...

A random sample of n1 = 10 regions in New
England gave the following violent crime rates (per million
population).
x1: New England Crime
Rate
3.5
3.9
4.0
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
4.3
4.7
5.3
3.3
4.8
3.5
2.4
3.1
3.5
5.2
2.8
Assume that the crime rate distribution...

A random sample of n1 = 12 in Chicago gave the
following information concerning violent crime rates (per million):
x-bar1 = 3.45 and s1 = 0.81. Another random
sample of n2 = 15 in Las Vegas gave the following
information concerning violent crime rates: x-bar2=
3.81 and s2 = 0.95. Do the data indicate that violent
crime rate in Chicago is higher than that in Las Vegas? Use a 5%
level of significance.
Show all work and calculations.

A random sample of n1 = 16 communities in western Kansas gave
the following information for people under 25 years of age.
x1: Rate of hay fever per 1000
population for people under 25
100
92
121
126
94
123
112
93
125
95
125
117
97
122
127
88
A random sample of n2 = 14 regions in
western Kansas gave the following information for people over 50
years old.
x2: Rate of hay fever per 1000
population for...

A random sample of n1 = 16 communities in
western Kansas gave the following information for people under 25
years of age.
x1: Rate of hay fever per 1000
population for people under 25
98
92
120
126
94
123
112
93
125
95
125
117
97
122
127
88
A random sample of n2 = 14 regions in
western Kansas gave the following information for people over 50
years old.
x2: Rate of hay fever per 1000
population for...

A random sample of n1 = 10 winter days in
Denver gave a sample mean pollution index x1 =
43. Previous studies show that σ1 = 21. For
Englewood (a suburb of Denver), a random sample of
n2 = 12 winter days gave a sample mean
pollution index of x2 = 36. Previous studies
show that σ2 = 13. Assume the pollution index
is normally distributed in both Englewood and Denver.
(a) Do these data indicate that the mean population...

Data were gathered from a simple random sample of cities. The
variables are Violent Crime (crimes per 100,000 population),
Police Officer Wage (mean $/hr), and Graduation Rate (%). Use the
accompanying regression table to answer the following questions
consider the coefficient of Graduation Rate. Complete parts a
through e.
Dependent variable is: Violent Crime
R squared=38.1 R squared (adjusted)=40.3 s=129.6 with 37
degrees of freedom
Variable
Coeff
SE (Coeff)
t-ratio
P-value
Intercept
1388.65
183
.9
7.55
<
0.0001
Police Officer...

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