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
3.5 | 3.9 | 4.0 | 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
3.7 | 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 is approximately normal in both regions. Use a calculator to calculate x_{1}, s_{1}, x_{2}, and s_{2}. (Round your answers to two decimal places.)
x_{1} | = |
s_{1} | = |
x_{2} | = |
s_{2} | = |
What is the value of the sample test statistic? Compute the
corresponding z or t value as appropriate. (Test
the difference μ_{1} − μ_{2}. Do not use rounded
values. Round your answer to three decimal places.)
(b) Find a 98% confidence interval for
μ_{1} − μ_{2}.
(Round your answers to two decimal places.)
lower limit | |
upper limit |
a)
x1 =3.55
s1 =0.83
x2 =3.87
s2 =0.95
std error =√(S21/n1+S22/n2)= | 0.380 | |
test stat t =(x1-x2-Δo)/Se = | -0.832 |
b)
Point estimate of differnce =x1-x2 = | -0.317 | |
for 98 % CI & 9 df value of t= | 2.821 | |
margin of error E=t*std error = | 1.073 | |
lower bound=mean difference-E = | -1.39 | |
Upper bound=mean differnce +E = | 0.76 |
Get Answers For Free
Most questions answered within 1 hours.