Answer:
Step-by-step explanation:
Hello!
There was a special program funded, designed to reduce crime in 8 areas of Miami.
The number of crimes per area was recorded before and after the program was established in each area. This is an example of a paired data situation. For each are in Miami you have recorded a pair of values:
X₁: Number of crimes recorded in one of the eight areas of Miami before applying the special program.
X₂: Number of crimes recorded in one of the eight areas of Miami after applying the special program.
Area: (Before; After)
A: (14; 2)
B: (7; 7)
C; (4; 3)
D: (5; 6)
E: (17; 8)
F: (12; 13)
G: (8; 3)
H; (9; 5)
To apply a paired sample test you have to define the variable "difference":
Xd= X₁ - X₂
I'll define it as the difference between the crime rate before the program and after the program.
If the original populations have a normal distribution, we can assume that the variable defined from them will also have a normal distribution.
Xd~N(μd; σd²)
If the crime rate decreased after the special program started, you'd expect the population mean of the difference between the crime rates before and after the program started to be less than zero, symbolically μd<0
The hypotheses are:
H₀: μd≥0
H₁: μd<0
α: 0.01
![t= \frac{X[bar]_d-Mu_d}{\frac{S_d}{\sqrt{n} } } ~~t_{n-1}](https://tex.z-dn.net/?f=t%3D%20%5Cfrac%7BX%5Bbar%5D_d-Mu_d%7D%7B%5Cfrac%7BS_d%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20~~t_%7Bn-1%7D)
To calculate the sample mean and standard deviation of the variable difference, you have to calculate the difference between each value of each pair:
A= 14 - 2= 12
B= 7 - 7= 0
C= 4 - 3= 1
D= 5 - 6= -1
E= 17 - 8= 9
F= 12 - 13= -1
G= 8 - 3= 5
H= 9 - 5= 4
∑Xdi= 12 + 0 + 1 + (-1) + 9 + (-1) + 5 + 4= 29
∑Xdi²= 12²+0²+1²+1²+9²+1²+5²4²= 269
X[bar]d= 29/8= 3.625= 3.63
![S_d=\sqrt{\frac{1}{n-1}[sumX_d^2-\frac{(sumX_d)^2}{n} ] } = \sqrt{\frac{1}{7}[269-\frac{29^2}{8} ] } = 4.84](https://tex.z-dn.net/?f=S_d%3D%5Csqrt%7B%5Cfrac%7B1%7D%7Bn-1%7D%5BsumX_d%5E2-%5Cfrac%7B%28sumX_d%29%5E2%7D%7Bn%7D%20%5D%20%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B1%7D%7B7%7D%5B269-%5Cfrac%7B29%5E2%7D%7B8%7D%20%5D%20%7D%20%3D%204.84)
This test is one-tailed to the left and so is the p-value, under a t with n-1= 8-1=7 degrees of freedom, the probability of obtaining a value as extreme as the calculated value is:
P(t₇≤-2.11)= 0.0364
The p-value is greater than the significance level, so the decision is to not reject the null hypothesis. Then at a 1% significance level, you can conclude that the special program didn't reduce the crime rate in the 8 designated areas of Miami.
I hope it helps!
