Question

The federal government recently granted funds for a special program designed to reduce crime in high-crime areas. A study of the results of the program in eight high-crime areas of Miami, Florida, yielded the following results.
Number of Crimes by Area
A B C D E F G H
before 14 7 4 5 17 12 8 9
after 2 7 3 6 8 13 3 5
Has there been a decrease in the number of crimes since the inauguration of the program? Use the .01 significance level. Estimate the p-value

Answer 1

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

$t= \frac{X[bar]_d-Mu_d}{\frac{S_d}{\sqrt{n} } } ~~t_{n-1}$

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

$S_d=\sqrt{\frac{1}{n-1}[sumX_d^2-\frac{(sumX_d)^2}{n} ] } = \sqrt{\frac{1}{7}[269-\frac{29^2}{8} ] } = 4.84$

$t_{H_0}= \frac{3.63-0}{\frac{4.86}{\sqrt{8} } } = 2.11$

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!

Answer 2

Using the t-distribution, it is found that since the p-value of the test is 0.048 > 0.01, there is not enough evidence to conclude that there has been a decrease in the number of crimes since the inauguration of the program.

At the null hypothesis, it is tested if there has been no reduction, that is, the subtraction of the mean after by the mean before is at least 0, hence:

$H_0: \mu_A - \mu_B \geq 0$

At the alternative hypothesis, it is tested if there has been a reduction, that is, the subtraction of the mean after by the mean before is negative, hence:

$H_1: \mu_A - \mu_B < 0$

For both before and after, the mean, standard deviation of the sample(this is why the t-distribution is used) and sample sizes are given by:

$\mu_B = 9.5, s_B = 4.504, n_B = 8$

$\mu_A = 5.875, s_A = 3.5632, n_A = 8$

The standard errors are given by:

$s_A = \frac{3.5632}{\sqrt{8}} = 1.2596$

$s_B = \frac{4.504}{\sqrt{8}} = 1.5924$

For the distribution of differences, the mean and standard error are given by:

$\overline{x} = \mu_A - \mu_B = 5.875 - 9.5 = -3.625$

$s = \sqrt{s_A^2 + s_B^2} = \sqrt{1.2596^2 + 1.5924^2} = 2.0304$

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

• In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-3.625 - 0}{2.0304}$

$t = -1.7854$

The p-value is found using a t-distribution calculator, with t = -1.7854, 8 + 8 - 2 = 14 df and a left-tailed test with a significance level of 0.01, as we are tested if the mean is less than a value.

• Using the calculator, the p-value is given by 0.048.

Since the p-value of the test is 0.048 > 0.01, there is not enough evidence to conclude that there has been a decrease in the number of crimes since the inauguration of the program.

You can learn more about the use of the t-distribution to test an hypothesis at https://brainly.com/question/13873630