Hypothesis Testing, paired mean

Question

I am kind of stuck on this question, I don't really know how to approach it and I would appreciate some help.

The number of crime incidents during a period of 10 years were recorded in a certain village. A new security system was installed to reduce the number of crime incidents. The number of incidents for the next 10 years after the security system was setup were collected. The people responsible wanted to make sure that this system was effective in reducing the number of crime.

• For the first ten years respectively: 430, 266, 567, 531, 707, 716, 651, 589, 469, 723

• And for the next ten years respectively, 415, 238, 390, 410, 605, 609, 632,523, 411, 612

Set your hypothesis and determine your decision criteria for this paired analysis. using appropriated distribution and test statistics, and provide statistical proof for the result by explaining the technical reasons for all your steps.

What I thought of doing so far is to set by hypothesis by saying Ho is the mean of the first group

• Ho : μ1 - μ2 = 0

• Ha : μ1 - μ2 ≠ 0, Ha : μ1 - μ2 < 0, Ha : μ1 - μ2 > 0

If this start is correct, do I continue to apply the t statistic and how could I do so?

Answer 1

Crime Rate Before:
Crime Rate After: $n = 10$

$s_d = 52.27321388$ $\mu_d =-80.4 $

                       $alpha = .05$

Hypothesis $H_0 : \mu_d \geq 0$ $H_a : \mu_d <0$

Test Statistic:

$$ t = \dfrac{\bar d - 0}{\frac{s_d}{\sqrt n}} = -4.863812745 $$

The test statistic is distributed as Student's t distribution with 9 degrees of freedom.

    (b) Decision rule

With alpha = .05 and df = 9, the critical value of t is -1.8331. We reject H0 if t < -1.8331.

(6) Statistical decision

We reject $H_0$ because -4.863812745 < -1.8331 and conclude that the crime rate has fallen after the installation of security system.

Remark:

Initially, I wanted to compute population means and then compare their means and now I changed the analysis becuase the question asks for a paired analysis and that they would want to compare 1991 - 2001 and 1992 - 2002. If then intent of the question is not paired then you may want to compute the population means and and set hypothesis such as $\mu_1 < \mu_2$. Compute pooled variance and go about find the t-statistic and then decide accordingly. Good luck

Thanks Satish

Answer 2

Usually this type of comparisons is performed by calculating the rate of the event (e.g., number of events/person-years). In this case the denominator could be inhabitant-years. Clearly the frequency of crimes is affected by the number of inhabitants (which may change over time) and the comparison should take it into account.

Once you have calculated the crime rate in the first decade and that in the second one by averaging corresponding values, you can compare them using one of the standard tests available on most statistical softwares (e.g., based on Poisson distribution, etc...).

Answer 3

Crime Rate Before: $$\bar X_1 = 564.9, n_1 = 10, s_1^2 = 21544.76667$$

Crime Rate After: $$\bar X_2 = 484.5, n_2 = 10, s_2^2 = 17221.16667$$

$s_p^2 = \dfrac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{(n_1 + n_2-2)} = 19382.96667$

                       $alpha = .05$

Hypothesis $H_0 : \mu_2 \geq \mu_1$ $H_a : \mu_2 < \mu_1$

Test Statistic:

$$ t = \dfrac{\bar x_2- \bar x_1}{\sqrt{s_p^2/n}} = - 1.82619 $$

The test statistic is distributed as Student's t distribution with 18 degrees of freedom.

    (b) Decision rule

With alpha = .05 and df = 18, the critical value of t is -1.8331. We reject H0 if t < -1.7341.

(6) Statistical decision

We reject $H_0$ because -1.82619 < -1.7341 and conclude that the crime rate has fallen after the installation of security system.