Question

A researcher is interested whether number of arrests vary across cities. Two random samples were collected. In city 1, the researcher sampled 185 individuals who had an average of 7.3 arrests with a standard deviation of 2.3 arrests. In city 2, the researcher sampled 160 individuals who had an average of 8.2 arrests with a standard deviation of 2.4 arrests. Test the null hypothesis at the .01 level of significance that the number of arrests does not vary across cities. In so doing, identify: (1) the research and null hypothesis, (2) the critical value needed to reject the null, (3) the decision that you made upon analyzing the data, and (4) the conclusion you have drawn based on the decision you have made.

Answer 1

Solution

Final answers in the stipulated format are given below. Back-up Theory and Details of calculations follow at the end.

Part (1)

Research hypothesis:

The number of arrests does not vary across the two cities. Answer 1

Null hypothesis: H₀: µ1 = µ2 [Vs Alternative: H_A: µ1 ≠ µ2] Answer 2

Part (2)

Critical value: 2.5902 Answer 3

Part (3)

Decision:

Since the calculated value of the test statistic [3.5521] is greater than the critical value,

H₀ is rejected. Answer 4

Part (4)

Conclusion: There is sufficient evidence to suggest that the null hypothesis is valid and hence we conclude that the mean number of arrests varies across the two cities. Answer 5

Back-up Theory and Details of calculations

Let X = number of arrests in City 1

      Y = number of arrests in City 2

Let mean and standard deviation of X be respectively µ₁ and σ₁ and those of Y be µ₂ andσ₂, where σ₁² = σ₂² = σ², say and σ² is unknown.

Hypotheses:

Null: H₀: µ1 = µ2 Vs Alternative: H_A: µ1 ≠ µ2

Test Statistic:

t = (Xbar - Ybar)/[s√{(1/n₁) + (1/n₂)}]

where

s² = {(n₁ – 1)s₁² + (n₂ – 1)s₂²}/(n₁ + n₂ – 2);

Xbar and Ybar are sample averages and

s₁,s₂ are sample standard deviations based on n₁ observations on X and n₂ observations on Y respectively.

Calculations

Summary of Excel calculations is given below:

n1	185
n2	160
Xbar	7.3
Ybar	8.2
s1	2.3
s2	2.4
s2	5.5079
s	2.3469
tcal	3.5521
α	0.01
tcrit	2.590239

Distribution, Significance Level, α Critical Value

Under H₀, t ~ t_{n1 + n2 - 2}. Hence, for level of significance α%,

Critical Value = upper (α/2)% point of t_{n1 + n2 - 2} and

Using Excel Function: Statistical TINV this is found to be as shown in the above table.

Decision:

Since | tcal | > tcrit, H₀ is rejected/.

Conclusion:

There is sufficient evidence to suggest that the null hypothesis is valid and hence we conclude that the mean number of arrests vary across the two cities.

DONE