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 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

For sample 1 :

x̅1 = 7.3, s1 = 2.3, n1 = 185

For sample 2 :

x̅2 = 8.2, s2 = 2.4, n2 = 160

Let, α = 0.05

1) Null and Alternative hypothesis:

Ho : µ1 = µ2

H1 : µ1 ≠ µ2

df = ((s1²/n1 + s2²/n2)²)/[(s1²/n1)²/(n1-1) + (s2²/n2)²/(n2-1) ] = 331.2872 = 331

2) Critical value :

Two tailed critical value, t_c = T.INV.2T(0.05, 331) = 1.967

Test statistic:

t = (x̅1 - x̅2)/√(s1²/n1 + s2²/n2) = (7.3 - 8.2)/√(2.3²/185 + 2.4²/160) = -3.5412

3) Decision:

As |t| = 3.5412 > 1.967 , Reject the null hypothesis.

4) Conclusion:

There is enough evidence to conclude that the number of arrests vary across cities.

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