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9. The police department in Madison, Connecticut released the following numbers of calls for the different days of the week f
b. Draw the graph and shade the critical area(s). Place the critical value(s) on the graph. C. Fill in the following: Critica
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Answer #1

a)

null hypothesis is(claim)

7 Ho: p1 = P2 = P3 = P4 = ps = po = pz = The alternative hypothesis states the opposite of the null hypothesis: H.: At least

b)

The x(chi-square) distribution with v = 6 degrees of freedom 0.16 p=0.0000 v=6 0.14 0.12 0.10 f(x) 0.08 0.06 0.04 0.02 0.00 0

c)

Categories Observed Expected (fo-fe)2/fe Category 1 114 1095*0.142857=156.428 (114-156.428)2/156.428 11.508 Category 2 152 10

Based on the information provided, the significance level is a = 0.01, the number of degrees of freedom is df = 7-1=6, so the

The Chi-Squared statistic is computed as follows: 72 x? (O; - E) Ei 11.508 + 0.125 + 0.082 +0.366 +3.257 + 10.01 + 4.465 = 29

If X is a random variable having a distribution with v = 6 degrees of freedom, then p = Pr[X > 29.814] = 0.0000.

d)

Since it is observed that x = 29.814 > xě = 16.812, it is then concluded that the null hypothesis is rejected.

e)

There is sufficient evidence to reject the claim that the difference days of the week have the same frequency.

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