Question

The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community of a village are shown below. Age (years) Percent of Canadian Population Observed Number in the Village Under 5 7.2% 43 5 to 14 13.6% 84 15 to 64 67.1% 278 65 and older 12.1% 50 Use a 5% level of significance to test the claim that the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village. (a) What is the level of significance? 0.05 State the null and alternate hypotheses. H0: The distributions are different. H1: The distributions are different. H0: The distributions are the same. H1: The distributions are different. H0: The distributions are different. H1: The distributions are the same. H0: The distributions are the same. H1: The distributions are the same. (b) Find the value of the chi-square statistic for the sample. (Round your answer to three decimal places.) Are all the expected frequencies greater than 5? Yes No What sampling distribution will you use? uniform normal Student's t binomial chi-square What are the degrees of freedom? (c) Estimate the P-value of the sample test statistic. P-value > 0.100 0.050 < P-value < 0.100 0.025 < P-value < 0.050 0.010 < P-value < 0.025 0.005 < P-value < 0.010 P-value < 0.005 (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis that the population fits the specified distribution of categories? Since the P-value > α, we fail to reject the null hypothesis. Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis. Since the P-value ≤ α, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, the evidence is insufficient to conclude that the village population does not fit the general Canadian population. At the 5% level of significance, the evidence is sufficient to conclude that the village population does not fit the general Canadian population.

Answer 1

p	Oi	Ei	(Oi-Ei)^2/Ei
0.072	43	32.76	3.2008
0.136	84	61.88	7.9071
0.671	278	305.305	2.4420
0.121	50	55.055	0.4641
1	455	455	14.0141
		TS	14.0141
		critical value	7.8147
		p-value	0.0029

a) level of significance = 0.05

Hypothesis

option B) is correct

H0: The distributions are the same. H1: The distributions are different.

b)

TS = 14.014

yes, all the expected frequencies are greater than 5

sampling distribution is chi-square

df =r-1 = 3

c)

p-value = 0.0029

P-value < 0.005

d)

we reject the null hypothesis as p-value < alpha

Since the P-value ≤ α, we fail to reject the null hypothesis

e)

option B)

At the 5% level of significance, the evidence is sufficient to conclude that the village population does not fit the general Canadian population.

Formulas in Excel

p	Oi	Ei	(Oi-Ei)^2/Ei
0.072	43	=$B$7*A2	=(B2-C2)^2/C2
0.136	84	=$B$7*A3	=(B3-C3)^2/C3
0.671	278	=$B$7*A4	=(B4-C4)^2/C4
0.121	50	=$B$7*A5	=(B5-C5)^2/C5
=SUM(A2:A5)	=SUM(B2:B5)	=SUM(C2:C5)	=SUM(D2:D5)
		TS	=D7
		critical value	=CHISQ.INV(0.95,3)
		p-value	=CHISQ.DIST.RT(D7,3)

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