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%	83
15 to 64	67.1%	284
65 and older	12.1%	45

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?


State the null and alternate hypotheses.

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

(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?

YesNo     

What sampling distribution will you use?

uniformbinomial     normalStudent's tchi-square

What are the degrees of freedom?


(c) Estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100     0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-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

(a) Level of significance = 0.05

Null and alternate hypotheses:

H₀: The distributions are the same.

H₁: The distributions are different.

Category	Observed(O)	Proportion, p	Expected Frequency (E)	(O-E)²/E
Under 5	43	0.072	455 * 0.072 = 32.76	(43 - 32.76)²/32.76 = 3.2008
5 to 14	83	0.136	455 * 0.136 = 61.88	(83 - 61.88)²/61.88 = 7.2084
15 to 64	284	0.671	455 * 0.671 = 305.305	(284 - 305.305)²/305.305 = 1.4867
65 and older	45	0.121	455 * 0.121 = 55.055	(45 - 55.055)²/55.055 = 1.8364
Total	455	1.00	455	13.7323

(b)

Test statistic:

χ² = ∑ ((O-E)²/E) = 13.732

Yes, the expected frequencies are greater than 5.

Sampling distribution to use: chi-square

df = n-1 = 3

(c)

p-value = CHISQ.DIST.RT(13.7323, 3) = 0.0033

P-value < 0.005

(d)

Decision:

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

(e) Conclusion in the context of the application:

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