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%	49
5 to 14	13.6%	70
15 to 64	67.1%	286
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?


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 different.    H₀: The distributions are different.
H₁: The distributions are different.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?

YesNo    

What sampling distribution will you use?

binomialStudent's t    normaluniformchi-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 < 0.050

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

(a) 5% is the level of significance

Null and Alternative Hypotheses

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

(b) Chi-square statistic

The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
Under 5	49	455*0.072=32.76	(49-32.76)2/32.76 = 8.051
5 to 14	70	455*0.136=61.88	(70-61.88)2/61.88 = 1.066
15 to 64	286	455*0.671=305.305	(286-305.305)2/305.305 = 1.221
65 and older	50	455*0.121=55.055	(50-55.055)2/55.055 = 0.464
Sum =	455	455	10.801

Test Statistics

The Chi-Squared statistic is computed as follows:

Yes, all the expected frequencies greater than 5

Chi-square sampling distribution will be used.

The number of degrees of freedom is

(c)

p-value =

0.01285

Hence

0.010 < P-value < 0.025

(d)

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

(e)

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

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