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%	47
5 to 14	13.6%	72
15 to 64	67.1%	295
65 and older	12.1%	41

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

uniformStudent's t    binomialnormalchi-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) e level of significance =0.05

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

b)

applying chi square test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
<5	0.072	47.000	32.760	2.49	6.190
5-14	0.136	72.000	61.880	1.29	1.655
15-64	0.671	295.000	305.305	-0.59	0.348
>65	0.121	41.000	55.055	-1.89	3.588
total	1.000	455	455		11.7808

value of the chi-square statistic for the sample =11.781

expected frequencies greater than 5 :Yes

chi-square

degrees of freedom =3

c)

0.005 < P-value < 0.010

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