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%	73
15 to 64	67.1%	297
65 and older	12.1%	42

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

normalbinomial     Student's tuniformchi-square

What are the degrees of freedom?


(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)


(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)

The level of significance is 0.05

Null and alternative hypothesis are

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

(b)

Expected number of Canadian population in sample of 455 residents.

Under 5,  0.072 * 455 = 32.76

5 to 14,   0.136 * 455 = 61.88

15 to 64, 0.671 * 455 = 305.305

65 and older, 0.121 * 455 = 55.055

Chi square test statistic $\chi^2 = \sum (O_i - E_i)^2 / E_i$

where Oi and Ei are the observed and expected frequency for each age distribution.

$\chi^2$ = (43 - 32.76)² / 32.76 + (73 - 61.88)² / 61.88 + (297 - 305.305)² / 305.305 + (42 - 55.055)² / 55.055

= 8.52

Yes, all the expected frequencies are greater than 5

We will use chi-square distribution.

Degree of freedom = number of groups - 1 = 4 - 1 = 3

(c)

P-value = P($\chi^2$ > 8.52) = 0.0364

(d)

As, p-value is less than the 0.05, we reject null hypothesis H0.

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.