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% 42

5 to 14 13.6% 77

15 to 64 67.1% 288

65 and older 12.1% 48

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 the same. 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?

normal uniform chi-square binomial Student's t

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

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

b)

Chi square test for Goodness of fit  
total frequency=   455
expected frequncy,E = expected proportions*total frequency  

category	observed frequencey, O	expected proportion	expected frequency,E	(O-E)	(O-E)²	(O-E)²/E
<5	42	0.072	32.76	9.24	85.38	2.606
5-14	77	0.136	61.88	15.12	228.61	3.694
15-64	288	0.671	305.31	-17.31	299.46	0.981
≥65	48	0.121	55.06	-7.06	49.77	0.904

chi square test statistic,X² = Σ(O-E)²/E =   8.186

Are all the expected frequencies greater than 5? Yes

What sampling distribution will you use? chi square

What are the degrees of freedom? = Degree of freedom=k-1=   4   -   1   =   3

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c)

0.025 < P-value < 0.050

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.