Question

Please show forumals if completed in excel

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%	44
5 to 14	13.6%	79
15 to 64	67.1%	290
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 different.
H₁: The distributions are the same.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.

(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?

chi-squareuniform    binomialnormalStudent's t

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

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

b)

applying chi square goodness of fit test:

	relative	observed	Expected	residual	Chi square
category	frequency(p)	Oi	Ei=total*p	R2i=(Oi-Ei)/√Ei	R2i=(Oi-Ei)2/Ei
1	0.0720	44.0000	32.76	1.96	3.856
2	0.1360	79.0000	61.88	2.18	4.736
3	0.6710	290.0000	305.31	-0.88	0.767
4	0.1210	42.0000	55.06	-1.76	3.096
total	1.000	455	455		12.4559
test statistic X2 =		12.456

Are all the expected frequencies greater than 5? :Yes
What sampling distribution will you use? chi-square
degrees of freedom =categories-1=3

c)

0.005 < P-value < 0.01

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.