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. 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 O Ho: The distributions are different. H1: The distributions are the same. 参H0: The distributions are different. H1: The distributions are different. O Ho: The distributions are the same. H1: The distributions are the same. O Ho: 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 S? o Yes O No What sampling distribution will you use? O binomial o normal o chi-square O uniform O Student's t 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.)

Answer 1

(a) The level of significance is 0.05 indeed.

The null and alternate hypothesis would be indeed as below.

• $\dpi{80} H_0 :$ The distributions are the same and $\dpi{80} H_1 :$ The distributions are different.

(b) The observed and expected data is as below.

Age	Percentage	Observed	Expected
<5	7.2%	46	$\dpi{80} 455*0.072=32.76$
5-14	13.6%	80	$\dpi{80} 455*0.136=61.88$
15-64	67.1%	288	$\dpi{80} 455*0.671=305.305$
65<	12.1%	41	$\dpi{80} 455*0.121=55.055$

Sum of observed and expected is both 455, which is the number of data in the sample. The chi square statistic is $\dpi{80} \chi^2 = \sum \frac{(O - E)^2}{E}$ , where O is observed and E is expected. The calculation is as below.

O	E	(O-E)^2 / E
46	32.76	5.350964591
80	61.88	5.305985779
288	305.305	0.980865118
41	55.055	3.58810326

The chi square statistic would be hence $\dpi{80} \chi^2 = 5.350964591 + 5.305985779 + 0.980865118 + 3.58810326$ or $\dpi{80} \chi^2 = 15.226$ .

The expected frequency in this case is the relative frequency of expected data itself, which is higher than 5% for all age group.

The sampling distribution will be chi square.

The degree of freedom for goodness of fit is k-1, for k be the number of category. In this case, there are 4 categories, and the degree of freedom is hence 3.

(c) The following R-command or any online calculator would give us the p-value as below.

> pchisq(15.226,df=3)
[1] 0.9983666

The p-value is hence 0.9984.