In: Math
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% | 73 |
15 to 64 | 67.1% | 295 |
65 and older | 12.1% | 40 |
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 the
same.H0: The distributions are different.
H1: The distributions are
different. 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?
YesNo
What sampling distribution will you use?
uniformbinomial chi-squarenormalStudent'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.
a) 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 | Oi | Ei=total*p | R2i=(Oi-Ei)/√Ei | R2i=(Oi-Ei)2/Ei |
1 | 0.072 | 47 | 32.76 | 2.49 | 6.190 |
2 | 0.136 | 73 | 61.88 | 1.41 | 1.998 |
3 | 0.671 | 295 | 305.31 | -0.59 | 0.348 |
4 | 0.121 | 40 | 55.06 | -2.03 | 4.117 |
total | 1.000 | 455 | 455 | 12.653 |
value of the chi-square statistic =12.653
Are all the expected frequencies greater than 5? :Yes
What sampling distribution will you use? Chi square
degree of freedom =categories-1= | 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.