Question

# The age distribution of the Canadian population and the age distribution of a random sample of...

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% 80 15 to 64 67.1% 289 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 different.H0: The distributions are the same.
H1: The distributions are different.     H0: The distributions are different.
H1: The distributions are the same.H0: The distributions are the same.
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?

uniformStudent's t     normalbinomialchi-square

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.

Solution:

1. The Level of Significance is α=0.05

The Null and Alternative Hypotheses are as follows:

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

1. The above test is the Test for Goodness of Fit where we need to check whether the age distribution of the general Canadian population fits the age distribution of the residents of Red Lake Village.

Let the population have k mutually exclusive classes and let us assume that according to the hypothesis, the population proportion of the ith class is pi. If the frequency of the ith class in random samples of size n from this population be fi, we have the following test statistic for the above hypothesis (Part (a))

Chi-squared = ∑ (fi, - n pi)2/ n pi. = ∑ (fi,2/ n pi.) – n …..(1)

The above Test statistic ~ Chi-squared with (k-1) degrees of freedom

For Computation we use the following table:

 Class Observed freq (fi) % of Canadian Population Expected freq (npi) fi^2/npi < 5 44 7.2% 7.2/100*455=32.76 44^2/32.76=59.0965 5 - 14 80 13.6% 13.6/100*455=61.88 80^2/61.88=103.4260 15 - 64 289 67.1% 67.1/100*455=305.305 289^2/305.305=273.5658 > 65 42 12.1% 12.1/100*455=55.055 42^2/55.055=32.04068 Total 455 100% 455 468.1289

So the using the formula (1) we have,

• Chi-square = ∑ (fi,2/ n pi.) – n =468.1289-455 (From the above table)

=13.129 (Round to three decimal places)

• From the above table we can see all the expected frequencies are greater than 5. Thus it meets the assumption of using a chi-squared test.
• We will use the Chi-squared test
• The degrees of freedom=k-1, where k is the number of classes, Here k=4

So degrees of freedom =4-1=3.

1. The p-value for the corresponding chi-squared test statistic =13.129 with degrees of freedom 3 and level of significance 0.05 is

=.004366

So, P-value < 0.005

1. Now at level of significance 0.05 we have,

P-value <α

So, we have - > Since the P-value ≤ α, we reject the null hypothesis.

1. So, we can have the following conclusion:

At the 5% level of significance, the evidence is sufficient to conclude that the village population does not fit the general Canadian population.

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