Question

Question 4

It has been hypothesized that the distribution of seasonal colds in Canada is as follows:

Season	Percentage
Fall	35%
Winter	25%
Spring	30%
Summer	10%

A random sample of 200 Canadian citizens provided the following results:

Season	Observed Frequency
Fall	80
Winter	40
Spring	70
Summer	10

10 marks     Do the observed data contradict the hypothesis? Formulate and test the appropriate hypotheses at the 5% level of significance. Use the critical value approach.

Answer 1

The following cross tablulation have been provided. The row and column total have been calculated and they are shown below:

	Column 1	Column 2	Total
	40	35	75
	20	25	45
	35	30	65
	5	10	15
Total	100	100	200

The expected values are computed in terms of row and column totals. In fact, the formula is ​​, where corresponds to the total sum of elements in row i, ​ corresponds to the total sum of elements in column j, and TT is the grand total. The table below shows the calculations to obtain the table with expected values:

Based on the observed and expected values, the squared distances can be computed according to the following formula: . The table with squared distances is shown below:

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0​: The distribution of seasonal colds in Canada is the same/independent

Ha​: The distribution of seasonal colds in Canada is different./ dependent

Rejection Region

Based on the information provided, the significance level is , the number of degrees of freedom is , so then the rejection region for this test is R = { χ2 : χ2 > 7.815}.

Test Statistics

The Chi-Squared statistic is computed as follows:

\

=

Decision about the null hypothesis

Since it is observed that χ2 = 2.94 ≤χc2 ​=7.815, it is then concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that the two variables are dependent, at the 0.05 significance level.

The corresponding p-value for the test is p=Pr(χ32​>2.94)=0.4009.