Question

Do male and female skiers differ in their tendency to use a ski helmet? Ruzic and Tudor (2011) report a study in which 710 skiers completed a survey about aspects of their skiing habits. Suppose the results from the question on the survey about ski helmet usage were as follows:

	Never	Occasionally	Always
Male	243	71	180
Female	105	33	78

Part a)
Which of the following null hypotheses could sensibly be tested by the data presented above?

A. There is no relationship between Gender and Helmet usage.
B. The mean number of male skiers who never wear a ski helmet is the same as the mean number of female skiers who never use one.
C. The proportions of males and females skiers are equal.
D. Male and female skiers are as likely to never use a ski helmet as always use one.


Part b)
Under the null hypothesis, what is the expected number of men in the survey who never wear a ski helmet?
Give your answer to 2 decimal places.


Part c)
Perform a suitable test on the data above to test the null hypothesis.
Provide the value of your test statistic to 2 decimal places.


Part d)
Under the null hypothesis, the test statistic should be an observation from which probability distribution?

A. The F_3,707 distribution.
B. The tt distribution on two degrees of freedom.
C. The Chi-squared distribution on five degrees of freedom.
D. The Chi-squared distribution on four degrees of freedom.
E. The standard Normal distribution.
F. The Chi-squared distribution on two degrees of freedom.


Part e)
Would you reject or not reject your null hypothesis at the 5 % significance level?

A. Not reject
B. Reject

Answer 1

Part a)
Which of the following null hypotheses could sensibly be tested by the data presented above?

A. There is no relationship between Gender and Helmet usage.

Part b)
Under the null hypothesis, what is the expected number of men in the survey who never wear a ski helmet?

Expected Values	Never	Ocasionally	Always	Total
Male				494
Female				216
Total	348	104	258	710

Part c)
Perform a suitable test on the data above to test the null hypothesis.

Chi square test for independence

Part d)
Under the null hypothesis, the test statistic should be an observation from which probability distribution?

F. The Chi-squared distribution on two degrees of freedom.

Part e)
Would you reject or not reject your null hypothesis at the 5 % significance level?

A. Not reject

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

	Never	Ocasionally	Always	Total
Male	243	71	180	494
Female	105	33	78	216
Total	348	104	258	710

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

Expected Values	Never	Ocasionally	Always	Total
Male				494
Female				216
Total	348	104	258	710

Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - O)^2/E. The table with squared distances is shown below:

Squared Distances	Never	Ocasionally	Always
Male
Female

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H_0 : The two variables are independent

H_a​: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df = (2 - 1)*(3 - 1) = 2.

Test Statistics

The Chi-Squared statistic is computed as follows:

  

Decision about the null hypothesis

Since it is observed that = 0.099 < = 5.991, 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.