Chapter 11, Problem 4CRE

(a)

To determine

To find: Type of chi square to be used.

Answer to Problem 4CRE

Chi square test of association will be used.

Explanation of Solution

Given information:

Total number of people $=1509$

The two way table of gender and sexual classification is as follows

	Readers
	Men	Female	General
Sexual	$105$	$225$	$66$
Not sexual	$514$	$351$	$248$

Chi square test of association is used where the concern is to check the independency between two variables or any kind of relation or association.

Chi square goodness of fit is to test the fit of a certain distribution to the data.

Here, only one variable can be used.

As the data in the question randomly selected, have two variables and the concern is to check the association so, chi square test for association will be used.

(b)

To determine

To find: The null and the alternate hypotheses

Answer to Problem 4CRE

  $H_0:$The gender of audience and the sexuality of the magazine ads are independent.

  $H_1:$ The gender of audience and the sexuality of the magazine ads are not independent.

Explanation of Solution

The null hypothesis is the natural one and is the one on the assumption of which the test statistic is calculated.

As the chi square test of association will be used so,

The null hypothesis will be that the gender of audience and the sexuality of the magazine ads are independent.

The alternate hypothesis will be that the gender of audience and the sexuality of the magazine ads are not independent.

(c)

To determine

Explanation of Solution

Calculation:

Total number of women readers = number of sexual ads read by women $+$ number of not sexual ads read by women

  $=225+351=576$

For $60.94$

The percentage of not sexual ads among the women readers can be calculated as follows

  $\frac{\text{Number of not sexual ads read by women}}{\text{Number of women}}=\frac{351}{576}=60.94$

For $424.8$

The sum total not sexual will be $=$ number of not sexual ads read by men $+$ number of not sexual ads read by women $+$ number of not sexual ads read by general

  $=514+351+248=1113$

The expected count for not sexual women cell will be

  $\begin{array}{l}\frac{\text{not sexual total }\times \text{ women readers total}}{\text{total number of ads}}\\ =\frac{1113\times 576}{1509}=424.843\left(=424.8\right)\end{array}$

For $12.835$

The contribution of the not sexual women reader cell to chi statistic will be

  $\frac{{\left(\text{observed - expected}\right)}^2}{\text{expected}}$

Observed value $=351$

So, $\frac{{\left(351-424.843\right)}^2}{424.843}=12.835$

(d)

To determine

To find: The conclusion based on the results given.

Answer to Problem 4CRE

The gender of the audience and the sexuality of the ads are not independent

Explanation of Solution

Given information:

Chi square value $80.874$

P value $0.00$

Concept used:

Reject null hypothesis if P value is less than the level of significance.

Here, assume the level of significance as $0.05$

So, $0.00<0.05$

Conclusion:

Using the above concept, there is sufficient evidence to reject null hypothesis at $5\%$ level of significance and thus conclude that the gender of the audience and the sexuality of the ads are not independent.