Q.R11.4 - The Practice of Statistics for AP

Question

Researchers looked at a random sample of $1509$ full-page ads that show a model in magazines aimed at young men, at young women, or at young adults in general. They classified the ads as “not sexual” or “sexual,” depending on how the model was dressed (or not dressed). Here are the data: 

The figure below displays Minitab output for a chi-square test using these data 

a) Which type of chi-square test should be used in this case? Justify your answer. 

(b) State an appropriate pair of hypotheses for the test you chose in part (a). 

(c) Show how each of the numbers $60.94, 424.8,$and $12.835$ was obtained for the “notsexy, Women” cell

(d) Assuming that the conditions for performing inference are met, what conclusion would you draw? Explain 

Step-by-Step Solution

Answer

(a) The Chi square association test will be employed.

(b) 

$H_0$:The audience's gender and the magazine advertising' sexuality are unrelated.

$H_1$ : The gender of the audience and the sexiness of magazine advertisements are not mutually exclusive.

(c) By using two equations

$\frac{Number of not sexual ads read by women}{ Number of women}$

$\frac{not sexual total \times women readers total}{ total number of ads}$

(d) That the audience's gender and the advertising' sexuality are not independent. 

Part (a)Step 1: Given information

Given in the question that,   Researchers looked at a random sample of $1509$ full-page ads that show a model in magazines aimed at young men, at young women, or at young adults in general. They classified the ads as “not sexual” or “sexual,” depending on how the model was dressed (or not dressed).

We need to find that the type of chi-square test used in this case.

Part(a) Step 2: Explanation

The total number of individuals=$1509$

$\begin{array}{|llll|}\hline & \text{ Readers }& & \\ & \text{ Men }& \text{ Female }& \text{ General }\\ \text{ Sexual }& 105& 225& 66\\ \text{ Not sexual }& 514& 351& 248\\ \hline\end{array}$

The following is a two-way table of gender and sexual classification.

When checking for independency between two variables or any form of relationship or association, the Chi square test of association is used.

The Chi square goodness of fit test is used to determine how well a distribution fits the data.

Only one variable can be used here.

Because the data in the question was chosen at random and has two variables, the chi square test for association will be utilised to determine the relationship.

Part(b) Step 1: Given information

Given in the question that,   Researchers looked at a random sample of 1509 full-page ads that show a model in magazines aimed at young men, at young women, or at young adults in general. 

We need to state appropriate hypotheses for performing the type of test  chose in part (a) .

Part (b) Step 2: Explanation

The null hypothesis is the most natural, and it is based on it that the test statistic is generated.

Because the chi square association test will be utilised,

The null hypothesis is that the audience's gender and the sexuality of magazine advertisements are unrelated.

The alternative idea is that audience gender and magazine ad sexuality are not mutually exclusive.

$H_0$:The audience's gender and the magazine advertising' sexuality are unrelated.

$H_1$: The gender of the audience and the sexiness of magazine advertisements are not mutually exclusive.

Part(c) Step 1: Given information

Given in the question that,   Researchers looked at a random sample of 1509 full-page ads that show a model in magazines aimed at young men, at young women, or at young adults in general.

We need to show that how each of the numbers 60.94, 424.8, and 12.835 was obtained for the “not sexy, Women” cell

Part(c) Step 2: Explanation

The null hypothesis is the most natural, and it is based on it that the test statistic is generated.

Total number of female readers = number of women who read sexual advertising + number of women who do not read sexual ads

$=225+351=576$ For $60.94$

The following formula can be used to calculate the percentage of non-sexual adverts among female readers.

Total number of female readers Equals total number of female sexual advertisements read Women read $+$ of non-sexual advertisements.

$=225+351=576$For $60.94$

The following formula can be used to calculate the percentage of non-sexual adverts among female readers.

The following formula can be used to calculate the percentage of non-sexual adverts among female readers.

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

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 predicted cell count for nonsexual women is

$\frac{not sexual total \times women readers total }{total number of ads}$

$$\begin{gathered} =\frac{1113\times 576}{1509}=424.843 \\ =424.8 \end{gathered}$$

The not sexual ladies reader cell will contribute $12.835$ to the chi statistic.

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

Observed value =351 

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

Part(d) Step 1: Given information

Given in the question that,   Researchers looked at a random sample of 1509 full-page ads that show a model in magazines aimed at young men, at young women, or at young adults in general. 

We need to provide a conclusion if the conditions for performing inference are met.

Part (d) Step 2: Explanation

Information provided:

value of chi square $80.874$ $P$ value $0.00$

If the $P$ value is less than the level of significance, reject the null hypothesis.

Assume that the significance threshold is $0.05$.

So, $0.00<0.05$

Conclusion: Using the above notion, there is sufficient evidence to reject the null hypothesis at the 5% level of significance, implying that the audience's gender and the advertising' sexuality are not independent.