In what ways do advertisers in magazines use sexual imagery to appeal to youth?
ID: 3358142 • Letter: I
Question
In what ways do advertisers in magazines use sexual imagery to appeal to youth? One study classified each of 1509 full-page or larger ads as "not sexual" or "sexual," according to the amount and style of the dress of the male or female model in the ad. The ads were also classified according to the target readership of the magazine. Here is the two-way table of counts.
(a) Summarize the data numerically and graphically. (Compute the conditional distribution of model dress for each audience. Round your answers to three decimal places.)
Conclusion
a. Reject the null hypothesis. There is significant evidence of an association between target audience and model dress.
b. Fail to reject the null hypothesis. There is not significant evidence of an association between target audience and model dress.
c. Fail to reject the null hypothesis. There is significant evidence of an association between target audience and model dress.
d. Reject the null hypothesis. There is not significant evidence of an association between target audience and model dress.
(c) All of the ads were taken from the March, July, and November issues of six magazines in one year. Discuss this fact from the viewpoint of the validity of the significance test and the interpretation of the results.
a. This is not an SRS. This gives us no reason to believe our conclusions are suspect.
b. This is an SRS. This gives us reason to believe our conclusions might be suspect.
c. This is an SRS. This gives us no reason to believe our conclusions are suspect.This is not an SRS.
d. This gives us reason to believe our conclusions might be suspect.
Magazine readership Model dress Women Men General interest Total Not sexual 358 529 258 1145 Sexual 201 90 73 364 Total 559 619 331 1509Explanation / Answer
Given table data is as below
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calculation formula for E table matrix
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expected frequecies calculated by applying E - table matrix formulae
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calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above
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set up null vs alternative as
null, Ho: no relation b/w target audience and model dress OR target audience and model dress are independent
alternative, H1: exists a relation b/w target audience and model dress OR target audience and model dress are dependent
level of significance, = 0.01
from standard normal table, chi square value at right tailed, ^2 /2 =9.21
since our test is right tailed,reject Ho when ^2 o > 9.21
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 74.605
critical value
the value of |^2 | at los 0.01 with d.f (r-1)(c-1)= ( 2 -1 ) * ( 3 - 1 ) = 1 * 2 = 2 is 9.21
we got | ^2| =74.605 & | ^2 | =9.21
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0
ANSWERS
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null, Ho: no relation b/w target audience and model dress OR target audience and model dress are independent
alternative, H1: exists a relation b/w target audience and model dress OR target audience and model dress are dependent
test statistic: 74.605
critical value: 9.21
p-value:0
decision: reject Ho
There is significant evidence of an association between target audience and model dress.
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