In what ways do advertisers in magazines use sexual imagery to appeal to youth?
ID: 3181835 • Letter: I
Question
In what ways do advertisers in magazines use sexual imagery to appeal to youth? One study classified each of 1500 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 age group of the intended readership. Here is a summary of the data.
Model Dress
Young Adult
Mature Adult
Not sexual (percent)
72.6%
75.4%
Sexual (percent)
27.4%
24.6%
Number of ads
1000
500
Perform the significance test that compares the model dress for the age groups of magazine readership. Summarize the results of your test. (Use = 0.05. Round your 2 to three decimal places and round your P-value to four decimal places.)
2 =
P-value =
Model Dress
Young Adult
Mature Adult
Not sexual (percent)
72.6%
75.4%
Sexual (percent)
27.4%
24.6%
Number of ads
1000
500
Explanation / Answer
The formula for chi-square test statistics:
X^2 = S (Oi –Ei)^2 / Ei
Oi are the given values (observed values)
Ei are the expected values
E = (row total * column total )/ Total
Oi:
726
377
274
123
1000 * 0.726 = 726
1000 *0.274 = 274
500 *0.754 = 377
500*0.246 = 123
Total
726
377
1103
274
123
397
Total
1000
500
1500
Ei
735.3333
367.6667
264.6667
132.3333
(Oi – Ei)
-9.3333
9.3333
9.3333
-9.3333
(Oi – Ei)^2
87.1105
87.1105
87.1105
87.1105
(Oi – Ei)^2 / Ei
0.118
0.237
0.329
0.658
Now we add them all
X^2 = 1.343
Degrees of freedom = ( no. of rows – 1 ) * ( no. of columns – 1) = (2-1)*(2-1) = 1
We use excel to find the p-value
=CHIDIST(1.343,1)
= 0.2465
P-value = 0.2465
Answer:
X^2 = 1.343
P-value = 0.2465
Here the p-value is greater than the level of significance (.05), we fail to reject the null hypothesis. There is no relationship.
726
377
274
123
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