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In what ways do advertisers in magazines use sexual imagery to appeal to youth?

ID: 3359578 • 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.)

2 =

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.

(Part 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.

d. This is not an SRS. 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 1509

Explanation / Answer

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set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.01
from standard normal table, chi square value at right tailed, ^2 /2 =9.2103
since our test is right tailed,reject Ho when ^2 o > 9.2103
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 74.6052
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 9.2103
we got | ^2| =74.6052 & | ^2 | =9.2103
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0
a. there is significant evidence of association between target audience and model dress

part.c

All of the ads were taken from the March, July, and November issues of six magazines in one year.

This is an SRS. This gives us reason to believe our conclusions might be suspect
ANSWERS
---------------
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 74.6052
critical value: 9.2103
p-value:0
decision: reject Ho

Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 358 529 258 1145 row 2 201 90 73 364 TOTALS 559 619 331 N = 1509 ------------------------------------------------------------------

calculation formula for E table matrix E-TABLE col1 col2 col3 row 1 row1*col1/N row1*col2/N row1*col3/N row 2 row2*col1/N row2*col2/N row2*col3/N ------------------------------------------------------------------

expected frequecies calculated by applying E - table matrix formulae E-TABLE col1 col2 col3 row 1 424.1584 469.6852 251.1564 row 2 134.8416 149.3148 79.8436 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 358 424.1584 -66.1584 4376.9339 10.3191 529 469.6852 59.3148 3518.2455 7.4906 258 251.1564 6.8436 46.8349 0.1865 201 134.8416 66.1584 4376.9339 32.4598 90 149.3148 -59.3148 3518.2455 23.5626 73 79.8436 -6.8436 46.8349 0.5866 ^2 o = 74.6052

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set up null vs alternative as
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
level of significance, = 0.01
from standard normal table, chi square value at right tailed, ^2 /2 =9.2103
since our test is right tailed,reject Ho when ^2 o > 9.2103
we use test statistic ^2 o = (Oi-Ei)^2/Ei
from the table , ^2 o = 74.6052
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 9.2103
we got | ^2| =74.6052 & | ^2 | =9.2103
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0
a. there is significant evidence of association between target audience and model dress

part.c

All of the ads were taken from the March, July, and November issues of six magazines in one year.

This is an SRS. This gives us reason to believe our conclusions might be suspect
ANSWERS
---------------
null, Ho: no relation b/w X and Y OR X and Y are independent
alternative, H1: exists a relation b/w X and Y OR X and Y are dependent
test statistic: 74.6052
critical value: 9.2103
p-value:0
decision: reject Ho

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