webassign.net Topic Five Quiz Practice Another Version For Each Of The Following

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webassign.net Topic Five Quiz Practice Another Version For Each Of The Following Situations Give The. 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. Magazine readership Model dress Women Men General interest Total 256 82 Not sexua 359 519 1134 Sexual 205 88 375 Total 564 607 338 1509 (a) Summarize the data numerically and graphically. (Compute the conditional distribution of model dress for each audience. Round your answers to three decimal places.) Women Men General Not sexual Sexual (b) Perform the significance test that compares the model dress for the three categories of magazine readership. Summarize the results of your test and give your conclusion. (Use = 0.01. Round your value for to two decimal places, and round your P-value to four decimal places.) x2 P-value- Conclusion Reject the null hypothesis. There is not significant evidence of an association between target audience and model dress. Fail to reject the null hypothesis. There is not significant evidence of an association between target audience and model dress Fail to reject the null hypothesis. There is significant evidence of an association between target audience and model dress.

Explanation / Answer

there is significant evidence os associatio between targeted audience and model dress

Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 359 519 256 1134 row 2 205 88 82 375 TOTALS 564 607 338 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 1 423.841 456.1551 254.004 row 2 140.159 150.8449 83.996 ------------------------------------------------------------------

calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 359 423.841 -64.841 4204.3553 9.9197 519 456.1551 62.8449 3949.4815 8.6582 256 254.004 1.996 3.984 0.0157 205 140.159 64.841 4204.3553 29.997 88 150.8449 -62.8449 3949.4815 26.1824 82 83.996 -1.996 3.984 0.0474 ^2 o = 74.8204 ------------------------------------------------------------------

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 = 74.8204
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| =74.8204 & | ^2 | =9.2103
make decision
hence value of | ^2 o | > | ^2 | and here we reject Ho
^2 p_value =0


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: 74.8204
critical value: 9.2103
p-value:0
decision: reject Ho

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