ID webassign.net As Part Or The Topic Five Quiz Explain With Why Do BIO-202L: An
ID: 3360293 • Letter: I
Question
ID webassign.net As Part Or The Topic Five Quiz Explain With Why Do BIO-202L: Anatomy Quiz 3 F Mail Portal | Main Self Reg Bachelor of Sci -10 points MintroStat .E.507xP My Notes Ask Your In what ways do advertisers in magazines use sexual imagery to appeal to youth? One study classified cach of 1509 full-page or larger ads as "nat 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 Generol interest Total 1099 Not scxua 352 506 Sexual 211 90 109 410 Total 563 596 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.) Men Not sexual Sexual (b Per orm the significance test hat compares the model dress or he three categories of magazine readership. Summarize he results o your test and give your condusion. Use -0.01. Round your value or places, and round your P-value to four decimal places.) to two decimal P-value = C Reject the null hypothesis. There is 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 C 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 (c) All of the ads were taken from the March, July, and November issucs 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. This is an SRS. This gives us no reason to believe our conclusions are suspect. This is not an SRS. This gives us reason to believe our conclusions might be suspect. C This is not an SRS. This gives us no reason to believe our conclusions are suspect This is an SRS. This gives us rcason to belicve our conclusions might be suspoct.Explanation / Answer
we have significant association between target audience and model dress
c.
this ia an SRS,this give us reason to believe our conclusions might be suspect
Given table data is as below MATRIX col1 col2 col3 TOTALS row 1 352 506 241 1099 row 2 211 90 109 410 TOTALS 563 596 350 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 1 410.031 434.065 254.904 row 2 152.969 161.935 95.096 ------------------------------------------------------------------
calculate chisquare test statistic using given observed frequencies, calculated expected frequencies from above Oi Ei Oi-Ei (Oi-Ei)^2 (Oi-Ei)^2/Ei 352 410.031 -58.031 3367.597 8.213 506 434.065 71.935 5174.644 11.921 241 254.904 -13.904 193.321 0.758 211 152.969 58.031 3367.597 22.015 90 161.935 -71.935 5174.644 31.955 109 95.096 13.904 193.321 2.033 ^2 o = 76.895 ------------------------------------------------------------------
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 =9.21
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 = 76.895
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| =76.895 & | ^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 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: 76.895
critical value: 9.21
p-value:0
decision: reject Ho