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. Magazine readership age group Model dress Young adult Mature adult Not sexual (percent) 72.4% 75.6% Sexual (percent) 27.6% 24.4% 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 α = 0.05. Round your χ2 to three decimal places and round your P-value to four decimal places.) χ2 = 1.537 Incorrect: Your answer is incorrect. P-value = 1.836

Answer 1

Data:

	Young adult	Mature Adult	Marginal Row Totals
Not Sexual	724	378	1102
Sexual	276	122	398
Marginal Column Totals	1000	500	1500    (Grand Total)

Solved Data:

The contingency table below provides the following information: the observed cell totals, (the expected cell totals) and [the chi-square statistic for each cell].

	Young adult	Mature Adult	Marginal Row Totals
Not Sexual	724   (734.67)   [0.15]	378   (367.33)   [0.31]	1102
Sexual	276   (265.33)   [0.43]	122   (132.67)   [0.86]	398
Marginal Column Totals	1000	500	1500    (Grand Total)

H_o: Model Dress and Age Group are independent. (No association)

H_a: Model Dress and Age Group are not independent. (Association)

Formulae:

DF = (r - 1) * (c - 1) = (2 - 1) * (2 - 1) = 1

E_r,c = (n_r * n_c) / n

where r is for row and c is for column in the 2*2 contingency table.

Expected total for 1st cell = (1000*1102)/1500

= 734.67

Expected total for 2nd cell in 1st row = (500*1102)/1500

= 367.33

All other values can be calculated in the same way.

Χ² = Σ [ (O_r,c - E_r,c)² / E_r,c ] .......2

A chi square value will be generated for each cell e.g.

Χ²_1,1 = (724-734.67)²/734.67

= 113.849/734.67

= 0.15

All other values can be calculated in the same way.

Using the above formula in eqn 2:

Χ² = 0.15+0.31+0.43+0.86

= 1.751

with df = 1

p-value (for two-tailed test) =  0.1857

Since p-value = 0.1857 > 0.05 i.e. we fail to reject the null hypothesis. There is not significant evidence of an association between model dress and age group.

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