Question

The authors of a paper classified characters who were depicted smoking in movies released between a certain range of years. The smoking characters were classified according to sex and whether the character type was positive, negative, or neutral. The resulting data is given in the accompanying table.

Assume that it is reasonable to consider this sample of smoking movie characters as representative of smoking movie characters. Do the data provide evidence of an association between sex and character type for movie characters who smoke? Use

α = 0.05.

	Character Type
Sex	Positive	Negative	Neutral
Male	255	108	131
Female	86	13	51

State the null and alternative hypotheses.

H₀: Sex and character type are independent.
H_a: Sex and character type are not independent. H₀: Sex and character type are not independent.
H_a: Sex and character type are independent.     H₀: The character type category proportions are not all the same for male and female characters.
H_a: The character type category proportions are the same for male and female characters. H₀: The character type category proportions are the same for male and female characters.
H_a: The character type category proportions are not all the same for male and female characters.

Calculate the test statistic. (Round your answer to two decimal places.)
χ² =

What is the P-value for the test? (Round your answer to four decimal places.)
P-value =

What can you conclude?

Reject H₀. There is convincing evidence to conclude that the proportions falling into the character type categories are not all the same for the two gender groups. Reject H₀. There is convincing evidence to conclude that there is an association between sex and character type for movie characters who smoke.     Do not reject H₀. There is not enough evidence to conclude that there is an association between sex and character type for movie characters who smoke. Do not reject H₀. There is not enough evidence to conclude that the proportions falling into the character type categories are not all the same for the two gender groups.

You may need to use the appropriate table in Appendix A to answer this question.

Answer 1

Expected Values positive negative Total neutral 121x 49492.817 644 341x494 - 261.575 18294139.609 male 494 121x 150 28.183 644 341x150 182x16042.391 female 150 6 79.425 Total 341 121 182 644 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E - O)2/E. The table with squared distances is shown below Squared Distances negative positive neutral (255-261.575) 261.575 (108-92.817) 92.817 (131-139.600) 139.609 0.531 male 0.165 2.484 (86-79.425) 79.425 (13-28.183) 28.183 (51-42.391)2 42.391 female = 0.544 = 8.18 = 1.748 Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: The two variables are independent Ha: The two variables are dependent This corresponds to a Chi-Square test of independence Rejection Region Based on the information provided, the significance level is a 0.05, the number of degrees of freedom is df (2 - 1) x (3 1) region for this test is R {x2:x2> 5.991}. 2, so then the rejection Test Statistics The Chi-Squared statistic is computed as follows: (Oij - Eij )2 Eij 0.1650.544+2.484 + 8.18+0.531 1.748 13.652 x2= i-1 Decision about the null hypothesis Since it is observed that x2 13.652 x = 5.991, it is then concluded that the null hypothesis is rejected. Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the two variables are dependent, at the 0.05 significance level The corresponding p-value for the test is p Pr(xi > 13.652) 0.0011.