Question

Each person in a large sample of German adolescents was asked to indicate which of 50 popular movies they had seen in the past year. Based on the response, the amount of time (in minutes) of alcohol use contained in the movies the person had watched was estimated. Each person was then classified into one of four groups based on the amount of movie alcohol exposure (groups 1, 2, 3, and 4, with 1 being the lowest exposure and 4 being the highest exposure). Each person was also classified according to school performance. The resulting data is given in the accompanying table.

Assume it is reasonable to regard this sample as a random sample of German adolescents. Is there evidence that there is an association between school performance and movie exposure to alcohol? Carry out a hypothesis test using

α = 0.05.

		Alcohol Exposure Group
		1	2	3	4
School Performance	Excellent	112	93	50	66
	Good	328	324	316	295
	Average/Poor	240	258	313	318

State the null and alternative hypotheses.

H₀: Alcohol exposure and school performance are not independent.
H_a: Alcohol exposure and school performance are independent.

H₀: Alcohol exposure and school performance are independent.
H_a: Alcohol exposure and school performance are not independent.   

H₀: The proportions falling into the alcohol exposure categories are the same for the three school performance groups.
H_a: The proportions falling into the alcohol exposure categories are not all the same for the three school performance groups.

H₀: The proportions falling into the alcohol exposure categories are not all the same for the three school performance groups.
H_a: The proportions falling into the alcohol exposure categories are the same for the three school performance groups.

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?

Do not reject H₀. There is not enough evidence to conclude that there is an association between alcohol exposure and school performance.

Do not reject H₀. There is not enough evidence to conclude that the proportions falling into the alcohol exposure categories are not all the same for the three school performance groups.    

Reject H₀. There is convincing evidence to conclude that there is an association between alcohol exposure and school performance.

Reject H₀. There is convincing evidence to conclude that the proportions falling into the alcohol exposure categories are not all the same for the three school performance groups.

Answer 1

Since, we test for association between school performance and movie exposure to alcohol, the appropriate hypothesis are

H₀: Alcohol exposure and school performance are independent.
H_a: Alcohol exposure and school performance are not independent.   

The observed frequencies Oij are,

	1	2	3	4	Total
Excellent	112	93	50	66	321
Good	328	324	316	295	1263
Average/Poor	240	258	313	318	1129
Total	680	675	679	679	2713

The Expected frequencies are given as,

Eij = (Ri * Cj) / Total

where Eij is the expected frequency of the row i and column j

Ri is the total for row i

Cj is the total for column j

	1	2	3	4	Total
Excellent	(321 * 680)/2713 = 80.45706	(321 * 675)/2713 = 79.86546	(321 * 679)/2713 = 80.33874	(321 * 679)/2713 = 80.33874	321
Good	(1263 * 680)/2713 = 316.5647	(1263 * 675)/2713 = 314.237	(1263 * 679)/2713 = 316.0992	(1263 * 679)/2713 = 316.0992	1263
Average/Poor	(1129 * 680)/2713 = 282.9783	(1129 * 675)/2713 = 280.8975	(1129 * 679)/2713 = 282.5621	(1129 * 679)/2713 = 282.5621	1129
Total	680	675	679	679	2713

Chi Square Test statistic, $\chi^2$ = $\sum (O_{ij} - E_{ij})^2 / E_{ij}$

= (112 - 80.45706)² / 80.45706 + (93 - 79.86546)² / 79.86546 + (50 - 80.33874)² / 80.33874 + (66 - 80.33874)² / 80.33874 + (328 - 316.5647)² / 316.5647 + (324 - 314.237)² / 314.237 + (316 - 316.0992)² / 316.0992 + (295 - 316.0992)² / 316.0992 + (240 - 282.9783)² / 282.9783 + (258 - 280.8975)² / 280.8975 + (313 - 282.5621)² / 282.5621 + (318 - 282.5621)² / 282.5621

= 46.7846

Degree of freedom = (r - 1) * (c - 1) = (3 - 1) * (4 - 1) = 6

P-value = P($\chi^2$ > 46.7846, df = 6) = 0.0000

Since, p-value is less than the significance level of 0.05,

Reject H₀. There is convincing evidence to conclude that there is an association between alcohol exposure and school performance.