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	111	93	49	67
	Good	329	326	317	297
	Average/Poor	239	259	314	319

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 = ___

Answer 1

Solution;-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

H₀: School performance and movie exposure to alcohol are independent.
H_a: School performance and movie exposure to alcohol are not independent.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square test for independence.

Analyze sample data. Applying the chi-square test for independence to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = (r - 1) * (c - 1) = (4 - 1) * (3 - 1)
D.F = 6
E_r,c = (n_r * n_c) / n


Χ² = 47.07

where DF is the degrees of freedom.

The P-value is the probability that a chi-square statistic having 6 degrees of freedom is more extreme than 47.07.

We use the Chi-Square Distribution Calculator to find P(Χ² > 47.07) = 0.0000

Interpret results. Since the P-value (0.0000) is less than the significance level (0.05), we cannot accept the null hypothesis.

Thus, we conclude that there is a relationship between school performance and movie exposure to alcohol.