Question

A new school district superintendent preparing to reallocate resources for physically impaired students wanted to know if the schools in the district differed in the distribution of physically impaired. The superintendent tested samples of 20 students from each of the five schools and found 5 physically impaired (and 15 unimpaired) students at School 1, 5 physically impaired (and 15 unimpaired) at School 2, 6 (and 14) at School 3, 4 (and 16) at School 4, and 7 (and 13) at School 5. Using the .05 significance level, test whether the distribution of physically impaired students is different at different schools. Figure the chi-square for this data set yourself (round to two decimal places). What is the chi-square obtained?

Answer 1

Solution:-

State the hypotheses.

Null hypothesis: The distribution of physically impaired students is different at different schools.

Alternative hypothesis: At least one of the null hypothesis statements is false.

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

Analyze sample data. Applying the chi-square test for homogeneity 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) = (2 - 1) * (5 - 1)
D.F = 4
E_r,c = (n_r * n_c) / n

	Impaired	Expected	[(Or,c - Er,c)2/ Er,c]	Unimpaired	Expected	[(Or,c - Er,c)2/ Er,c]	Total
School 1	5	5	0.02962963	15	15	0.010958904	20
School 2	5	5	0.02962963	15	15	0.010958904	20
School 3	6	5	0.066666667	14	15	0.024657534	20
School 4	4	5	0.362962963	16	15	0.134246575	20
School 5	7	5	0.474074074	13	15	0.175342466	20
Total	27	27	0.96	73	73	0.36	100

Χ² = 1.32

where DF is the degrees of freedom, r is the number of populations, c is the number of levels of the categorical variable, n_r is the number of observations from population r, n_c is the number of observations from level c of the categorical variable, n is the number of observations in the sample, E_r,c is the expected frequency count in population r for level c, and O_r,c is the observed frequency count in population r for level c.

The P-value is the probability that a chi-square statistic having 4 degrees of freedom is more extreme than 1.32.

We use the Chi-Square Distribution Calculator to find P(Χ² > 1.32) = 0.858.

Interpret results. Since the P-value (0.858) is greater than the significance level (0.05), hence we failed to reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that the distribution of physically impaired students is different at different schools.