A new school district superintendent was preparing to reallocate resources for p

Question

A new school district superintendent was preparing to reallocate resources for physically impaired students. He wanted to know if the schools in his district differed in the distribution of physically impaired students. He tested samples of 20 students from each of his five schools. He found the following: School 1: 4 impaired and 16 unimpaired students School 2: 1 impaired and 19 unimpaired students School 3: 6 impaired and 14 unimpaired students School 4: 3 impaired and 17 unimpaired students School 5: 7 impaired and 13 unimpaired students a) Conduct a Chi-Square Test for Independence to determine if “School Location” and “Physical Impairment” are dependent on each other. (Use a .01 significance level and make sure you show all 5 steps of the hypothesis test)

Explanation / Answer

The two variables are:

School location (r) = 5

Physical impairment (c) = 2

Thus, degrees of freedom = (5-1) ( 2-1) = 4

Thus,

Chi-squared value = 6.87

P ( X2 > 6.87) for 4 degrees of freedom = 0.14

We have used a significance value of 0.01

Hence, since P-value (0.14) > 0.01,

we fail to reject the null hypothesis.

Thus, the two variables are dependent.

Hope this helps.

Expected Values table Impaired Unimpaired Total Impaired Unimpaired School 1 4 16 20 School 1 4.2 15.8 School 2 1 19 20 School 2 4.2 15.8 School 3 6 14 20 School 3 4.2 15.8 School 4 3 17 20 School 4 4.2 15.8 School 5 7 13 20 School 5 4.2 15.8 Total 21 79 100 Chi-squared 0.009524 0.002532 2.438095 0.648101 0.771429 0.205063 0.342857 0.091139 1.866667 0.496203 Total 5.428571 1.443038 6.871609

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.