In: Statistics and Probability
Are teacher evaluations independent of grades? After the midterm, a random sample of 284 students were asked to evaluate teacher performance. The students were also asked to supply their midterm grade in the course being evaluated. In this study, only students with a passing grade were included in the summary table. Use a 5% level of significance to test the claim that teacher evaluations are independent of midterm grades.
Midterm Grade
Teacher Evaluation |
A |
B |
C |
Row total |
Positive |
35 |
33 |
28 |
96 |
Neutral |
25 |
46 |
35 |
106 |
Negative |
20 |
22 |
40 |
82 |
Column Total |
80 |
101 |
103 |
284 |
(A) State the null and alternate hypotheses.
(B) Identify the appropriate sampling distribution: Chi-square test of independence, Chi-square goodness of fit, or Chi-square for testing or estimating σ2 or σ.
(C) What is the value of the sample test statistic?
(D) Find or estimate the P-value.
(E) Based on your answers for parts (a) through (d), will you reject or fail to reject the null hypothesis?
A) Ho:teacher evaluations are independent of midterm grades.
Ha:teacher evaluations are dependent of midterm grades.
B)
Chi-square test of independence
C)
Applying chi square test:
Expected | Ei=row total*column total/grand total | A | B | C | Total |
positive | 27.04 | 34.14 | 34.82 | 96 | |
neutral | 29.86 | 37.70 | 38.44 | 106 | |
negative | 23.10 | 29.16 | 29.74 | 82 | |
total | 80 | 101 | 103 | 284 | |
chi square χ2 | =(Oi-Ei)2/Ei | A | B | C | Total |
positive | 2.3417 | 0.0381 | 1.3347 | 3.715 | |
neutral | 0.7908 | 1.8287 | 0.3085 | 2.928 | |
negative | 0.4157 | 1.7589 | 3.5401 | 5.715 | |
total | 3.548 | 3.626 | 5.183 | 12.3571 |
alue of the sample test statistic =12.357
d)
degree of freedom(df) =(rows-1)*(columns-1)= | 4 |
for 4 degree of freedom and above test statistic , p value =0.0149
e)
as p value is less than 0.05 level , we reject null hypothesis and conclude that teacher evaluations are dependent of midterm grades.