In: Statistics and Probability
Answer :
Given data is :
Sample size n = 89
A | B | C | D |
23 | 22 | 34 | 10 |
Based on these,
Null hypothesis is :
Correct answers for all multiple choice questions are evenly distributed.
Alternative hypothesis is ;
Correct answers for all multiple choice questions are evenly distributed.
Now,
Choice | Probability (P) |
Observed (O) |
Expected value (E) E = n * p |
|
A | 1 / 4 = 0.25 | 23 |
= 89 * 0.25 = 22.25 |
= = = =0.0253 |
B | 1 / 4 = 0.25 | 22 | 22.25 | 0.0028 |
C | 1 / 4 = 0.25 | 34 | 22.25 | 6.205 |
D | 1 / 4 = 0.25 | 10 | 22.25 | 6.744 |
= 23 + 22 + 34 +10 = 89 |
= 0.0253 + 0.0028 + 6.205 + 6.744 = 12.977 |
a)Test statistics = 12.977
b)Given significance level
Degree of freedom df = n - 1 = 4 - 1 = 3,
So critical value form chi square table is ;
Critical value = 6.251.
We get Critcal value < test statistics.
c)YES
We have sufficient evidence to support the claim.