250+ TOP MCQs on Chi-Squared Distribution and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Chi-Squared Distribution”.

1. A dice is tossed 120 times with the following results

No. turned up	1	2	3	4	5	6
Frequency	30	25	18	10	22	15

Test the hypothesis that the dice is unbiased (X² = 11.7). Calculate the frequency observed for Chi Square distribution.
a) Dice is unbiased, 11.3
b) Dice is biased, 12.9
c) Dice is unbiased, 10.9
d) Dice is biased, 12.3
Answer: b
Clarification: Step 1: Null Hypothesis: dice is unbiased.
Step 2: Calculation of Expected frequency:
Since the dice is unbiased P(r) = 1/6.
r = 1, 2, 3, 4, 5, 6
Expected frequency f(r) = N*r = 120* 1/6 = 20
Step3: Calculation of X²
X² (=∑frac{[(fe-fo)]^2}{fe} )
Hence X² = 12.90 > 11.90.
Thus dice is biased.

2. Consider a set of 18 samples from a standard normal distribution. We square each sample and sum all the squares. The number of degrees of freedom for a Chi Square distribution will be?
a) 17
b) 18
c) 19
d) 20
Answer: b
Clarification: In Chi Square Distribution the number of standard normal derivatives or samples equals the number of degrees of freedom.
Here total number of standard normal derivatives = 18.
Hence the number of degrees of freedom for a Chi Square distribution = 18.

3. What is the mean of a Chi Square distribution with 6 degrees of freedom?
a) 4
b) 12
c) 6
d) 8
Answer: c
Clarification: By the property of Chi Square distribution, the mean corresponds to the number of degrees of freedom.
Degrees of freedom = 6.
Hence mean = 6.

4. Which Chi Square distribution looks the most like a normal distribution?
a) A Chi Square distribution with 4 degrees of freedom
b) A Chi Square distribution with 5 degrees of freedom
c) A Chi Square distribution with 6 degrees of freedom
d) A Chi Square distribution with 16 degrees of freedom
Answer: d
Clarification: When the number of degrees of freedom in Chi Square distribution increases it tends to correspond to normal distribution. The option with a maximum number of degrees of freedom is 16.

5. A bag contains 80 chocolates. This bag has 4 different colors of chocolates in it. If all four colors of chocolates were equally likely to be put in the bag, what would be the expected number of chocolates of each color?
a) 12
b) 11
c) 20
d) 9
Answer: c
Clarification: If all four colors were equally likely to be put in the bag, then the expected frequency for a given color would be 1/4th of the chocolates.
N = 80, r = 1/4
So, the expected frequency = N*r = (1/4)*(80) = 20.

6. Suppose a person has 8 red, 5 green, 12 orange, and 15 blue balls. Test the null hypothesis that the colors of the balls occur with equal frequency. What is the Chi Square value you get?
a) 5.6
b) 5.68
c) 5.86
d) 5.8
Answer: d
Clarification: By Chi Square Test we get,
Observed frequency f₀= (8+5+12+15)/4 = 10
X² =(∑frac{[(fe-fo)]^2}{fe} )
The sum of each (expected – observed)²/expected = (10-8)²/10 + (10-5)²/10 + (10-12)²/10 + (10-15)²/10 = 5.8.

7. A faculty is interested in whether there is a relationship between gender and subject at his college. He tabulated some men and women on campus and asked them if their subject was Mathematics (M), Geography (G), and Science (S). What would be the expected frequency of women in Geography based on this table?

	M	G	S	Total
Women	10	14	10	34
Men	11	22	14	23
Total	21	36	24	57

a) 31.12
b) 11.32
c) 12.13
d) 13.12
Answer: d
Clarification: The expected value of women in social sciences is the product of the total number of women and the total number of social science majors divided by the total number of participants. (22*34)/57 = 13.12.

8. In a sample survey of public opinion answer to the question:
1) Do you drink?
2) Are you in favor of local option sale of Liquor

	Yes	No	Total
Yes	56	31	87
No	18	6	24
Total	74	37	111

Infer or not the local option on the sale of liquor is dependent on individual drinker? Find the value of X² for degrees of freedom at level of significance 3.841.
a) 0.957
b) 0.975
c) 0.759
d) 0.795
Answer: a
Clarification: Step 1: Null hypothesis: The option on the sale of liquor is not dependent with the individual drinking.
Step 2: Calculation of theoretical frequencies (Expected)
Expected frequency of (1,1) cell
f_e11 = 87*74/111 = 58
f_e12 = 87*37/111 = 29
f_e21 = 24*74/111 = 16
f_e22 = 24*32/111 = 8

Step 3: calculation of X² distribution we know that
X² =( ∑frac{[(fe-fo)]^2}{fe} )
X² = 0.957 < 3.841.
Hence the null hypothesis is accepted
Thus the sale of liquor does not depend on the individual drinker.