MAT 202 Exam 4 | Score 115/125 | Straighterline

6/14/25, 10:38 AM Graded Exam 4
Topic 8: Analysis of Variance

Started on Saturday, June 14, 2025, 11:13 AM
State Finished
Completed on Saturday, June 14, 2025, 11:38 AM
Time taken 24 mins 32 secs
Grade 115.00 out of 125.00 (92%)




Question 1
The chi-square distribution is a statistical distribution commonly used to test hypotheses about
Partially correct categorical data. It is particularly useful in determining whether there is a significant difference
10.00 points out between observed frequencies and expected frequencies in one or more categories. The chi-square
of 15.00 distribution is positively skewed, but as the degrees of freedom increase, it approaches a normal
distribution. This distribution is only composed of positive values because it is derived from the sum
of squared standard normal variables.


1. Which of the following statements about the chi-square distribution is true?

The chi-square distribution is used for testing hypotheses involving categorical data.
The chi-square distribution is symmetrical regardless of the degrees of freedom.
The chi-square distribution becomes more skewed as the degrees of freedom increase.
The chi-square distribution can take on negative values.

2. As the degrees of freedom increase in a chi-square distribution, what happens to the shape of the distribution?
It becomes bimodal.
It approaches a normal distribution.
It remains the same regardless of degrees of freedom.
It becomes more left-skewed.

3. Why are all values of the chi-square distribution non-negative?
Because it is based on the squares of normally distributed variables, which are always positive or zero.
Due to the subtraction of expected frequencies from observed frequencies.
Because it only measures positive relationships between variables.
Because negative frequencies are not possible in any statistical distribution.

, 6/14/25, 10:38 AM Graded Exam 4




Question 2
A study was conducted to investigate the relationship between being a music student and being on
Correct the honor roll. In a sample of 300 students:
15.00 points out - Music Students: 50 students were music students, and 250 were not.
of 15.00
- Honor Roll Status: 97 students were on the honor roll, and 203 were not.
Researchers wanted to determine the expected number of music students who are also on the
honor roll, assuming independence between the two events. They calculated:
50 1 50 1
Probability of being a music student, P(Music) = = 𝑃(Music) = = .
300 6 300 6
97 97
Probability of being on the honor roll, P(Honor Roll) = 𝑃(Honor Roll) = .
300 300
Assuming independence:
P(Music and Honor Roll) = P(Music) × P(Honor Roll) 𝑃(Music and Honor Roll) = 𝑃(Music) × 𝑃(Honor Roll).
Expected number:
1
E = Total Students × P(Music and Honor Roll) ≈ 300 × × 97 ≈ 16.17 𝐸 = Total Students × 𝑃(Music and Honor Roll)
6 300

1. What is the probability that a randomly selected student is a music student?
1 1
5 5
1 1
6 6
250 250
300 300
97 97
300 300


2. Assuming independence, approximately how many music students are expected to be on the honor roll?
50 students
16 students
97 students
203 students

3. Which calculation correctly computes the expected number of music students on the honor roll?
E = (50 × 97) × 300 𝐸 = ( 50 × 97 ) × 300
(50+97) ( 50 + 97 )
E= 𝐸=
300 300
(97−50) ( 97 - 50 )
E= 𝐸=
300 300

(50×97) ( 50 × 97 )
E= 𝐸=
300 300