Question

Problem 1

The following table lists data on the participation scores and midterm exam scores for 23 students who took the Accounting Theory class in Spring 2019.

Student Number	Participation Score	Midterm Score
1	5	75
2	6.5	84
3	6.5	73
4	7	96
5	7.5	83
6	7.5	88
7	7.5	75
8	8	75
9	8.5	82.5
10	8.5	89
11	8.5	90
12	8.5	91
13	9	92
14	9	81.5
15	9	95
16	9	88
17	9	89.5
18	9	93
19	9.5	90
20	10	99
21	10	87.5
22	10	88
23	10	81

1. Using the example above as a guide, construct a regression model that captures the relationship between participation scores and midterm scores.
2. Using the coefficients from your model, calculate the expected midterm scores for students whose participation scores are 5, 7 and 8 respectively.
3. Using the expected scores that you computed in (2) above, calculate the unexpected scores for student numbers 1, 4 and 8.
4. Give three reasons why the expected midterm scores for these students does not equal their actual midterm scores.

Answer 1

Solution

Part (1)

NOTE: Final answers are given below. Back-up Theory and Details of calculations follow at the end.

Regression model:

Mid-term score (y) = 62.6070 + (2.8292 x Participation score) (x) Answer 1

Part (2)

Substituting x = 5, 7 and 8 in the above regression equation we get:

Expected midterm scores for students whose participation scores are 5 is: 76.75 Answer 2

Expected midterm scores for students whose participation scores are 7 is: 82.41 Answer 3

Expected midterm scores for students whose participation scores are 8 is: 85.24 Answer 4 Part (3)

Since participation scores for student numbers 1, 4 and 8 are 5, 7 and 8 respectively,

Expected mid-term scores for student numbers 1, 4 and 8 are respectively:

76.75, 82.41 and 85.24 Answer 5

Part (4)

Comparison

Student #	Expected score (E)	Actual score (A)	Difference E – A
1	76.75	75	1.75
4	82.41	96.	- 13.59
8	85.24	75	10.24

In two cases, E > A and in one case A > E.

The difference is quite significant for students 4 and 8.

Explanations for the difference

1. The model assumed (simple linear) is not adequate to predict. This is quite possible since the correlation coefficient is only 0.51.

2. Assumption made for the linear model such as error terms are iid N(0, σ²), independence of x with error term, etc are not fulfilled.

3. Data recording is not reliable.

Answer 6

Back-up Theory and Details of calculations

The linear regression model: Y = β₀ + β₁X + ε, ……………………..............................................................…………..(1)

where ε is the error term, which is assumed to be Normally distributed with mean 0 and variance σ².

Estimated Regression of Y on X is given by: Y_hat = β_0hat + β_1hatX, …………......................................................…….(2)

β_1cap = Sxy/Sxx and β_0cap = Ybar – β_1cap.Xbar.............………………...............................................................….…..(3)

where

[Mean X = Xbar = (1/n) Σ(i = 1 to n)x_i ; Mean Y = Ybar = (1/n) Σ(i = 1 to n)y_i Sxx = Σ(i = 1 to n)(x_i – Xbar)²

Syy = Σ(i = 1 to n)(y_i – Ybar)² ; Sxy = Σ(i = 1 to n){(x_i – Xbar)(y_i – Ybar)} ] ................................................................ (4)

n	23
Xbar	8.3913
ybar	86.3478
Sxx	36.9783
Syy	1127.2174
Sxy	104.6196
β1cap	2.8292
β0cap	62.6070
r	0.5124

DONE