Question

Human Resources department at a manufacturing company is concerned about unauthorized absences of its employees. As a part of their research, they decided to investigate the relationship between the number of unauthorized days that employees are absent per year and the distance (miles) between home and work for the employees. A sample of 10 employees was chosen, and the following data were collected. Distance to Work (miles) Number of Days Absent 1 8 3 5 4 8 6 7 8 6 10 3 12 5 14 2 14 4 18 2 Problem 1k) What is the conclusion for this problem based on Hypothesis test results? Problem 1l) Provide an interpretation for the “Coefficient of Determination” for this problem. Problem 1m) State why the estimated regression equation is “good” or “no good”. Problem 1n) What is the value of sample correlation coefficient in this problem? Problem 1o) Would you recommend using the estimated Regression equation? Why or why not? (Provide your reasoning in a sentence). Problem 1p) What is numerical value of the Slope Coefficient in this problem? Interpret the “numerical value” of slope coefficient for this problem. Problem 1q) Calculate the “expected number of days absent” for an employee who lives 10 miles away from work.

Answer 1

Solution

Let x = Distance (miles) to work and y = Number of days absent.

NOTE

Direct answers are given below. Back-up Theory and Details of Calculations follow at the end.

Problem (1k)

Conclusion for this problem based on Hypothesis test results

Test for significance of the correlation coefficient, r, shows that r is significant. So, the conclusion is that the Distance (miles) to work and Number of days absent are linearly related and the former can be used to predict the latter. ANSWER

Problem (1l)

Interpretation for the “Coefficient of Determination” for this problem.

Coefficient of Determination, r², is found to be 0.7108. This implies that 71% of the variation in dats absent is accounted by the distance to work.ANSWER

Problem (1m)

The estimated regression equation is “good”because

the predictor variable, distance to work, is able to account for a sizable percentage, 71%, of the variation in the days absent. ANSWER

Problem (1n)

The value of sample correlation coefficient = - 0.8431 ANSWER

Problem (1o)

It is recommend using the estimated Regression equation for the reason already given under (1m) ANSWER

Problem (1p)

Numerical value of the Slope Coefficient, β_1cap = - 0.3442

Interpretation: For every increase of one mile in distance to work, days absent would on an average reduce by 0.34 days. ANSWER

Problem (1q)

The “expected number of days absent” for an employee who lives 10 miles away from work.

= 4.66 days. ANSWER

Now to work out the solution,

Back-up Theory

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 = β_0cap + β_1capX, …………………….(2)

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

Mean X = Xbar = (1/n)sum of x_i ………………………………………….……….….(4)

Mean Y = Ybar = (1/n)sum of y_i ………………………………………….……….….(5)

Sxx = sum of (x_i – Xbar)² ………………………………………………..…………....(6)

Syy = sum of (y_i – Ybar)² ……………………………………………..………………(7)

Sxy = sum of {(x_i – Xbar)(y_i – Ybar)} ……………………………………………….(8)

All above sums are over i = 1, 2, …., n,n = sample size ……………………………..(9)

Correlation coefficient, r = Sxy/sqrt(Sxx. Syy) …………………………………….. (10)

To test for significance of correlation coefficient

Hypotheses:

Null: H₀: ρ = 0 Vs Alternative H₁: ρ ≠ 0

Test Statistic:

t = r√{(n - 2)/(1 – r²)}

Distribution, Significance Level, α, Critical Value

t ~ t_{n – 2}

So, Critical Value = upper (α/2) % point of t_{n – 2}

Decision

H₀ is rejected if | t_cal | > t_crit

Details of Calculations

Data

i	xi	yi
1	1	8
2	3	5
3	4	8
4	6	7
5	8	6
6	10	3
7	12	5
8	14	2
9	14	4
10	18	2

Summary of Calculations

n	10
Xbar	9.00
ybar	5
Sxx	276
Syy	46
Sxy	-95
β1cap	-0.3442029
β0cap	8.09782609
r	-0.8431215
r^2	0.71085381

Test for significance of r		t = r√{(n - 2)/(1 - r^2)}
1- r^2	0.28914619
(n-2)/(1-r^2)	27.6676655
sqrt	5.26000622
tcal	-4.4348242
tcrit	2.30600413

DONE