Question

A regression analysis can be used to derive a relationship for input to another CER. Derive a relationship for converting size units, plug in a value and use its output for the size input in a CER for systems engineering effort.

You are tasked to perform an early systems engineering cost estimate during the conceptual stage of a project. The only measure of size available from the architects is an estimated 22 capabilities, but your CER is based on more detailed measurements for system requirements.

Derive a relationship that converts capabilities to system requirements using linear regression with the data below from past projects. System requirements is the dependent variable and capabilities is the independent variable in the regression. What is your assessment of the strength of the regression relationship?

Project	# Capabilities	# System Requirements
1	14	129
2	25	277
3	9	122
4	21	248
5	9	35
6	18	240
7	19	262
8	26	259
9	3	36
10	15	147
11	8	119
12	16	141
13	7	108
14	14	157
15	12	70
16	1	10
17	3	67
18	17	244
19	22	196
20	15	186

1. What is the estimated number of System Requirements for 22 Capabilities using your model?
2. Also approximate the 95% confidence interval for the number of requirements using the regression standard error.
3. Then use the CER below to derive an estimate of systems engineering effort for 22 capabilities using the requirements point estimate.

Effort (Person-Months) = .25 * System Requirements^1.06

Answer 1

Solution

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

Let x = capability and y = system requirement.

Part (a)

Estimated regression equation is: System Requirements = 8.51 + 10.52Capabilities

Estimated number of System Requirements for 22 Capabilities = 239.98 Answer 1

Part (b)

95% confidence interval for the number of requirements (for 22 Capabilities)

= [212.92, 267.03] Answer 2

Part (c)

Systems engineering effort for 22 capabilities = 0.25 x 239.98 = 59.995 Answer 3

Back-up Theory and Details of calculations

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_hat = β_0hat + β_1hatX, ………….........................................................………….(2)

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

Estimate of σ² is given by s² = (Syy – β_1cap²Sxx)/(n - 2)…………………………………...................................................…..(4)

100(1 - α)% Confidence Interval (CI) for y_cap at x = x₀ is: (y_cap at x = x₀) ± t_{n – 2, α/2}xs√[(1/n) + {(x₀ – Xbar)²/Sxx}] ........... (5)

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)} ] .......................................................................... (6)

Details of calculations

n	20
Xbar	14
ybar	153
Sxx	982.2
Syy	133564.55
Sxy	10333.9
β1cap	10.5212
β0cap	8.5099
s^2	1379.9866
s	37.1482
α	0.05
n-2	18
tn-2,α/2	2.1009
x0	22
ycap at x0	239.9758
CIYcapLB	212.9245
CIYcapUB	267.0271

DONE