HI6007 T1 2018 Group Assignment: Statistical Analysis & Reporting

This document presents a solved group assignment for HI6007, focusing on statistical data analysis and interpretation. It includes frequency distributions, relative frequency distributions, and histograms for a given dataset. The assignment further involves hypothesis testing using t-tests to determine the relationship between demand and unit price, along with calculating the coefficient of determination and correlation. An ANOVA test is performed to compare treatment means, and a regression model is developed and analyzed, including F-tests and t-tests to assess the significance of variables. The solution provides detailed calculations and interpretations for each question, offering a comprehensive understanding of statistical concepts and their application. Desklib provides more solved assignments and past papers for students.

HI6007 T12018 Group Assignment 1
HI6007 T12018 Group Assignment
Name
Course
Instructor
Date

HI6007 T12018 Group Assignment 2
Question 1
a)
Class
Frequenc
y
Relative
Frequency
Percent
Frequency
100 149 3 0.06 6%
150 199 15 0.3 30%
200 249 14 0.28 28%
250 299 6 0.12 12%
300 349 4 0.08 8%
350 399 3 0.06 6%
400 449 3 0.06 6%
450 499 2 0.04 4%
SUM 50 1 100%
b)
100-
149 150-
199 200-
249 250-
299 300-
349 350-
399 400-
449 450-
499
0%
5%
10%
15%
20%
25%
30%
35%
Histogram
Percent frequency
distribution
Class
Percent Relative distribution

HI6007 T12018 Group Assignment 3
The distribution is skewed to the right.
c) The most appropriate measure of location will be median. This is because the data is
skewed and we need to combat outliers (Draper & Smith, 2015).
Question 2
a) Using t test:
H0 : β1=0
H1 : β1 ≠ 0
Level of significance , α=0.05.
W h en p−value<0.05 , demand price isrelated ¿ t h e unit price
t= b1−β1
se ( β1 ) ∽tn−2
Approximated gradient , b1=−2.137
Standard Error of gradient , se ( β1 ) =0.248
df =46
t=−2.137−0
0.248 =−8.617
Using Excel , p−value is given by :
p−value=2 P ¿
¿ 3.70425 E−11
¿ 0.0000000004 T h e computed p−value is less t h an 0.05 ,
T h us we reject t h e null h ypot h esis at 5 % significance∧¿
conclude t h at demand∧unit price are related d ,

HI6007 T12018 Group Assignment 4
b) C oefficient of determination, R2= SSR
SST = 5048.818
818.479 =0.61703 .
Helence t h e coefficient of detrmination is 0.61703.
T h is means t h at 61.703 % of t h e variability observed∈demand can be explained by unit pric e
c) Coefficient of correlation= √ Coefficient of determination= √0.61703
¿ ∓ 0.7855
Give t h at multiple R=0.7855 .
T h e sign of regression coefficient =negative . T h us t h e coefficient of correlation=−0.7855
Consequently , t h ereis a negative relationbetween demand∧unit price .
W h en unit price increase , t h e demand decreases ,∧vice versa .
Question 3
N=24 ,
K=3
Source of
Variation SS
Degree of
Freedom
Mean Square F
Between
treatments
390.58 k-1=2
195.29
25.891
Within
Treatments
158.40 N-k=21 7.5428
Total 548.98 N-1=23
H0 : μ1=μ2=μ3 because t h e t h ree treatment means are equal

HI6007 T12018 Group Assignment 5
H1 : μ1 ≠ μ2 ≠ μ3
if F= MSST
MSSE < Ftab value at [ ( k−1 ) , ( n−k ) ] degree of freedom ,
we reject our hypot h esis H0 . ¿ Ftable at ( 2,21 ) degreesof freedom we get 3.467 .
h ence t h e calculated F value of 25.89is more t h an F table value of 3.467 .
T h us we reject null hypot h essis∧conclude t h at t h e da t a provided is sufficient evidence ¿
conlude t h at t h e means are same for eac h treatment at α=0.05
.
Question 4
a) F rom t h e estimated regression output ¿ t h e tables ,
Intercept=0.8051
Coefficient of x1 =0.4977
x2=0.4733
Hence y=0.8051+0.4977 x1 +0.4733 x2
b) D egree of Regression=k−2
Degree of residual=n−k −1=7−2−1=4
MS= SS
df
F= MSR
MSE
Degree of SS MS F

HI6007 T12018 Group Assignment 6
freedom
Regression 2 40.700 20.35 80.118
Residual 4 1.016 0.254
Thus p-value is 0.00593
T h us by conducting F−test , t h e obtained p−value isless t h atn0.05 .
Hence t h ere ∃ a significant relations h ipbetween all t h e dependent∧¿
independent variables
c) H0 : β1 against H2 : β1 ≠ 0
Thus we test the claim at the α=005 level of signifcance .
T h e test statistic is t= b1
sb1
= 0.4977
0.4617 =1.079375
2 P ( t >t0 )=2 (1−P ( t<t0 ) )
¿ 2 ( 1−P ( t<1.079375 ) )
¿ 2 ( 1−0.8276 )
¿ 2 ( 0.1724 )
¿ 0.3448 Because 0.3448>0.05 ,
we fail¿ reject H0∧conclude t h at price is not significant
d) If x2 increase by aunit , t h e y valueincreases by 0.4733
e) y=0.8051+0.4977 x1 +0.4733 x2
y=0.8051+ 0.4977∗20+ 0.4733∗10=15.4922
Thus on average, the expectated number of mobile phones is equal 15.4922.

HI6007 T12018 Group Assignment 7
References
Draper, N. & Smith, H., 2015. Applied Regression Analysis (Wiley Series in Probability and
Statistics). New York: Wiley-Interscience.
Montgomery, D., peck, E. & Vining, G., 2006. Introduction to Linear Regression Analysis. 1st
ed. New York: Wiley-Interscience.