STA2601 ASSIGNMENT 4 Solutions 2026

STA2601 ASSIGNMENT 4 Solutions 2026
UNISA
Unique Nr.: 135223
Opening date: 03 August 2026
Due date: 13 August 2026

,Question 1

The observed table is:

Parent Colour down White down Total

A 210 50 260

B 146 54 200

C 34 6 40

Total 390 110 500



(a) Mosaic plot and interpretation

The proportions of white chicks are:

• Parent A: 50/260 = 19.23%

• Parent B: 54/200 = 27.00%

• Parent C: 6/40 = 15.00%

A mosaic plot would therefore show that Parent B has the largest proportion of white-
coloured chicks, while Parent C has the smallest proportion.

The widths of the parent sections also represent their sample sizes: Parent A has the
largest number of chicks, followed by Parent B and then Parent C.

Interpretation: The mosaic plot suggests that the colour distribution may differ
between the three parent types, particularly because Parent B has a noticeably higher
proportion of white chicks.

(b) Hypotheses

𝐻0 : Type of parent and colour of down are independent.
𝐻1 : Type of parent and colour of down are associated.


1(c) Test statistic

Expected frequencies are calculated using:
(row total)(column total)
𝐸𝑖𝑗 =
grand total


For Parent A, coloured:

, 260(390)
𝐸= = 202.8
500


Parent A, white:

260(110)
𝐸= = 57.2
500


Similarly:

Parent Colour Expected White Expected

A 202.8 57.2

B 156.0 44.0

C 31.2 8.8

The chi-square statistic is

2
(𝑂 − 𝐸)2
𝜒 =∑
𝐸
(210 − 202.8)2 (50 − 57.2)2
𝜒2 = +
202.8 57.2
(146 − 156)2 (54 − 44)2
+ +
156 44
(34 − 31.2)2 (6 − 8.8)2
+ +
31.2 8.8
2
𝜒 = 5.218


Degrees of freedom:

𝑑𝑓 = (3 − 1)(2 − 1) = 2


At 𝛼 = 0.05:
2
𝜒0.05,2 = 5.991


Since

5.218 < 5.991


we fail to reject 𝐻0 .