Dr. Suren Phansalker
ADM2304 X
Assignment#1
Solution
115 Marks
May 28, 2011
Qu.#1 (15 Marks)
Here:
657
ˆ
0.8760
750
X
p
n
=
=
=
, Assume LS =
α
= 0.05, CC = 1-
α
= 0.95
S1: 4 Marks
0
0
:
(
0.9)
H
p
p
=
=
0
:
(
0.9)
a
H
p
p
≠
=
: Two Tailed Test
{Not accurate
Î
Less than or More than}
(You must test ‘p’ and not ‘
’)
ˆ
p
S2: 7 Marks
N.B.: Here,
n p = 50 (0.125) = 6.25 > 5.0, and
0
n q
0
= 50 (0.95) = 43.75 > 5.0.
Thus “Normal” distribution is, strictly speaking
reasonable.
{3 Marks}
0
0
0.9(0.1)
ˆ
(
)
(Pr
)
0.0110
750
p q
p
SD
oportion
n
σ
=
=
=
=
{2 Marks}
0
ˆ
0.876
0.900
2.1909
ˆ
(
)
0.0110
Calc
p
p
z
p
σ
−
−
=
=
= −
{2 Marks}
S5: 4 Marks
p-Val = P[Z > {|z
Calc
| = 2.1909] x 2
= P[Z > 2.1909] x 2
= (1.0 – 0.9857) x 2
= 0.0286
Since {p-Val = 0.0286} < {
α
= 0.05}
Î
Reject H0
You could also use the equivalent critical value approach shown below.
S3:
LS =
α
= 0.05 and with the Two Tailed Test,
/ 2
0.025
1.96
Crit
z
z
z
α
=
=
=
S4:
Since {|z
Calc
| = 2.1909} > {z
Crit
= 1.96}
Î
Reject H . {Just as before}
0
Qu.#2 (38 Marks)
a. -- 6 Marks: 4 for one of the plots, 2 for comments.
Descriptive Statistics: BMI
Variable
N
N*
Mean
SE Mean
StDev
Minimum
Q1
Median
Q3
BMI
2500
0
28.006
0.120
5.993
8.230
23.960
28.000
32.040
Variable
Maximum
BMI
46.650
1

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