1
STA2023 – Inferential Statistics Practice Problems – Answer Key
1.
Use 1sample tprocedures (chapter 23) because
σ
is unknown.
See the assumptions/conditions on
pg. 591592.
All conditions have been met for assuming that tprocedures give close enough
approximations (note that n = 200 is large).
n = 200,
_
y
= 13.55,
s = 3.1
1.
H
0
:
μ
= 14
H
A
:
< 14
(a lefttailed test)
3.
Test Statistic:
_
13.55
14
2.053
/
3.1/ 200
y
t
s
n


=
=
= 
4.
pvalue = tcdf(999, 2.053, 199) = 0.021
From Table T (with df = 180), the pvalue is between 0.01 and 0.025
5.
Reject H
0
at
α
= 0.10 and at
= 0.05, but not at
= 0.01
6.
With a pvalue of 0.021 and a 5% level of significance, the sample data provides enough evidence
to conclude that, on average, adults now spend fewer than 14 hours per weekend on chores.
95% Confidence Interval (t* = 1.973) – formula on page 590:
(13.12, 13.98) hours
Based on this sample, we are 95% confident that, on average, adults now spend between 13.12 hours
and 13.98 hours per weekend on chores.
2.
Use 2proportion zprocedures (chapter 22) because the sample data was collected from two
independent populations (all men & all women) and the researcher wishes to compare the proportion
of success between these populations.
See the assumptions/conditions on pg. 560.
All conditions
have been met for assuming the distribution of
^
^
1
2
p
p

is approximately normal.
Men:
^
^
1
165
0.33
500
M
p
p
=
=
=
Women:
^
^
2
71
0.237
300
W
p
p
=
=
=
1.
H
0
:
p
M
= p
W
or
H
0
:
p
M
 p
W
= 0
H
A
:
p
M
> p
W
H
A
:
p
M

p
W
> 0
(a righttailed test)
3.
Test Statistic:
^
^
^
^
^
^
0.33 0.237
2.877
0.33 0.67
0.237 0.763
500
300
M
W
W
W
M
M
M
W
p
p
z
p
q
p
q
n
n


=
=
=
⋅
⋅
⋅
⋅
+
+
4.
From ZTable:
p(z
≤
2.88) = 0.9980, so the pvalue = 1 – 0.9980 = 0.0020
5.
Reject H
0
at
= 0.10,
= 0.05, and
= 0.01
6.
With a pvalue of 0.0020, the sample data provides very strong evidence to conclude that the
percent of all men who regularly play the lottery is higher than the percent of all women who
regularly play the lottery.
95% Confidence Interval (z* = 1.960) – formula on page 561:
(0.030, 0.156), or 3.0% to 15.6%
Based on this sample, we can say, with 95% confidence, the percent of all men who regularly play the
lottery is between 3.0% and 15.6% higher than the percent of all women who regularly play the
lottery.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document2
3.
Use 1sample tprocedures (chapter 23) because the data is a SRS from a single population (all
packages of Sunlight Vanilla Wafers), the population has a normal distribution, and
σ
is unknown.
Since the population’s distribution is known to be normal, tprocedures will give perfectly accurate
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 Ripol
 Statistics, Inferential Statistics, Normal Distribution, Statistical hypothesis testing

Click to edit the document details