STAT 409
Fall 2008
Version A
Name ANSWERS .
Quiz 2
(10 points)
Be sure to show all your work, your partial credit might depend on it.
No credit will be given without supporting work.
1. (6) This year Doug played 14 complete rounds of golf (18 holes), his (sample) average
score was 85 with sample standard deviation 6.5. Brian played 10 complete rounds
of golf, and his (sample) average score was 82 with sample standard deviation 6.
Assume that their golf scores are approximately normally distributed, and that the
overall variances are equal. Construct a 90% confidence interval for the difference
between their overall average scores on a complete round of golf.
(
)
(
)
2
10
14
0.61105.6114
22
2
pooled
−+
−+−
=
⋅⋅
s
=
39.693
pooled
s
=
6.3
( )
21
pooled
2
11
YX
nn
st
+
α
±−
⋅⋅
14 + 10 – 2 =
22
degrees of freedom
α
= 0.10
t
0.05
(
22
)
=
1.717
( )
10
1
14
1
3.6717.18285
+±−
⋅⋅
3 ±
±±
± 4.5
(
–
1.5
, 7.5
)
2.
(4) A survey of 16
Initech
employees revealed that the average number of minutes an
employee talks on the phone per day while in the office is 56 minutes, with a sample
standard deviation of 7 minutes. Assume that the time the employees spend on the
phone is normally distributed. Construct a 95% confidence interval for the overall
average number of minutes an employee talks on the phone per day while in the office.
σ
is unknown
n
= 16 – small The confidence interval: X
t
s
±
α2
n
.
n
− 1 = 16 − 1 =
15
degrees of freedom
α
= 0.05
t
0.025
(
15
)
=
2.131
16
7
131256 ⋅±
.
56 ±
±±
± 3.73
( 52.27 ; 59.73 )