Sample Size is = 16

Sample Mean = 24.1
Sample Standard Deviation = 4.8
Degree of freedom is 16-1 = 15
σ
is unknown
t
=
x
−
μ
0
σ
/
√
n
=
24.1
−
21.6
4.8
/
√
16
=
2.0833
For degree of freedom 15 the t value 2.08 is between 1.753 and 2.131 which is p-value between 0.025 and 0.05
The exact p-value from excel formula calculation is
0.0274
c)
At
α
= 0.05, test for a significant difference. What is your conclusion?
The p-value of 0.0274 is <
α
= 0.05
We reject
H
0
at 5% level of significance and conclude that the per capita milk consumption Webster City is higher
than the annual per capita consumption of milk of 21.6 gallons.

Chapter 10
Question 1

Question 5

Question 7

Question 9

Question 13:
FedEx and United Parcel Service (UPS) are the world's two leading cargo carriers by volume and revenue (
The Wall Street
Journal
, January 27th, 2004). According to the Airports Council International, the Memphis International Airport (FedEx)
and the Louisville International Airport (UPS) are 2 of the 10
largest cargo airports in the world. The following random
samples show the tons of cargo per day handled by these airports. Data are in thousands of tons.
a.
Compute the sample mean and sample standard deviation for each airport
Sample size for Memphis airport
n
1
=
12
Sample mean for Memphis airport
x
1
=
∑
x
i
n
1
=
111.6
12
=
9.3
Standard deviation for Memphis airport
s
1
=
√
∑
(
x
i
−
x
1
)
2
n
1
−
1
=
√
71.1
12
−
1
=
2.54
Sample size for Louisville airport
n
2
=
10
Sample mean for Louisville airport
x
2
=
∑
x
i
n
2
=
42
10
=
4.2
Standard deviation for Louisville airport
s
2
=
√
∑
(
x
i
−
x
2
)
2
n
2
−
1
=
√
18.4
10
−
1
=
1.43

b.
What is the point estimate of the difference between the two population means? Interpret this value in terms of
the higher-volume airport and a comparison of the volume difference between the two airports.
Point estimator of the difference between two population means
=
x
1
−
¿
x
2
=
9.3
−
4.2
=
5.1
Memphis airport is the higher volume airport and handles a mean of 5.1 tons of cargo per day more than
Louisville airport.
c.
Develop a 95% confidence interval of the difference between the daily population means for the two airports
df
=
(
s
1
2
n
1
+
s
2
2
n
2
)
2
1
n
1
−
1
(
s
1
2
n
1
)
2
+
1
n
2
−
1
(
s
2
2
n
2
)
2
=
(
2.54
2
12
+
1.43
2
10
)
2
1
12
−
1
(
2.54
2
12
)
2
(
2.54
2
12
)
+
1
10
−
1
(
1.43
2
10
)
2
=
17.8
Rounding down, we will use a t distribution with 17 degrees of freedom,
At 95% confidence,
α
= 0.05,
t
α
2
=
t
0.025
=
2.110
,
Interval Estimate of the difference between the population mean,
¿
(
x
1
−
x
2
)
±t
α
2
√
s
1
2
n
1
+
s
2
2
n
2
=
5.1
±
√
2.54
2
12
+
1.43
2
10
=
5.1
±
1.82
Thus the 95% confidence interval estimate of the difference between population means is
3.28 to 6.92.
Question 15:
a.
1
= population mean for annual lease rate for commercial properties in Hong Kong
2
= population mean for annual lease rate for commercial properties in Paris
H
0
:
μ
1
−
μ
2
≤
0
H
a
:
μ
1
−
μ
2
>
0
b
.
Sample size (Hong Kong)
n1
= 30
Sample mean
x
1
=
$
1114
Standard deviation
s
1
=
$230
Sample size (Paris)
n1 =
40
Sample mean
x
2
=
$
989
Standard deviation
s
2
=
$195

Point estimator of the difference between two population means,
=
x
1
−
x

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- Fall '15
- Statistics, Normal Distribution, Standard Deviation, Statistical hypothesis testing