Chapter 7
Estimation and Statistical Intervals
Section 7.1
1.
A single randomly selected item forms a sample of size n = 1.
If x denotes the value of this item, then the
sample average is also equal to x.
Since the sample average, based on any sample size, is an unbiased
estimator or the population mean, then the length x is indeed an unbiased estimator of the population mean.
Of course, this estimator is not as precise as the sample average based on larger sample sizes (i.e., it has a
larger standard error), but it is still unbiased.
2.
(a) An unbiased point estimate of
μ
, the average amount of gas used by all houses in the area is:
therms
x
6
.
120
10
206
,
1
=
⎟
⎠
⎞
⎜
⎝
⎛
=
(b)
An unbiased point estimate of
π
, the proportion of all homes that use over 100 therms is:
80
.
10
8
=
⎟
⎠
⎞
⎜
⎝
⎛
=
p
3.
(a)
Because the population is known to be normal and its standard deviation is known to be
σ
= 5, the
sampling distribution of
x
is also normal.
Standardizing,
P(
μ
-1
≤
x
≤
μ
+ 1) =
P(
n
σ
μ
μ
−
−
)
1
(
≤
z
≤
n
σ
μ
μ
−
+
)
1
(
) =
P(-
σ
n
≤
z
≤
σ
n
)
=
P(-.63
≤
z
≤
.63)
=
.7357 - .2643 =
.4714.
(b)
In each case, as in part (a), the desired probability is P(-
σ
n
≤
z
≤
σ
n
):
For n = 50,
P(-
σ
n
≤
z
≤
σ
n
) = P(-1.41
≤
z
≤
1.41)
= .9207 - .0793
= .8414.
For n = 100,
P(-
σ
n
≤
z
≤
σ
n
) = P(-2.00
≤
z
≤
2.00)
= .9772 - .0228 = .9544.
For n = 1000,
P(-
σ
n
≤
z
≤
σ
n
) = P(-6.32
≤
z
≤
6.32)
≈
1.0000 - .0000 = 1.0000.
In other words, as n increases, it becomes more and more likely that the sample mean will fall within
±
1 standard deviation from the population mean.

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