Download Example: A SRS of 1000 high school students gains an average of

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Example: A SRS of 1000 high school students gains
an average of x̄ = 22 points in their second attempt at
the SAT mathematical exam. The change in score has
a normal distribution with standard deviation σ = 50.
(1) Find the 95% CI for µ.
(2) Find the margin of error for 99% confidence.
Solution:
(1) The 95% CI for µ is
σ
50
σ
50
[x̄ − z ∗ √ , x̄ + z ∗ √ ] = [22 − 1.96 √
, 22 + 1.96 √
]
n
n
1000
1000
= [18.9, 25.1]
(2) The the margin of error for 99% confidence is
σ
50
= 4.073.
z ∗ √ = 2.576 √
n
1000
Example: Acid rain, caused by the reaction of certain pollution with rainwater, appear to be a growing
problem in the northeastern United States. Pure rain
falling through clean air registers a pH value of 5.7 (pH
is a measure of acidity:0 is acid; 14 is alkaline). We
knew that water pH standard deviation is 0.5. Suppose water SRS from 15 rainfalls are analyzed for pH,
and X̄ = 3.7.
(a) Find a 99% CI for the mean pH in rainfall.
(b) What assumption must be made for the confidence
interval to be valid?
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Solution (a) Since t∗ = 2.57, we have
σ
σ
[x̄ − z ∗ √ , x̄ + z ∗ √ ]
n
n
0.5
0.5
= [3.7 − 2.57 √ 3.7 + 2.57 √ ]
15
15
= [3.368, 4.317].
(b) We must assume that water pH has a normal distribution.
Choosing the sample size
√
We know that σ/ n → 0 as n → ∞. Thus the margin
error of any CI tends to zero as n → ∞. Thus, the bigger the sample size, the closer the CI to the estimated
parameter. However, the bigger the sample size, the
more money we need to spend. To save money, time,
manpower, and so on, and same time to attend our
required accuracy, we can find suitable sample size for
us.
Sample size for desired margin of error:
The confidence interval for the mean of a normal population will have a specified margin of error m when
the sample size is
µ ∗ ¶2
z σ
.
n=
m
If we require that the margin of error within m, then,
we have to choose n such that
µ ∗ ¶2
z σ
n≥
.
m
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Example: Some high school students will try their
second attempt at the SAT mathematical exam. Suppose the change in score has a normal distribution
with standard deviation σ = 50.
How large a SRS of high school students would be
needed to estimate the mean change µ in SAT score
to within ±2 points with 95% confidence?
Solution: According to the formula, we have
µ ∗ ¶2 µ
¶2
z σ
(1.96)(50)
n=
=
= 2401.
m
2
Example: We are concerned with the daily adult intake of selenium in a region of the United States. Suppose that the daily intakes have a normal distribution
with standard deviation of 24.3 micrograms and you
wish to estimate the mean daily intake of this region
correct to within 5 micrograms, with probability equal
to 0.9.How large should n be?
Solution: According to the question, we have
σ
24.3
1.645 √ = 1.645 √ ≤ 5.
n
n
So
7.9947 ≤
√
n
or
n ≥ 63.9152.