Download Math 144 tutorial 7

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Question 2
Question 3
Question 4
Math 144 tutorial 7
12 April, 2010
Question 5
Question 6
Question 1
Question 2
1. Let us define S02 =
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1
n
Pn
i=1 (Xi
Question 4
− X̄)2 . Show that
E(S02 ) =
n−1 2
σ .
n
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Question 6
Question 1
Question 2
1. Let us define S02 =
Question 3
1
n
Pn
i=1 (Xi
Question 4
− X̄)2 . Show that
E(S02 ) =
n−1 2
σ .
n
Question 5
Question 6
Question 1
Question 2
1. Let us define S02 =
Question 3
1
n
Pn
i=1 (Xi
Question 4
Question 5
− X̄)2 . Show that
E(S02 ) =
n−1 2
σ .
n
This result also indicate that S02 is a biased estimator of the σ 2 .
Question 6
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2. If X is a binomial random variable, show that
(a). p̂ = X/n√is an unbiased estimator of p;
n/2
√
is a biased estimator of p;
(b). p0 = X+
n+ n
(c). Show that the estimator p0 becomes unbiased as n → ∞.
Question 6
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2. If X is a binomial random variable, show that
(a). p̂ = X/n√is an unbiased estimator of p;
n/2
√
is a biased estimator of p;
(b). p0 = X+
n+ n
(c). Show that the estimator p0 becomes unbiased as n → ∞.
Solution: (a). Since
E[p̂] = E[X]/n = np/n = p,
this shows that p̂ is an unbiased estimator of p.
(b). E[p0 ] =
√
E[X]+ n/2
√
n+ n
=
√
np+ n/2
√
n+ n
6= p.
√
np + n/2
√
(c).
lim
n∞
n+ n
p
1
√
√
= lim
+ lim
n∞ 1 + n/n
n∞ 2(1 + n)
= p.
Question 6
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3. An electrical firm manufactures light bulbs that have a length of life
that is approximately normally distributed with a standard deviation
of 40 hours. If a sample of 30 bulbs has an average life of 780 hours.
(a). Find a 96% confidence interval for population mean of all bulbs
produced by this firm;
(b). How large a sample is needed if we wish to be 96% confident that
our sample mean will be within 10 hours of the true mean?
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3. An electrical firm manufactures light bulbs that have a length of life
that is approximately normally distributed with a standard deviation
of 40 hours. If a sample of 30 bulbs has an average life of 780 hours.
(a). Find a 96% confidence interval for population mean of all bulbs
produced by this firm;
(b). How large a sample is needed if we wish to be 96% confident that
our sample mean will be within 10 hours of the true mean?
Solution: The sample mean x̄ is 780, and the standard deviation of
population is 40, using the Normal table, we can get z0.02 = 2.05.
Hence the 96% confidence interval for µ is
σ
σ
[x̄ − z0.02 × ( √ ), x̄ + z0.02 × ( √ )] = [765.029, 794.971].
30
30
(b). We can be 96% confident that this error will less than
zα/2 × ( √sn ), i.e, n = ( z0.02e×40 )2 = 67.24. Therefore, a sample with
size 68 is needed if we wish be 96% confident that our sample mean
will be within 10 hours of the true mean.
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4. Many cardiac patients wear implanted pacemakers to control their
hear-beat. A plastic connector module mounts on the top of the
pacemaker. Assuming a standard deviation of 0.0015 and an
approximate normal distribution,
(a). Find a 95% confidence interval for the mean of all connector
modules made by a certain manufacturing company. A random
sample of 75 modules has an average of 0.310 inch.
(b). How large a sample is needed if we wish to be 95% confident that
our sample mean will be within 0.0005 inch of true mean?
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4. Many cardiac patients wear implanted pacemakers to control their
hear-beat. A plastic connector module mounts on the top of the
pacemaker. Assuming a standard deviation of 0.0015 and an
approximate normal distribution,
(a). Find a 95% confidence interval for the mean of all connector
modules made by a certain manufacturing company. A random
sample of 75 modules has an average of 0.310 inch.
(b). How large a sample is needed if we wish to be 95% confident that
our sample mean will be within 0.0005 inch of true mean?
Solution (a). From the problem, we obtain
n = 75, x̄ = 0.310, σ = 0.0015, hence the 95% confidence interval
for the mean of all modules is:
0.0015
0.0015
[0.310−z0.025 ×( √
), 0.310+z0.025 ×( √
)] = [0.309661, 0.310339].
75
75
×0.0015 2
(b). n = ( z0.0250.0005
) = 34.5744. That is, a sample with size 35 is
needed if we wish to be 95% confident that our sample mean will be
within 0.0005 inch of true mean.
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5.The heights of a random sample of 50 college students showed a
mean of 174.5 centimeters and a standard deviation of 6.9 centimeters.
(a). Construct a 98% confidence interval for the mean height of all
college students.
(b). What can we assert with 98% confidence about the possible size
of our error it we estimate the mean height of all college students to
be 174.5 centimeters?
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5.The heights of a random sample of 50 college students showed a
mean of 174.5 centimeters and a standard deviation of 6.9 centimeters.
(a). Construct a 98% confidence interval for the mean height of all
college students.
(b). What can we assert with 98% confidence about the possible size
of our error it we estimate the mean height of all college students to
be 174.5 centimeters?
Solution: Because standard deviation of the population is unknown,
so we should use the t distribution:
(a). Because s = 6.9, x̄ = 174.5, n = 50, so 98% confidence interval
for the mean height of all college students is:
[174.5.t0.01,49 × ( √6.9
), 174.5 + t0.01,49 ( √6.9
)] = [173.229, 175.771].
50
50
(b). We can be 98% confident that this error will less than tα/2 ( √sn ),
i.e,
6.9
e ≤ t0.01,49 × ( √ ) = 1.27148.
50
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6. A random sample of 100 automobile owners shows that, in the state
of Virginia, an automobile is driven on the average 23,500 kilometers
per year with a standard deviation of 3900 kilometers. Assume the
distribution of measurements to be approximately normal.
(a). Construct a 99% confidence interval for the average number of
kilometers an automobile is driven annually in Virginia.
(b). What can we assert with 99% confidence about the possible size
of our error if we estimate the average number of kilometers driven by
car owners in Virginia to be 23,500 kilometers per year?
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6. A random sample of 100 automobile owners shows that, in the state
of Virginia, an automobile is driven on the average 23,500 kilometers
per year with a standard deviation of 3900 kilometers. Assume the
distribution of measurements to be approximately normal.
(a). Construct a 99% confidence interval for the average number of
kilometers an automobile is driven annually in Virginia.
(b). What can we assert with 99% confidence about the possible size
of our error if we estimate the average number of kilometers driven by
car owners in Virginia to be 23,500 kilometers per year?
Solution: (a). We use the t distribution,
x̄ = 23500, s = 3900, n = 100, then the 99% confidence interval for
population mean µ is:
3900
3900
), 23500+t0.005,99 ×( √
)] = [22479.4, 24520.6].
[23500−t0.005,99 ×( √
100
100
3900
(b). e ≤ t0.005,99 × ( √
) = 1020.63.
100