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CONFIDENCE INTERVAL WORKSHEET
MTH 3210 SPRING 2015
1. Supplementary Notes
Confidence intervals are the first example we have of inferential statistics in this course.
Recall that the objective of inferential statistics is to make a correct statement about a population parameter based on information from a single sample. For this quiz, the population
parameter being estimated is the unknown population mean, µ. We do not know µ because
we do not have a census; that is, we do not have all the data in our entire population. In any
situation where the entire population is unattainable, especially in situations where only a
relatively small sample is realistically attainable, we must use a sample from our population
to build a precise interval estimate of µ with a prescribed reliability called the confidence
level.
There are 2 types of confidence intervals for the unknown mean, µ, of a single population.
Recipes for each of these confidence intervals follow. For both of these confidence intervals,
it is crucially important that the sample collected is a simple random sample, that is, every
experimental unit in the population has the same chance of being selected for the sample. The
foundational theorem that the following confidence intervals are built upon is the Central
Limit Theorem. In order to use the Central Limit Theorem reliably we must collect a
simple random sample and we must assume that the size of the sample is greater than 30 or
that the underlying population that we are working with is normally distributed.
1.1. Confidence interval for µ when σ is known. For this test, we assume that the
population standard deviation, σ, is known. If σ is not known (as is usually the case), then
we must use the next confidence interval, below. Precisely, we assume that
• We have a simple random sample
• The sample size is at least 30 or the underlying population has a normal distribution
• The population standard deviation, σ, is already known.
A 100(1 − α)% confidence interval for the population mean µ of the population is given
by,
σ
σ
x¯ − zα/2 √ , x¯ + zα/2 √
n
n
Here n is the sample size, x is the sample mean, and zα/2 is the z-value under the normal
curve for which the area under the curve, to the right of z, is α/2. The zα/2 values can be
inferred from Table II in the book.
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MTH 3210 SPRING 2015
1.2. Confidence interval for µ when σ is unknown. For this test, we do not need to
know the population standard deviation σ. The cost of not knowing σ is that we now need
to use the t-distribution, which depends on the number of degrees of freedom, df ; in this
case, df = n − 1, where n is our sample size. Precisely, we assume that
• We have a simple random sample
• The sample size is at least 30 or the underlying population has a normal distribution
A 100(1 − α)% confidence interval for the population mean µ of the population is given
by,
s
s
(¯
x − tα/2 √ , x¯ + tα/2 √ )
n
n
Here n is the sample size, x is the sample mean, s is the sample standard deviation and
tα/2 is the t-value under the t-distribution (with n − 1 degrees of freedom) curve for which
the area under the curve, to the right of t, is α/2. Such values can be inferred from Table
IV in the book.
2. Practice Problems
2.1. Practice Problem. For these exercises, σ is known. Find a 100(1 − α)% confidence
interval for µ in each case. Assume that the underlying population is normally distributed.
(1) x¯ = 20, σ = 3, n = 36, α = .05
(2) x¯ = 25, σ = 3, n = 36, α = .01
(3) x¯ = 30, σ = 3, n = 25, α = .1
2.2. Practice Problem. For these exercises, σ is NOT known. Find a 100(1 − α)% confidence interval for µ in each case. Assume that the underlying population is normally
distributed.
(1) x¯ = 35, s = 3.76, n = 25, α = .1
(2) x¯ = 50, s = 5.02, n = 16, α = .01
(3) x¯ = 50, s = 3.71, n = 12, α = .05
2.3. Practice Problem. A simple random sample of 36 St Bernard dog weights yields a
sample mean of 193 pounds. It is known that the standard deviation of the population of all
St Bernard dog weights is 18 pounds. Find a confidence interval for the true mean weight
of all St Bernard dogs with a 99% confidence level.
2.4. Practice Problem. Eleven regions in the Congolese rain forest are randomly sampled.
In each region rainfall was monitored for one year, and the following total yearly rainfalls,
in centimeters, were reported:
{276, 255, 255, 297, 213, 241, 269, 262, 145, 185, 209}
CONFIDENCE INTERVAL WORKSHEET
MTH 3210 SPRING 2015
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Assume that yearly rainfalls within the Congolese rain forest are distributed normally. Find
a confidence interval for the true mean rainfall in the Congolese rain forest with a 95%
confidence level.
3. Quiz Problems
3.1. Quiz Problem. For these exercises, σ is known. Find a 100(1−α)% confidence interval
for µ in each case. Assume that the underlying population is normally distributed.
(1) x¯ = 1.05, σ = 0.25, n = 9, α = .05
(2) x¯ = 1990, σ = 120, n = 36, α = .01
(3) x¯ = 31.3, σ = 2.67, n = 16, α = .1
3.2. Quiz Problem.
(1) x¯ = 43.1, s = 6.8, n = 26, α = .01
(2) x¯ = −10, s = 1.34, n = 9, α = .1
(3) x¯ = 3600, s = 94, n = 40, α = .05
3.3. Quiz Problem. A simple random sample of 21 chihuahua dog weights yields a sample mean of 5.6 pounds. It is known that the standard deviation of the population of all
chihuahua weights is 1.8 pounds. Find a confidence interval for the true mean weight of all
chihuahuas with a 95% confidence level.
3.4. Quiz Problem. Fifteen regions in an Amazon rain forest are randomly sampled. The
total yearly rainfalls, in centimeters, for these regions were reported as follows:
{181, 232, 230, 158, 242, 197, 228, 302, 242, 194, 200, 186, 146, 194, 202}
Assume that yearly rainfalls within the Amazon rain forest are distributed normally. Find a
confidence interval for the true mean rainfall in the Amazon rain forest with a 95% confidence
level.
Solution: Since we are not given the population standard deviation, σ, we need to compute
the sample standard deviation and use the latter of our two confidence interval recipes. We
find that x¯ = 208.9 and s = 38.51. Since we can assume that yearly rainfalls within the
Amazon rain forest are distributed normally, we can apply the t confidence interval method
despite the small sample size. Also, df = n − 1 = 14 Therefore, a 95% confidence interval
for the true mean rainfall in the Congolese rain forest is given by,
38.51
38.51
208.9 ± t.025,14 √
= 208.9 ± 2.144
= [187.582, 230.218]
3.873
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Appendix A. Solutions Practice Problems
A.1. Solution to Practice Problem 2.1.
(1) 20 ± z.025 √336 = 20 ± 1.96 63 = [19.02, 20.98]
(2) 25 ± z.005 √336 = 25 ± 2.58 63 = [23.71, 26.29]
(3) 30 ± z.05 √325 = 30 ± 1.64 35 = [29.01, 30.98]
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A.2. Solution to Practice Problem 2.2.
3.76
(1) 35 ± t.05,24 √
= 35 ± 1.711 3.76
= [33.71, 36.29]
5
25
5.02
5.02
(2) 50 ± t.005,15 √16 = 50 ± 2.95 4 = [46.3, 53.7]
3.71
3.71
(3) 50 ± t.025,11 √
= 50 ± 2.2 √
= [47.64, 52.36]
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A.3. Solution to Practice Problem 2.3. Since we are given the population standard
deviation, σ = 18, we us the first of our two confidence interval recipes. Therefore, a 99%
confidence interval for the true mean weight of all St Bernard dogs is given by,
18
193 ± z.005 √ = [185.27, 200.72]
36
A.4. Solution to Practice Problem 2.4. Since we are not given the population standard
deviation, σ, we need to compute the sample standard deviation and use the latter of our two
confidence interval recipes. We find that x¯ = 237 and s = 44.68. Since we can assume that
yearly rainfalls within the Congolese rain forest are distributed normally, we can apply the t
confidence interval method despite the small sample size. Also, df = n − 1 = 10 Therefore,
a 95% confidence interval for the true mean rainfall in the Congolese rain forest is given by,
44.68
= [206.98, 267.04]
237 ± t.025,10 √
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Appendix B. Solutions to Quiz Problems
Solution to 3.1:
.25
= [0.887, 1.213]
(1) 1.05 ± z.025 √
= 1.05 ± 1.96 .25
3
9
120
120
(2) 1990 ± z.005 √36 = 1990 ± 2.58 6 = [1938.4, 2041.6]
2.67
(3) 31.3 ± z.05 √
= 31.3 ± 1.64 2.67
= [30.21, 32.395]
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Solution to 3.2:
6.8
(1) 43.1 ± t.025,25 √6.8
= 43.1 ± 2.79 5.099
= [39.38, 46.82]
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1.34
1.34
(2) −10 ± t.005,8 √9 = −10 ± 1.86 3 = [−10.83, −9.17]
94
(3) 3600 ± t.05,39 √9440 = 3600 ± 2.02 6.325
= [3570, 3630]
Solution to 3.3: Since we are given the population standard deviation, σ = 1.8, we us the
first of our two confidence interval recipes. Therefore, a 95% confidence interval for the true
mean weight of all chihuahuas is given by,
1.8
1.8
5.6 ± z.025 √ = 5.6 ± 1.96
= [4.83, 6.38]
4.58
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Solution to 3.4: Since we are not given the population standard deviation, σ, we need to
compute the sample standard deviation and use the latter of our two confidence interval
recipes. We find that x¯ = 208.9 and s = 38.51. Since we can assume that yearly rainfalls
within the Amazon rain forest are distributed normally, we can apply the t confidence interval
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MTH 3210 SPRING 2015
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method despite the small sample size. Also, df = n − 1 = 14 Therefore, a 95% confidence
interval for the true mean rainfall in the Congolese rain forest is given by,
38.51
38.51
208.9 ± t.025,14 √
= [187.582, 230.218]
= 208.9 ± 2.144
3.873
15