Estimating Mean Stand. Dev. Known Worksheet Answers
1. Suppose 𝑥 has a mound – shaped distribution with 𝜎 = 9. A random sample size of
36 has a sample mean of 20.
a) Is it appropriate to use a normal distribution to compute a confidence interval for
the population mean 𝜇 and why or why not.
Yes, we have a random sample, 𝜎 is known, and sample size ≥ 30
b) Find a 75% confidence interval for 𝜇
𝑧0.75 = 1.15 ( 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒 )
𝐸 = 𝑧𝑐 ∙
𝜎
√𝑛
= 1.15 ∙
9
√36
= (1.15)(1.5) = 1.73
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
20 − 1.73 < 𝜇 < 20 + 1.73
18.27 < 𝜇 < 21.73
c) Find a 90% confidence interval for 𝜇
𝑧0.90 = 1.645 ( 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒 )
𝐸 = 𝑧𝑐 ∙
𝜎
√𝑛
= 1.645 ∙
9
√36
= (1.645)(1.5) = 2.47
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
20 − 2.47 < 𝜇 < 20 + 2.47
17.53 < 𝜇 < 22.47
d) Find a 95% confidence interval for 𝜇
𝑧0.95 = 1.96 ( 𝑓𝑟𝑜𝑚 𝑡𝑎𝑏𝑙𝑒 )
𝐸 = 𝑧𝑐 ∙
𝜎
√𝑛
= 1.96 ∙
9
√36
= (1.96)(1.5) = 2.94
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
20 − 2.94 < 𝜇 < 20 + 2.94
17.06 < 𝜇 < 22.94
2. Suppose 𝑥 has a mound – shaped distribution with 𝜎 = 3
a) Find the minimal sample size required so that for a 95% confidence interval, the
maximal margin of error is 𝐸 = 0.4
𝑧𝑐 ∙ 𝜎 2
𝑛=(
)
𝐸
(1.96)(3) 2
𝑛=(
0.4
𝑛 = 217
5.88 2
) = ( 0.4 ) = 14.72 = 216.09
3. A random sample of size 36 is drawn from an 𝑥 distribution. The sample mean is
100.
a) Suppose the 𝑥 distribution has 𝜎 = 30. Compute a 90% confidence interval for 𝜇.
What is the value of the margin of error ?
𝑧0.90 = 1.645
𝐸 = 𝑧𝑐 ∙
𝜎
√𝑛
= (1.645) ∙
30
√36
= (1.645)(5) = 8.225
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
100 − 8.225 < 𝜇 < 100 + 8.225
91.775 < 𝜇 < 108.225
b) Suppose the 𝑥 distribution has 𝜎 = 20. Compute a 90% confidence interval for 𝜇.
What is the value of the margin of error ?
𝜎
20
𝑧0.90 = 1.645
𝐸 = 𝑧𝑐 ∙ = (1.645) ∙
= (1.645)(3.33) = 5.48
√𝑛
√36
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
100 − 5.48 < 𝜇 < 100 + 5.48
94.52 < 𝜇 < 105.48
c) Suppose the 𝑥 distribution has 𝜎 = 10. Compute a 90% confidence interval for 𝜇.
What is the value of the margin of error ?
𝜎
10
𝑧0.90 = 1.645
𝐸 = 𝑧𝑐 ∙ = (1.645) ∙
= (1.645)(1.67) = 2.75
√𝑛
√36
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
100 − 2.75 < 𝜇 < 100 + 2.75
97.25 < 𝜇 < 102.75
4. A random sample is drawn from a population with 𝜎 = 12. The sample mean is 30.
a) Compute a 95% confidence interval for 𝜇 based on a sample size 49. What is the
value of the margin of error ?
𝜎
12
𝑧0.95 = 1.96
𝐸 = 𝑧𝑐 ∙ = (1.96) ∙
= (1.96)(1.71) = 3.35
√𝑛
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
30 − 3.35 < 𝜇 < 30 + 3.35
26.65 < 𝜇 < 33.35
Margin of error = 𝐸 = 3.35
√49
#4 continued
b) Compute a 95% confidence interval for 𝜇 based on a sample size 100. What is the
value of the margin of error ?
𝜎
12
𝑧0.95 = 1.96
𝐸 = 𝑧𝑐 ∙ = (1.96) ∙
= (1.96)(1.2) = 2.35
√𝑛
√100
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
30 − 2.35 < 𝜇 < 30 + 2.35
27.65 < 𝜇 < 32.35
Margin of error = 𝐸 = 2.35
c) Compute a 95% confidence interval for 𝜇 based on a sample size 225. What is the
value of the margin of error ?
𝜎
12
𝑧0.95 = 1.96
𝐸 = 𝑧𝑐 ∙ = (1.96) ∙
= (1.96)(0.80) = 1.57
√𝑛
√225
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
30 − 1.57 < 𝜇 < 30 + 1.57
28.43 < 𝜇 < 31.57
Margin of error = 𝐸 = 1.57
5. Overproduction of uric acid in the body can be an indication of cell breakdown. This
may be an advance indication of illness such as gout, leukemia, or lymphoma. Over a
period of time, an adult has taken eight blood tests for uric acid. The mean
concentration was 𝑥̅ = 5.35 𝑚𝑑/𝑑𝑙. The distribution of uric acid in healthy adult
males can be assumed to be normal, with 𝜎 = 1.85 𝑚𝑙/𝑑𝑙.
a) Find a 95% confidence interval for the population mean concentration of uric
acid in this patient’s blood. What is the margin of error ?
𝜎
1.85
b) 𝑧0.95 = 1.96
𝐸 = 𝑧𝑐 ∙ = (1.96) ∙
= (1.96)(0.65) = 1.27
√𝑛
√8
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
5.35 − 1.27 < 𝜇 < 5.35 + 1.27
4.08 < 𝜇 < 6.62
Margin of error = 𝐸 = 1.27
c) Find the sample size necessary for a 95% confidence level with maximal margin
of error 𝐸 = 1.10 for the mean concentration of uric acid in this patient’s blood.
𝑧𝑐 ∙ 𝜎 2
(1.96)(1.85) 2
3.626 2
𝑛=(
) =(
) =(
) = 3.32 = 10.89
𝐸
1.10
1.1
𝑛 = 11 patients
6. A small group of 15 Allen’s hummingbirds has been under study in Arizona. The
average weight for these birds is 𝑥̅ = 3.15 grams. Based on previous studies, we can
assume that the weights of Allen’s hummingbirds have a normal distribution, with
𝜎 = 0.33 grams.
a) Find an 80% confidence interval for the average weights of Allen’s
hummingbirds in the study region. What is the margin of error ?
𝜎
0.33
𝑧0.80 = 1.28
𝐸 = 𝑧𝑐 ∙ = (1.28) ∙
= (1.28)(0.09) = 0.12
√𝑛
√15
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
3.15 − 0.12 < 𝜇 < 3.15 + 0.15
3.03 < 𝜇 < 3.27
Margin of error = 𝐸 = 0.12
b) Find the sample size necessary for an 80% confidence level with maximal margin
of error 𝐸 = 0.08 for the mean weights of the hummingbirds.
𝑧𝑐 ∙ 𝜎 2
(1.28)(0.33) 2
0.422 2
𝑛=(
) =(
) =(
) = 5.2752 = 27.82
𝐸
0.08
0.08
𝑛 = 28 hummingbird
7. The following data shows annual profits per employee ( in thousands of dollars)
for representative companies in financial services. Companies such as Wells Fargo,
First Bank Systems, and Key Banks were included.
42.9
36.9
28.8
43.8
43.8
27.0
29.3
36.9
48.2
47.1
31.5
31.9
60.6
33.8
31.7
25.5
54.9
28.1
31.1
23.2
55.1
28.5
38.0
29.8
52.9
29.1
32.0
22.3
54.9
36.5
31.7
26.5
42.5
36.1
32.9
26.7
a) Find the mean 𝑥̅ and 𝜎 standard deviation for the data set
** I used a TI-83 calculator to find the statistics below :
𝑥̅ = 36
𝜎 = 10.1
𝑛 = 42
33.0
26.9
23.1
33.6
27.8
54.9
#7 continued
b) Let us say that the data are representative of the entire sector of successful
financial services corporations. Find a 75% confidence interval for 𝜇, the average
annual profit per employee for all successful banks.
𝑧0.75 = 1.15
𝐸 = 𝑧𝑐 ∙
𝜎
√𝑛
𝑥̅ − 𝐸 < 𝜇 < 𝑥̅ + 𝐸
36 − 1.79 < 𝜇 < 36 + 1.79
34.21 < 𝜇 < 37.79
= (1.15) ∙
10.1
√42
= (1.15)(1.558) = 1.79