Solution for Problem Set 8

HANU - Faculty of Information and Technology

MAT204: Probability & Statistics
May 8, 2015

Problem 1: [1, Exercise 9.2]

An electrical firm manufactures light bulbs that have a length of life that is approxi-
mately normally distributed with a standard deviation of 40 hours. If a sample of 30
bulbs has an average life of 780 hours, find a 96% confidence interval for the population
mean of all bulbs produced by this firm.
Solution:
n = 30, x = 780, σ = 40
1 − α = 96% = 0.96 ⇒ α = 0.04 ⇒ zα/2 = z0.02 = 2.05
Therefore, a 96% confidence interval for the population mean can be calculated as:
σ σ
x − zα/2 √ < µ < x + zα/2 √
n n
40 40
⇒ 780 − (2.05) √ < µ < 780 + (2.05) √
30 30
⇒ 765 < µ < 795

Problem 2: [1, Exercise 9.5]

A random sample of 100 automobile owners in the state of Virginia shows that an au-
tomobile is driven on 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 auto-
mobile 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:
n = 100, x = 23, 500, σ = 3900
1 − α = 99% = 0.99 ⇒ α = 0.01 ⇒ zα/2 = z0.005 = 2.575
(a) A 99% confidence interval for the population mean is:
σ σ
x − zα/2 √ < µ < x + zα/2 √
n n

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 3900 3900
⇒ 23, 500 − (2.575) < µ < 23, 500 + (2.575)
10 10
⇒ 22, 496 < µ < 24, 504
(b) We can be 99% confident that the error will not exceed a specified amount e when
the sample size is:  z σ 2
α/2
n=
e
Therefore:
σ 3900
e < zα/2 √ = (2.575) = 1004
n 10
In another way, the possible size of our error must be smaller than 1004.

Problem 3: [1, Exercise 9.8]

An efficiency expert wishes to determine the average time that it takes to drill three
holes in a certain metal clamp. How large a sample will she need to be 95% confident
that her sample mean will be within 15 seconds of the true mean? Assume that it is
known from previous studies that σ = 40 seconds.
Solution:
e = 15, σ = 40
1 − α = 95% = 0.95 ⇒ α = 0.05 ⇒ zα/2 = z0.025 = 1.96
 z σ 2
α/2
n=
e
 2
(1.96)(40)
⇒n= = 28
15

Problem 4: [1, Exercise 9.12]

A random sample of 10 chocolate energy bars of a certain brand has, on average, 230
calories per bar, with a standard deviation of 15 calories. Construct a 99% confidence
interval for the true mean calorie content of this brand of energy bar. Assume that the
distribution of the calorie content is approximately normal.
Solution:
n = 10, x = 230, s = 15
υ = n − 1 = 10 − 1 = 9
1 − α = 99% = 0.99 ⇒ α = 0.01 ⇒ tα/2 = t0.005 = 3.25 with υ = 9 (degrees of f reedom)
A 99% confidence interval for the population mean is:
s s
x − tα/2 √ < µ < x + tα/2 √
n n
15 15
⇒ 230 − (3.25) √ < µ < 230 + (3.25) √
10 10
⇒ 214.58 < µ < 245.42

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Problem 5: [1, Exercise 9.19]
A random sample of 25 tablets of buffered aspirin contains, on average, 325.05 mg of
aspirin per tablet, with a standard deviation of 0.5 mg. Find the 95% tolerance limits
that will contain 90% of the tablet contents for this brand of buffered aspirin. Assume
that the aspirin content is normally distributed.
Solution:
n = 25, x = 325.05
s = 0.5, γ = 5%
1 − α = 90% & n = 25 ⇒ k = 2.208
(Look up the T olerance F actors f or N ormal Distribution table)
So:
x ± ks ⇒ 325.05 ± (2.208)(0.5) ⇒ (323.946, 326.151)
Therefore, we are 95% confident that the above tolerance interval will contain 90% of
the aspirin contents for this brand of buffered aspirin.

Problem 6: [1, Exercise 9.35]

A random sample of size n1 = 25, taken from a normal population with a standard
deviation σ1 = 5, has a mean x1 = 80. A second random sample of size n2 = 36, taken
from a different normal population with a standard deviation σ2 = 3, has a mean x2 =
75. Find a 94% confidence interval for µ1 − µ2 .
Solution:
n1 = 25, n2 = 36
x1 = 80, x2 = 75
σ1 = 5, σ2 = 3
α = 0.06 ⇒ zα/2 = z0.03 = 1.88
Therefore, a 94% confidence interval for µ1 − µ2 is:
s s
2 2
σ1 σ2 σ12 σ22
(x1 − x2 ) − zα/2 + < µ1 − µ2 < (x1 − x2 ) + zα/2 +
n1 n2 n1 n2
r r
25 9 25 9
⇒ (80 − 75) − (1.88) + < µ1 − µ2 < (80 − 75) + (1.88) +
25 36 25 36
⇒ 2.9 < µ1 − µ2 < 7.1

Problem 7: [1, Exercise 9.36]

Two kinds of thread are being compared for strength. Fifty pieces of each type of thread
are tested under similar conditions. Brand A has an average tensile strength of 78.3
kilograms with a standard deviation of 5.6 kilograms, while brand B has an average
tensile strength of 87.2 kilograms with a standard deviation of 6.3 kilograms. Construct
a 95% confidence interval for the difference of the population means.
Solution:
n1 = nA = 50, n2 = nB = 50

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 x1 = xA = 78.3, x2 = xB = 87.2
σ1 = σA = 5.6, σ2 = σB = 6.3
1 − α = 95% = 0.95 ⇒ α = 0.05 ⇒ zα/2 = z0.025 = 1.96
Therefore, a 95% confidence interval for the difference of the population means is:
s s
σ12 σ22 σ12 σ22
(x1 − x2 ) − zα/2 + < µ1 − µ2 < (x1 − x2 ) + zα/2 +
n1 n2 n1 n2
r
5.62 6.32
(87.2 − 78.3) ± 1.96 + = 8.9 ± 2.34
50 50
⇒ 6.56 < µ1 − µ2 < 11.24

Problem 8: [1, Exercise 9.43]

A taxi company is trying to decide whether to purchase brand A or brand B tires for its
fleet of taxis. To estimate the difference in the two brands, an experiment is conducted
using 12 of each brand. The tires are run until they wear out. The results are:
Brand A: x1 = 36,300 kilometers, s1 = 5000 kilometers.
Brand B: x2 = 38,100 kilometers, s2 = 6100 kilometers.
Compute a 95% confidence interval for µA − µB assuming the populations to be ap-
proximately normally distributed. You may not assume that the variances are equal.
Solution:
n1 = n2 = nA = nB = 12
x1 = xA = 36, 300, x2 = xB = 38, 100
s1 = sA = 5, 000, s2 = sB = 6, 100
(s21 /n1 + s22 /n2 )2
υ=
[(s21 /n1 )2 /(n1 − 1)] + [(s22 /n2 )2 /(n2 − 1)]
50002 /12 + 61002 /12
= = 21
[(50002 /12)2 )(12 − 1)] + [(61002 /12)2 )(12 − 1)]
α = 0.05 ⇒ tα/2 = t0.025 = 2.080 with υ = 21 (degrees of f reedom)
Therefore:
s s
s21 s22 s21 s2
(x1 − x2 ) − tα/2 + < µ1 − µ2 < (x1 − x2 ) + tα/2 + 2
n1 n2 n1 n2
r
50002 61002
(36, 300 − 38, 100) ± (2.080) + = −1, 800 ± 4, 736
12 12
⇒ −6, 536 < µ1 − µ2 < 2, 936

Problem 9: [1, Exercise 9.48]

An automotive company is considering two types of batteries for its automobile. Sam-
ple information on battery life is collected for 20 batteries of type A and 20 batteries

4
of type B. The summary statistics are xA = 32.91, xB = 30.47, sA = 1.57, and sB =
1.74. Assume the data on each battery are normally distributed and assume σA = σB .
(a) Find a 95% confidence interval on µA − µB .
(b) Draw a conclusion from (a) that provides insight into whether A or B should be
adopted.
Solution:
n1 = n2 = nA = nB = 20
x1 = xA = 32.91, x2 = xB = 30.47
s1 = sA = 1.57, s2 = sB = 1.74
s
(n1 − 1)s21 + (n2 − 2)s22
sp = = 1.657
n1 + n2 − 2
(s21 /n1 + s22 /n2 )2
υ= = 38
[(s21 /n1 )2 /(n1 − 1)] + [(s22 /n2 )2 /(n2 − 1)]
(a)
α = 0.05 ⇒ tα/2 = t0.025 ≈ 2.042 with υ = 38 (degrees of f reedom)
Therefore:
r r
1 1 1 1
(x1 − x2 ) − tα/2 sp + < µ1 − µ2 < (x1 − x2 ) + tα/2 sp +
n1 n2 n1 n2
r
1 1
(32.91 − 30.47) ± (2.042)(1.657) + = 2.44 ± 1.07
20 20
⇒ 1.37 < µA − µB < 3.51
(b) Since it is apparent that type A battery has longer life, it should be adopted.

Problem 10: [1, Exercise 9.50]

Two levels (low and high) of insulin doses are given to two groups of diabetic rats to
check the insulin-binding capacity, yielding the following data:
Low dose: n1 = 8, x1 = 1.98, s1 = 0.51
High dose: n2 = 13, x2 = 1.30, s2 = 0.35
Assume that the variances are equal. Give a 95% confidence interval for the difference
in the true average insulin-binding capacity between the two samples.
Solution:
n1 = 8, n2 = 13
x1 = 1.98, x2 = 1.30
s1 = 0.51, s2 = 0.35
s
(n1 − 1)s21 + (n2 − 2)s22
sp = = 0.416
n1 + n2 − 2
(s21 /n1 + s22 /n2 )2
υ= = 19
[(s21 /n1 )2 /(n1 − 1)] + [(s22 /n2 )2 /(n2 − 1)]
α = 0.05 ⇒ tα/2 = t0.025 = 2.093 with υ = 19 (degrees of f reedom)

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Therefore:
r r
1 1 1 1
(x1 − x2 ) − tα/2 sp + < µ1 − µ2 < (x1 − x2 ) + tα/2 sp +
n1 n2 n1 n2
r
1 1
(1.98 − 1.30) ± (2.093)(0.416) + = 0.68 ± 0.39
8 13
⇒ 0.29 < µ1 − µ2 < 1.07

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References
[1] Walpole, R. E., Myers, R. H., Myers, S. L. and Ye, K., Probability &
Statistics for Engineers & Scientists, 9th ed., MA, USA: Prentice-Hall, 2012.

[2] DeGroot, M. H. and Schervish, M. J., Probability and Statistics, 4th ed., MA,
USA: Pearson Education, Inc., 2012.

[3] Murray, R. S., John, J. S. and R, A. Srinivasan, Probability and Statistics,

3rd ed., USA: McGraw-Hill, 2009.