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Math 201-337-77
Statistics
Test III (version a)
7XHVGD\ 1RYHPEHU (d59)
NAME: ___________________________________
20
For each of the following, make sure you carefully describe your solution by clearly identifying your data and
using the correct notation. You are required to state the general formula for the confidence interval before
you proceed. Keep at least 2 decimals in your calculations.
1.
2 points
According to the June 13, 1994, issue of Time magazine, the proportion of all workers who are union
members equals 15.8%. Find the probability that in a national survey of 2500 workers, at most 450 will be
union members.
x ~ > B( 2500,0.158)
σ ≈ 18.24
σ =1
n = 450 ; p = 0.158
np = 395 > 5 and nq
= 55 > 5
Therefore we use the normal
for p ≤ 450 =
( x )
µ = 395
x − µ
z =
σ
x = 450.5
µ = 0 z 450.5−395
460.5 = ≈3.04
18.24
µ = np = 395
σ = npq ≈18.24
⇒ 50% + 49.88% = 99.88%
The lifetimes of batteries produced by a firm are normally distributed with a mean of 110 hours and a
2.
standard deviation of 8 hours. What’s the probability that
6 points a) a battery will last less than 106 hours
µ = 110
σ = 8
σ = 8
x − µ
z =
σ
σ = 1
x = 106
µ = 110
µ = 0
z 106 110
106 = − =−0.5
8
⇒50% − 19.15% = 30.85%
b) a battery will last between 106 hours and 112 hours
µ = 110
σ = 8
σ = 8
x − µ
z =
σ
σ = 1
µ =110
x 1 = 106 x 2 = 112
z 106 110
106 = − =−0.5
8
µ = 0
z 112 110
112 = − = 0.25
8
⇒ 19.15% + 9.87 % = 29.
02%

c) The average of a sample of 36 batteries will last between 106 hours and 112 hours
µ x = 110
µ = 110
8
⇒
σ
8
x = ≈1.33
36
σ = 8 σ x = ≈1.33
x
36
− µ x
z =
σ x
σ = 1
x 1 = 106
µ x = 110 x 2 = 112
µ = 0
z 106 110
106 = − =−3.01
1.33
z 112 110
112 = − = 1.50
1.33
⇒ 49.87% + 43. 32% = 93.19%
3.
2 points
A random sample of 81 components is selected to determine the life expectancy of a certain type of
electronic part. The sample mean life expectancy was 495 hours with a standard deviation of 64 hours. At
the 95% level of confidence, estimate the true life expectancy of the components.
n = 81 x −Z σ ˆ ≤ µ ≤ x + Z σˆ
α
2 x
α
2
x = 495
s s
x −Z
≤ µ ≤ x + Z
=
α
α
s 64
2
n
2
n
64 64
495 −1.96⋅ ≤ µ ≤ 495 + 1.96⋅
95% C.I. ⇒ Zα
= 1.96 81 81
2
495 −13.94 ≤ µ ≤ 495 + 13.94
481.06 ≤ µ ≤508.94
x
4.
2 points
The Automated Automaton Assembly Plant has had a large number of employees quit their jobs soon
after they start employment. The personnel manager, Mr. Rowe Botts, would like to estimate the average
stay with the company of each employee who quits. A random sample of 15 former employees’ records
indicated that the average stay with the company was 54 days with a standard deviation of 16 days. If the
average stay is approximately normal, construct a 99% confidence interval of the true population mean µ.
n = 15 x −t σˆ
≤ µ ≤ x + t σˆ
α
2 x
α
2
x = 54
s s
x −t
≤ µ ≤ x + t
=
α
α
s 16
2
n
2
n
16 16
54 −2.977⋅ ≤ µ ≤ 54 + 2.977⋅
99% C.I. ⇒ tα
= 2.977 15 15
2
54 −12.30≤ µ ≤ 54+
12.30
41.70 ≤ µ ≤ 66.30
x
5.
2 points
The Badd-Bolt Manufacturing Company selected a random sample of 300 bolts from production to
estimate the percentage π of defectives. There were 30 defective bolts in the sample. What is the 90%
confidence interval estimate of the percentage π of defective bolts for the whole production
n = 300 p−Z σˆ
≤π ≤ p+
Z σˆ
α
2
α p
α
2 2
x = 30
10⋅
90 10⋅
90
30
10% −1.645⋅ ≤π
≤ 10% + 1.645⋅
p = = 10%
300
300 300
10% −2.85% ≤π
≤ 10% + 2.85%
90% C.I. ⇒ Z = 1.645 7.15% ≤ π ≤12. 8% 5
p

6.
2 points
The Badd-Bolt Manufacturing Company selected a second random sample of 61 bolts in order to
determine if a new process would improve the tolerance specification for the variation of bolt size. The
current process has a standard deviation σ = 0.09 mm. If the 61 bolts produced with the new process had a
variance of 0.0064 mm, construct a 90% confidence interval for the true production standard deviation σ
of the new process (assuming the population is normally distributed). Is the new process better Explain.
5%
2
χ L = 43.188
2
χ R = 79.082
5%
n = 61 2 2
( n−1) s 2 ( n−1)
s
σ
2 new
current = 0.09
≤σ
≤
2
χR
χL
2
Snew
= 0.0064 60⋅0.0064 2 60⋅0.0064
≤σ
n ew ≤
S = 0.08 79.082 43.188
new
90% C.I.
new
2
new
0.004855719 ≤σ
≤0.008891358
0.004855719 ≤σ
≤ 0.008891358
new
0.0697 ≤σ
≤0.
0943
Therefore we cannot guarantee that the new process is better since the confidence interval
includes the current standard deviation.
7.
You are asked to conduct a study to determine the percentage of defective hammers produced by your
plant. The study must be conducted so that there’s 95% probability that the results are within 3%.
a)
E = 3% 2
Zα
p 100 − p
2
95% C.I. ⇒ Z 1.96 n
α =
=
2
2
E
2
=
p( 100 − p)
2
E = Z
3
α
2 n
= 1067.11
2
Zα
p( 100
− p)
2
⇒ n =
2
E
Sample Size = 1068
( )
( 1.96) 50( 50)
b)
E = 3% 2
Zα
p 100 − p
2
95% C.I. ⇒ Z 1.96 n
α =
=
2
2
E
2
=
p( 100 − p)
2
E = Z
3
α
2 n
= 349.5856
2
Zα
p( 100
− p)
2
⇒ n =
2
E
Sample Size = 30 5
( )
( 1.96) 9( 91)
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