EF 507
PS-Chapter 8
FALL 2008
1. The amount of material used in making a custom sail for a sailboat is normally
distributed. For a random sample of 15 sails, you find that the mean amount material
is 912 square feet, with a standard deviation of 64 square feet. Which of the following
represents a 99% confidence interval for the population mean amount of material used
in a custom sail?
A)
B)
C)
D)
912  49.2
912  42.6
912  44.3
912  46.8
ANSWER: A
2. Let X 1 , X 2 , X 3 , and X 4 be a random sample of observations from a population
with mean  and variance  2 . Consider the following estimator of  : ˆ1 = 0.15 X 1 +
0.35 X 2 + 0.20 X 3 + 0.30 X 4 . What is the variance of ˆ1 ?
A)
0.275
B)
0.275 2
C)
0.55  2
D)
0.55
ANSWER: C
3. Which of the following statements is false?
A)
An estimator ˆ that is a function of the sample data X1 , X 2 ,....., X n is called
an unbiased estimator of the population parameter ˆ ; that is , E ˆ   
The statistic X is an unbiased estimator of the population mean 
The sample statistic s is an unbiased estimator of the population parameter
 when (n – 1) is used as a divisor in the calculation of the sample standard
deviation
D)
None of the above
ANSWER:
C
B)
C)
4. In developing an interval estimate for a population mean, the population standard
deviation  was assumed to be 8. The interval estimate was 45.82  2.36. Had 
equaled 16, the interval estimate would be
A)
45.82  4.72
B)
55.82  2.36
C)
45.85  9.44
D)
91.64  4.72
ANSWER: A
5. The lower limit of a 95% confidence interval for the population proportion P given
a sample size n = 100 and sample proportion P̂ = 0.62 is equal to
A)
B)
C)
D)
0.715
0.699
0.525
0.540
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ANSWER:
C
6. In order to estimate the average daily down time, a manufacturer randomly samples
40 days of production records and found a mean of 51.576 minutes and standard
deviation of 75.9 minutes. An 80% confidence interval is given by:
A)
51.675  1.28(12.0)
B)
51.675  1.28(75.9)
C)
51.675  1.304(12.0)
D)
51.675  1.304(8.7)
ANSWER: C
7. The Daytona Beach Tourism Commission is interested in the average amount of
money a typical college student spends per day during spring break. They survey 35
students and find that the mean spending is $63.57 with a standard deviation of
$17.32.
a) Develop a 95% confidence interval for the population mean daily spending.
ANSWER:
X  tn1 , / 2 s / n = 63.57  2.032(17.32) / 5.92 = 63.57 
35 = 63.57  5.95. Then,
UCL = 69.52 and LCL = 57.62.
b) Interpret the confidence level in 7 (a).
ANSWER: If independent random samples of size 35 are repeatedly selected from
the population and 95% confidence intervals for each of these samples are
determined, then over a very large number of repeated trials, 955 of these intervals
will contain the value of the true average amount of money a typical college student
spends per day during spring break.
8. What level of confidence is associated with an interval of $58.62 to $68.52 for the
population mean daily spending.
ANSWER:
 t34,  / 2 
17.32
35
tn1 , / 2 s / n =
–
(68.52
58.62)
/
2
=
4.95
 4.95  t34,  / 2  1.69 .
Hence,  / 2 = 0.05, or  = 0.10. Then the confidence level is 90%.
9. In a recent survey of 600 adults, 16.4% indicated that they had fallen asleep in front
of the television in the past month. Develop a 95% confidence interval for the
population proportion.
ANSWER:
pˆ  z / 2 pˆ 1  pˆ  / n = 0.164  1.96(0.0151) = 0.164  0.0296.
Hence, UCL = 0.1936 and LCL = 0.1344.
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QUESTIONS 10 AND 11 ARE BASED ON THE FOLLOWING
INFORMATION:
A researcher is interested in determining the percentage of all households in the U.S.
that have more than one home computer. In a survey of 492 households, 27%
indicated that they own more than one home computer.
10.
Develop a 90% confidence interval for the proportion of all households in the
U.S. with more than one computer.
ANSWER:
n = 492, p̂ =0.27, z / 2  z0.05 =1.645
pˆ  z / 2 pˆ 1  pˆ  / n  0.27  1.645
 0.27  0.73 / 492  0.27  0.033 .
Hence, UCL= 0.303 and LCL = 0.237
11. The researcher reports a confidence interval of 0.2312 to 0.3096 but neglects to
tell you the confidence level. What is the confidence level associated with this
interval?
ANSWER:
Width= z / 2 pˆ 1  pˆ  / n =0.3096-0.2312 = 0.0784
Then, 2 z / 2
 0.27  0.76  / 492 = 0.0784
0.04 z / 2 = 0.0784  z / 2 =1.96.
confidence level is 95%.
Hence,  / 2 = 0.025, or  = 0.05.
Then, the
QUESTIONS 12 AND 13 ARE BASED ON THE FOLLOWING
INFORMATION:
Suppose that the amount of time teenagers spend on the internet is normally
distributed with a standard deviation of 1.5 hours. A sample of 100 teenagers is
selected at random, and the sample mean computed as 6.5 hours.
12. Determine the 95% confidence interval estimate of the population mean.
ANSWER: x  z / 2   / n  6.5  (1.96)(1.5/ 100)  6.5  0.294  6.206 <  < 6.794
13. Interpret what the interval estimate in Question 102 tells you.
ANSWER: If we repeatedly draw independent random samples of size 100 from the
population of teenagers, and confidence interval for each of these samples are
determined, then over a very large number of repeated trials, 95% of these confidence
intervals will contain the value of the true population mean amount of time teenagers
spend on the internet.
QUESTIONS 14 THROUGH 17 ARE BASED ON THE FOLLOWING
INFORMATION:
A furniture mover calculates the actual weight as a proportion of estimated weight for
a sample of 31 recent jobs. The sample mean is 1.13 and the sample standard
deviation is 0.16.
14. Calculate a 95% confidence interval for the population mean using t tables.
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ANSWER: Given: n = 31, x  1.13, and s = 0.16. To calculate a 95% confidence
interval, we need the t table value for df = 30. With   .05, this value is t.025,30  2.042.
The confidence interval is x  t / 2  s / n  1.13  (2.042)(0.16 / 31)  1.13  .059 or
1.071    1.189.
15. Assume that the population standard deviation is known to be 0.16. Calculate a
95% confidence interval for the population mean using the z- table.
ANSWER: If  is known to be 0.16, we may use the z table value z.025  1.96 (rather
than
the
t
table
value)
in
the
95%
confidence
interval.
x  z / 2   / n  1.13  (1.96)(0.16 / 31)  1.13  .056 or 1.074    1.186.
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