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Exam 1 Outline
1) A ___________ is a measure that describes characteristics of a sample.
2) A ___________ is a measure that describes characteristics of a population.
3) The general practice of estimation theory consists of using ______________________
to estimate ____________________________
4) The general requirement for a sample statistic to be an unbiased point estimator of its
corresponding population parameter is that __________________________________
5) The sample mean x is an unbiased point estimator of __________________________
because __________________________
6) The sample median is a biased point estimator of the population median because ______
7) If a sufficiently large number of independent simple random samples of equal size 16
are selected from the same population of numerical scores having mean 45 and standard
deviation 6, (using the same sampling procedure), then the sampling distribution of the
sample mean x will have mean  x = _______ and standard deviation   ________.
x
If the population proportion P is known to be 0.34, then the sampling distribution of the
sample proportion p̂ will have mean  p̂ = _____ and standard deviation  p̂ =________
8) If a sufficiently large number of independent simple random samples of equal size 9 are
selected from the same normally distributed population of numerical scores having
mean 36 and standard deviation 4, (using the same sampling procedure), then the shape
of the sampling distribution of the sample mean x will be _______________________
and will have mean  x = _______ and standard deviation   ________.
x
If the population proportion P is known to be 0.34, then the sampling distribution of the
sample proportion p̂ will have mean  p̂ = _____ and standard deviation  p̂ =________
9) A population of scores has mean 102 and standard deviation 12. If nothing is known
about the shape of the distribution of this population and a sufficiently large number of
independent simple random samples of equal size 24 are selected from it, (using the same
sampling procedure), what will be the shape of the sampling distribution of the sample
mean x ? What will be the value of the mean of the sampling distribution  x ? What
will be the value of the standard error of the mean  ?
x
10) State the large-sample conditions for the sampling distribution of a sample proportion p̂
to be approximately the normal distribution.
11) Suppose that the proportion of left-handed people in the whole world is near 0.10. If a
sufficiently large number of independent simple random samples of equal size 100 are
selected from the population of people in the whole world, (using the same sampling
procedure), what will be the shape of the sampling distribution of p̂ ? What will be the
value of the mean of the sampling distribution  p̂ ? What will be the value of the
standard error of the proportion  p̂ ?
12) Suppose that if we collect all samples of size 3 (with replacement) from a population of
numerical scores, the resulting probability distribution of x is as follows:
1.0 1.5
x
1
P( x ) 641
16
2.0
2.5
3.0
3.5
4.0
3
16
15
32
3
16
1
16
1
64
a) Find the value of the population mean.
b) Find the value of the population variance.
13) Suppose that if we collect all samples of size 2 (with replacement) from a population of
numerical scores, the resulting probability distribution of the sample proportion of odd
numbers p̂ is as follows:
1
p̂
0
1 a) Find the value of the population proportion of odd numbers
2
P( p̂ )
4
25
12
25
9
25
b) Find the standard error of p̂
c) Compute the value P(1  P) compare to the result in part b)
n
14) A manufacturer of light bulbs wants to estimate the mean life of a large shipment of light
bulbs to be delivered out of her factory very soon. She takes a random sample of 50 light
bulbs and finds the mean life to be 1600 hours with standard deviation 12 hours.
A) Give the point estimate value of the mean life of all light bulbs in the shipment.
B) Describe the shape of the sampling distribution of the sample mean life of the light
bulbs. Explain your reasoning with excruciating details.
C) If the manufacturer selects a random sample of 24 light bulbs (instead of 50), describe
the shape of the sampling distribution of the sample mean life of the light bulbs.
Explain your reasoning with excruciating details.
15) A state insurance auditor needs to determine the proportion of claims that are paid by a
health insurance company within two months of receiving the claim. She takes a random
sample of 400 of such claims and finds that 90 were paid out within two months of
receiving the claim.
A) Give the point estimate value of the proportion of all claims paid out by this health
insurance company within two months of receiving the claim.
B) Describe the shape of the sampling distribution of the sample proportion of all claims
paid out by this health insurance company within two months of receiving the claim.
Explain your reasoning with excruciating details.
C) If the auditor finds that only 14 (instead of 90) were paid out within two months of
receiving the claim, describe the shape of the sampling distribution of the sample
proportion of all claims paid out by this health insurance company within two months
of receiving the claim. Explain your reasoning with excruciating details.
16) A) Find the correct value of  if z = 2.09
2
B) Find the critical value z for a 91% confidence interval
2
C) Find the critical value z if z = z0.013
2
2
17) A radial tire manufacturer claims that the tires she produces last an average of 66,000
miles. A state inspector takes a random sample of 38 tires from this manufacturer and
finds the mean life to be 65,000 miles with standard deviation 2,800 miles.
A) The 93% confidence interval estimate of the mean life of all radial tires produced by
this manufacturer is (64,177.86 ; 65,822.14). Interpret this interval with excruciating
details. B) Construct a 98% confidence interval estimate of the mean life of all radial
tires produced by this manufacturer. C) At the 93% confidence level, is the
manufacturer’s claim correct? Explain with excruciating details. D) At the 98%
confidence level, is the claim correct? Explain with excruciating details. E) Describe the
shape of the sampling distribution of the sample mean life of the radial tires from this
manufacturer. Explain with excruciating details.
18) A list of ages at retirement of randomly selected Victoria Secret’s Angels is shown here:
26, 28, 26, 29, 27, 28. A) Construct a 95% confidence interval estimate of the population
mean age of retirement of Victoria Secret’s Angels. B) Under which condition(s) is the
theory used for this confidence applicable? Explain with excruciating details. C) Many
networks claim that the mean age of retirement of Victoria Secret’s Angels is 28. Use
the 95% CI to explain with excruciating details whether this claim is true.
19) In a random sample of 984 U.S. adult consumers, 55% of the participants responded that
they feel vulnerable to identity theft. A) Verify the large-sample condition(s). Explain
with excruciating details. B) Construct a 96% CI estimate of the true proportion of all
U.S. adult consumers who feel vulnerable to identity theft. C) Interpret the result with
excruciating details. D) Many networks claim that the true proportion of all U.S. adult
consumers who feel vulnerable to identity theft is at least 0.6. Use the 96% CI to explain
with excruciating details whether this claim is true.
20) Many stores are interested in the proportion of adult consumers who purchase clothing
online. Estimate the number of adult consumers that must be surveyed in order to be 99%
confident that the sample percentage error is no more than 0.025.
21) It is believed that the percentage of men who leave it for the very last minute to purchase
a Valentines gift for their sweethearts on Valentines day is near 75%. Estimate the
number of men that must be surveyed in order to be 90% confident that the sample
percentage error is no more than 0.045.
22) Suppose that the population variance of IQ scores for normal adults is 196. Estimate the
number of adults that must sampled to estimate the mean IQ score for normal adults
with a 99% CI and with a maximum error of 3 IQ points.
Answers:
1) statistic 2) parameter 3) sample statistics; population parameters 4) the mean of the sampling
distribution of the statistic is identically equal to the corresponding population parameter. In other
words, using the correct probability distribution of the sample statistic, the expected value of the
sample statistic is identically equal to the corresponding population parameter. 5) the population
mean  ; the mean of the sample distribution of x is identically equal to the population mean  .
In other words, E( x ) =  . 6) the mean of the sample distribution of the sample median is not
equal to the population median. In other words, E(sample median)  (population median). 7)  x = 45;
(0.34)(1  0.34)
 0.118 ; 8) also a normal distribution;  x = 36;
 x  6  1.5 ;  pˆ  0.34 ;  pˆ 
16
16
(0.34)(1  0.34)
 0.158 ; 9) since the shape of the distribution
 x  4  1.333 ;  pˆ  0.34 ;  pˆ 
9
9
of the original population is unknown and n = 24 < 30, then the Central Limit Theorem does not
guarantee that the sampling distribution of x will be normally distributed (not even approximately).
So, we cannot determine the shape of the sampling distribution of x ;  x = 102 ;   12  2.449
x
24
10) n p̂  15 and n(1  p̂ )  15 ; 11) since the shape of the distribution of the original population is
unknown and n p̂ = 100(0.10) = 10 < 15, then the Central Limit Theorem does not guarantee that the
sampling distribution of p̂ will be normally distributed (not even approximately). So, we cannot
determine the shape of the sampling distribution of p̂ ;   0.10 ;   (0.10)(1  0.10)  0.030
pˆ
pˆ
100
  E( x ) =  xP( x )  1.0   1.5   2.0   2.5   3.0   3.5161   4.0641  = 2.391
2 3
2 1
2 1
2
b)  2   x 2 P( x )   2  1.0 2  641   1.52 161   2.0 2 163   2.52 15
32   3.0 16   3.5 16   4.0  64   2.391
x
1
64
12)a)
1
16
3
16
15
32
3
16
 x2  0.822 . Since  x2   , then  2  n x2  (3)(0.822)  2.466
n
  1259   1525  53
P  E( p̂ ) =  pˆ P( pˆ )  0254   12 12
25
2 9
3 2
b)  p̂ =  2   pˆ 2 P( pˆ )  P 2  0 2  254    12 2 12
25   1  25    5 
13)a)
p̂
 

3
25

3
5
 53 15   53 . The values are identically equal as expected.
2
n
14)a) 1600 hrs; b) the shape of the distribution of the original population is unknown. However, since
n = 50 > 30, then the Central Limit Theorem guarantees that the sampling distribution of x will be
approximately normally distributed (it will not be normally distributed, but it will be sufficiently
close to a normal distribution); c) since the shape of the distribution of the original population is
unknown and n = 24 < 30, then the Central Limit Theorem does not guarantee that the sampling
distribution of x will be normally distributed (not even approximately). So, we cannot determine
the shape of the sampling distribution of x ; 15)A) P  pˆ  x  90  0.225 ; B) examine the values
n 400
of n p̂ and n(1  p̂ ). That is: n p̂ = 400(0.225) = 90 and n(1  p̂ ) = 400(1  0.225) = 310. Since n p̂ > 15
and n(1  p̂ ) > 15, the Central Limit Theorem guarantees that the sampling distribution of p̂ will be
approximately normally distributed (it will not be normally distributed, but it will be sufficiently
c)  p̂ = P(1  P) 
3 1 3
5 5
close to a normal distribution); C) In this case, pˆ  x  14  0.035, so n p̂ = 14. Since the shape of
n
400
the distribution of the original population is unknown and n p̂ = 14 < 15, then the Central Limit
Theorem does not guarantee that the sampling distribution of p̂ will be normally distributed (not
even approximately). So, we cannot determine the shape of the sampling distribution of p̂
16)A) Since P(0 < z < 2.09) = 0.4817, then  = 0.5 – 0.4817 = 0.0183. Thus,  = 2(0.0183) = 0.0732
2
B) Notice that
to
0.91 = 0.455
2
and the closest area for P(0 < z < z ) = 0.455 is 0.4554. This corresponds
2
z 2 = 1.70 ; C) z 2 = z0.013 means that 2 = 0.013. So, P(z > z 2 ) = 0.013 and P(0 < z < z 2 ) = 0.487 by
subtracting 0.5 – 0.013 = 0.487. The area closest to 0.487 is 0.4871. Therefore,
z 2 = 2.23
17)A) We are 93% confident that the interval (64,177.86 ; 65,822.14) does contain the true mean-life
of all radial tires produced by this manufacturer. So, if we took all random samples of 38 tires (or
a sufficiently large number of random samples of size 38, all from this same manufacturer and all
using the same sampling method) and for each sample we constructed the corresponding
confidence interval, then 93% of the resulting intervals will contain the true mean -life of all radial
( 2.33)( 2800)
tires produced by this manufacturer. B) 65000 
= 65000  1058.33 = (63941.67; 66058.33)
38
C) Since the claimed 66,000 miles average life is not in the 93% CI, (64,177.86 ; 65,822.14), then at
the 93% confidence level, the sample data at hand did not provide sufficient evidence to affirm
that the manufacturer’s claim is correct; D) Since the claimed 66,000 miles average life is in the
98% CI, (63941.67; 66058.33), then at the 98% confidence level, the sample data at hand did
provide sufficient evidence to affirm that the manufacturer’s claim is correct; E) the shape of the
distribution of the original population is unknown. However, since n = 38 > 30, then the Central Limit
Theorem guarantees that the sampling distribution of x will be approximately normally distributed
(it will not be normally distributed, but it will be sufficiently close to a normal distribution)
 x 2
164 2
2
x

4490 
x


164
n
6

18)A) We must compute x  n  6  27.333 and s 
 1.211
n 1
6 1
So, 95% CI = 27.333 
( 2.5706)(1.211)
6
= 27.333  1.271 = (26.062; 28.604) ; B) Since n = 6 < 30, it is
assumed that the population of retirement ages of Victoria Secret’s Angels is normally distributed
or not too deviated from a normal distribution shape. C) Since the claimed mean age 28 is in the
95% CI, (26.062; 28.604), then at the 95% confidence level, the sample data at hand did provide
sufficient evidence to affirm that the networks’ claims are correct.
19)A) examine the values of n p̂ and n(1  p̂ ): n p̂ = 984(0.55) = 541.2; n(1  p̂ ) = 984(0.45) = 442.8
Since n p̂ > 15 and n(1  p̂ ) > 15, the Central Limit Theorem guarantees that the sampling distribution
of p̂ will be approximately normally distributed (it will not be normally distributed, but it will be
sufficiently close to a normal distribution); B) 0.55  2.05
( 0.55)(1 0.55)
984
= (0.534; 0.566); C) We are
96% confident that the interval (0.534; 0.566) does contain the true population proportion of all
adult consumers who feel vulnerable to identity theft. So, if we took all random samples of 984
U.S. adult consumers (or a sufficiently large number of random samples of size 984, all from the
same population of U.S. adult consumers and all using the same sampling method) and for each
sample we constructed the corresponding confidence interval, then 96% of the resulting intervals
will contain the true population proportion of all adult consumers feeling vulnerable to identity theft.
D) Since the claimed proportion 0.6 is not contained in the 96% CI, (0.534; 0.566), then at the 96%
confidence level, the sample data at hand did not provide sufficient evidence to affirm that the
networks’ claims are true. 20) n 
(2.575) 2 (0.5)(0.5)
(0.025)
21) n 
22) n 
(1.645) 2 (0.75)( 0.25)
(0.045) 2
( 2.575) 2 (196)
(3) 2
2
= 2652.25. Use n = 2653 adult consumers.
 250.558. Use n = 251 men.
 144.4. Use n = 145 normal adults.