Self assessment exam A - Confidence Intervals- answers
Name: __________________________
1. True-false test.
__T____ As n increases the student's t distribution approaches the standard normal
distribution.
___F___ s is an unbiased estimator of  .
__ T __ x is an unbiased estimator of µ .
___T___ For 7 degrees of freedom, t 0.01  2.998
___T__ z 0.005  2.576
___T___ pˆ is an unbiased estimator of p
___F___ Increasing the sample size will make the confidence interval for the mean wider.
___T___ The main objective of Statistics is to make inferences about a population based on
information contained in a sample.
__ T__ An estimator of a parameter is said to be unbiased if the mean
of its sampling distribution is equal to the parameter.
___F___ As n increases the standard error of the mean increases.
2. Complete:
a) According to the central limit theorem as n increases the sampling distribution of the mean
approaches a __normal_________ distribution.
b) In a sample of size 31 if 6 conditions must be satisfied, then we have
____25___ degrees of freedom.
c) If samples of size n=16 are taken from a population with mean µ =15 and standard
deviation then the sampling distribution of the mean has a mean  x = __15____ and a
standard deviation  x = __4___________.
d) As n increases the student’s t distribution approaches the __standard normal______
distribution.
e) As n increases the standard error of the mean ___decreases_________.





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
3. A cigarette manufacturer claims that the mean nicotine content in its king-size
cigarettes is 2.1 mg with a standard deviation equal to 0.35 mg. If this claim is valid,
what is the approximate probability that a sample of 25 cigarettes will yield a mean
nicotine content exceeding 2.03 mg?
  2.1 mg,   0.35
0.35 0.35
Distributi on of x : n  25,  x  2.1,  x 

 0.07
5
25
P( x  2.03)  normalcdf (2.03, 10 99 , 2.1, 0.07)  0.8413447404
answer: _0.8413447404____
4. The weight of food packed in certain containers is a random variable with a mean
weight of 16 ounces and a standard deviation of 0.6 ounces. If the containers are
shipped in boxes of 36, find , approximately, the probability that a randomly selected
box will weigh over 585 ounces?
  16 ounces,   0.6 ounces
0.6
Distributi on of x : n  36,  x  16,  x 
P( sum  585)  P( x 
36

0.6
 0.1,
6
answer:
585
 16.25)  normalcdf (16.25, 10 99 , 16, 0.1)  0.0062096799
36
answer: 0.0062096799

. 100 cars on a thruway are clocked at an average speed of 63 mph.
Construct a 88% confidence interval for the mean speed of all cars
on this thruway. Use s=4.2.
n  100, x  63, s  4.2,   0.12, z  2  invNorm(0.06)  1.554773593
formula : x  z  2
s
n
 63  (1.55477)
4.2
100
 63  0.65
Answer: __We are 88% confident that the mean speed of all cars on this thruway is between
__62.35 mph and 63.65 mph___________
6. We wish to determine the average length of time that an automobile is parked on the campus parking
lot, in order to improve the current situation. How large a sample is needed, in order to make a statement
with a 99% confidence that our mean is within
1
hour of the true mean. Assume the population standard
2
deviation is 2.2 hours.
  0.01, z  z 0.005  invNorm(0.005)  2.575829303,   2.2, E  0.5
2
 z 
n   2
 E
2
2

 2.5758  2.2 
  
  129
0.5



Answer: _n=129_________
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7.
A manufacturer of parts believes that approximately 6% of his products contains
flaws. If he wishes to estimate the true proportion to within 0.005 and to be certain with
a probability of 90% of being correct, how large a sample should he take?
  0.10, z  z 0.05  invNorm(0.05)  1.644853626, p  0.06, q  0.94, E  0.005
2
n
z22 pq
E2

1.644852 0.06(0.94)

 6104
(0.005) 2
Answer: _6104__________
8.
A random sample of 400 citizens in a community showed that 300 favored having
their water fluoridated. Use these data to find a 98% confidence for the proportion of
the population favoring fluoridation.
x 300 3
n  400, x  300, pˆ  
  0.75, qˆ  0.25
n 400 4
  0.02, z 2  z 0.01  invNorm(0.01)  2.326347877
pˆ qˆ
(0.75)(0.25)
 0.75  2.326347877
 0.75  0.0503666909
n
400
(0.70, 0.80) approximately.
answer: based on the random sample of 400 citizens we are 98% confident that the
proportion of the population favoring fluoridation is anywhere between 70% and 80%.
Formula : pˆ  z 2
9.
A set of 12 experimental animals was fed a special diet for 3 weeks and produced
the following gains in weight: 30, 25, 31, 26, 25, 40, 35, 32, 32, 33, 28, 30 pounds. Find
a 99% confidence interval for the population mean gain in weight, assuming that gain
in weight is a normal variable.
Solution:
 x  365  30.5833, s 2 
n  12, x 
x
2
n
12
s  30.0833  5.4848
 x 2

n 1
n

11433 
11
365 2
12  30.0833
  0.01, t  , 11  T , eq : 0  tcdf (10 9 , T ,11)  0.005 solve T  t   3.105806513
2
Formula : x  t  2
2
s
 30.5833  3.105806513 
5.4848
n
30.5833  4.9175, (25.6658, 35.5008)
12
answer: we are 99% confident that the population mean gain in weight is between
25.6658 and 35.5008 pounds.
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. In a study to estimate the proportion of residences in Stoney Creek and its suburbs
that subscribe to Writer's digest, it is found that 60 of 100 urban residences subscribe
while only 80 of 125 suburban residences subscribe. Find a 99% confidence interval
for the difference in the proportion of urban and suburban residences that subscribe
to writer's digest.
Solution:
Urban
n1  100
x1  60
pˆ1  60 / 100  0.60
10
n1  125
Suburban
ˆ 2  80 / 125 
p
x2  80
  0.01   / 2  0.005, z0.005  invNorm(0.005)  2.575829303
pˆ 1  pˆ 2  z / 2 
pˆ 1qˆ1 pˆ 2 qˆ 2

 0.60  0.64  2.5758
n1
n2
0.60(0.40)  0.64(0.36)
100
125
 0.04  2.5758(0.06513984960)  0.04  0.17
 0.21  p1  p 2  0.13
Conclusion :
With a 99% confidence we conclude that the proportion of urban residences that
subscribe to writer's digest could be as much as 21% lower or as much 13% higher
than the proportion of suburban residences who subscribe.
A sample of 50 men studying statistics shows an average grade of 80 points with a
standard deviation of 5 points. A sample of 40 women studying statistics shows an
average grade of 85 points with a standard deviation of 6 points. Use the given sample
to construct a 98% confidence interval for the difference in average grade between all
men and women who study statistics.
11.
Men
n1  50
1 =80
1  5
Women
n2  40
 2 =85
2  6
  0.02   / 2  0.01, z0.01  invNorm(0.01)  2.326347877
x1  x 2  z / 2 
 12
n1

 22
n2
 80  85  2.3263
25 36

50 40
 5  2.3263( 1.4 )  5  2.32631.1832  5  2.753
 7.753  1   2  2.247
Conclusion :
With a 98% confidence we conclude that the average grade in statistics for men s is
below the average grade of women by anywhere between 2.247 and 7.753 points.
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