1
Mann – Introductory Statistics, Fifth Edition, Solutions Manual
7.1The probability distribution of the population data is called the population distribution. Tables 7.1 and 7.2 on page 309 of the text provide an example of such a distribution. The probability distribution of a sample statistic is called its sampling distribution. Tables 7.3 to 7.5 on page 311 of the text provide an example of the sampling distribution of the sample mean.
7.2Sampling error is the difference between the value of the sample statistic and the value of the corresponding population parameter, assuming that the sample is random and no nonsampling error has been made. Example 7–1 on page 312 of the text exhibits sampling error. Sampling errors occur only in sample surveys.
7.3Nonsampling errors are errors that may occur during collection, recording, and tabulation of data. The second part of Example 7–1 on pages 312 and 313 of the text exhibits nonsampling error. Nonsampling errors occur both in sample surveys and censuses.
7.4a.
b.
Sampling error =
- Liza’s incorrect
Sampling error (from part b) = –1.83
Nonsampling error
Sample / /
15, 13, 8, 17 / 13.25 / .92
15, 13, 8, 9 / 11.25 / –1.08
15, 13, 17, 9 / 13.50 / 1.17
15, 8, 17, 9 / 12.25 / –.08
13, 8, 17, 9 / 11.75 / –.58
15, 13, 8, 12 / 12.00 / –.33
15, 13, 17, 12 / 14.25 / 1.92
15, 8, 17, 12 / 13.00 / .67
13, 8, 17, 12 / 12.50 / .17
15, 13, 9, 12 / 12.25 / –.08
15, 8, 9, 12 / 11.00 / –1.33
13, 8, 9, 12 / 10.50 / –1.83
15, 17, 9, 12 / 13.25 / .92
13, 17, 9, 12 / 12.75 / .42
8, 17, 9, 12 / 11.50 / –.83
7.5a.
b.
Sampling error =
c.Rich’s incorrect
Sampling error (from part b) = –.27
Nonsampling error
d.
Sample / /25, 13 19,9,15,11,7,17,30 / 16.22 / –.38
20, 13, 19, 9, 15, 11, 7, 17, 30 / 15.67 / –.93
20, 25 19, 9, 15, 11, 7, 17, 30 / 17.00 / .40
20, 25, 13, 9, 15, 11, 7, 17, 30 / 16.33 / –.27
20, 25, 13, 19, 15, 11, 7, 17, 30 / 17.44 / .84
20, 25, 13, 19, 9, 11, 7, 17, 30 / 16.78 / .18
20, 25, 13, 19, 9, 15, 7, 17, 30 / 17.22 / .62
20, 25, 13, 19, 9, 15, 11, 17, 30 / 17.67 / 1.07
20, 25, 13, 19, 9, 15, 11, 7, 30 / 16.56 / –.04
20, 25, 13, 19, 9, 15, 11, 7, 17 / 15.11 / –1.49
7.6
x / P(x) / x P(x) / x2P(x)70 / .20 / 14.00 / 980.00
78 / .20 / 15.60 / 1216.80
80 / .40 / 32.00 / 2560.00
95 / .20 / 19.00 / 1805.00
= 8.09
7.7a.
55 / 1/6=.16753 / 1/6=.167
28 / 1/6=.167
25 / 1/6=.167
21 / 1/6=.167
15 / 1/6=.167
b.
Sample /55, 53, 28, 25, 21 / 36.4
55, 53, 28, 25, 15 / 35.2
55, 53, 28, 21, 15 / 34.4
55, 53, 25, 21, 15 / 33.8
55, 28, 25, 21, 15 / 28.8
53, 28, 25, 21, 15 / 28.4
36.4 / 1/6=.167
35.2 / 1/6=.167
34.4 / 1/6=.167
33.8 / 1/6=.167
28.8 / 1/6=.167
28.4 / 1/6=.167
- The mean for the population data is:
Suppose the random sample of five family members includes the observations: 55, 28,25,21, and 15. The mean for this sample is:
Then the sampling error is:
7.8a.
14 / 1/5 = .208 / 2/5 = .40
7 / 1/5 = .20
20 / 1/5 = .20
b.
Sample /14, 8, 7, 7 / 9.00
14, 8, 7, 20 / 12.25
14, 8, 7, 20 / 12.25
14, 7, 7, 20 / 12.00
8, 7, 7, 20 / 10.50
9.00 / 1/5 = .20
12.25 / 2/5 = .40
12.00 / 1/5 = .20
10.50 / 1/5 = .20
c.The mean for the population data is:
Suppose the random sample of four faculty members includes the observations: 14, 8, 7 and 20. The mean for this sample is:
Then the sampling error is:
7.9a.Mean of
b.Standard deviation of where population standard deviation and n = sample size.
7.10A sample statistic used to estimate a population parameter is called an estimator. An estimator is unbiased when its mean is equal to the corresponding population parameter. The sample mean is an unbiased estimator of μ, because the mean of is equal to μ.
7.11An estimator is consistent when its standard deviation decreases as the sample size is increased. The sample mean is a consistent estimator of μ because its standard deviation decreases as the sample size increases. This is obvious from the formula of.
7.12Since, increasing n decreases . Hence, as nincreases, decreases.
7.13 and
a.and
b.and
7.14 and
a.and
b.and
7.15a.n / N = 300 / 5000 = .06 > .05
b.
7.16a. = .800
b.
7.17 and
a. so
b.
7.18and
a. so
b.approximately
7.19, and n = 80
and
7.20square feet;square feet; and n = 25
square feet and square feet
7.21, , and n = 25, and
7.22minutes, minutes, and
minutes andminutes
7.23 and
, so players
7.24 million and million
so
7.25a.
/ P() / P() / P()76.00 / .20 / 15.200 / 1155.200
76.67 / .10 / 7.667 / 587.829
79.33 / .10 / 7.933 / 629.325
81.00 / .10 / 8.100 / 656.100
81.67 / .20 / 16.334 / 1333.998
84.33 / .20 / 16.866 / 1422.310
85.00 / .10 / 8.500 / 722.500
∑P()=80.600 / ∑P() = 6507.262
∑P()=80.600 = same value found in Exercise 7.6 for μ.
b.
c. is not equal to in this case because which is greater than .05.
d.
7.26The population from which the sample is drawn must be normally distributed.
7.27The central limit theorem states that for a large sample, the sampling distribution of the sample mean is approximately normal, irrespective of the shape of the population distribution. Furthermore, and , where μ and σ are the population mean and standard deviation, respectively. A sample size of 30 or more is considered large enough to apply the central limit theorem to .
7.28The central limit theorem will apply in cases a and c since It will not apply in case b because
7.29a.Slightly skewed to the right
bApproximately normal because and the central limit theorem applies
c.Close to normal with perhaps a slight skew to the right
7.30In both cases the sampling distribution of would be normal because the population distribution is normal.
7.31In both cases the sampling distribution of would be normal because the population distribution is normal.
7.32μ = 6.7 minutes, σ = 2.1 minutes, and n = 16
minutes and minute
The sampling distribution of is normal because the population is normally distributed.
7.33miles per hour, miles per hour, and n = 20
miles per hour and mile per hour
The sampling distribution of is normal because the population is normally distributed.
7.34 and n = 25
and
The sampling distribution of is approximately normal because the population is approximately normally distributed.
7.35, , N = 5540 and n = 48
Since which is less than .05,
The sampling distribution of is approximately normal because the population is approximately normally distributed.
7.36 pounds, pounds, and
pounds and pounds
The sampling distribution of is approximately normal because the sample size is large (n > 30).
7.37, , and
and
The sampling distribution of is approximately normal because the sample size is large (n > 30).
7.38, and n = 50
and
The sampling distribution of is normal because the population distribution is normal.
7.39, and n = 625
and
The sampling distribution of is approximately normal because the sample size is large (n > 30).
7.40 or 98.76%.
7.41 or 86.64%.
7.42, , and n = 36
a.z =
b.z =
c.z =
d.z =
7.43, , and n = 49
a.z =
b.z =
c.z =
d.z =
7.44, , and
and
- For
For
= .4812+ .2673 =.7485
- For
7.45, , and
and
a.For
For
b.For
7.46, , and n = 40
and
a.For
bFor
For
=.4961 –.3133=.1828
7.47, , and
and
a.For
approximately
b.For
7.48 miles per hour, miles per hour, and
miles per hour
a.For
b.For
c.For
For
7.49, , and
a.For
b.For
- For
For
7.50 minutes, minutes, and
minute
a.For
For
=.4934–.2157=.2777
b.P(within 1 minute ofμ)
For
For
c.P(lower than μ by 1 minute or more)
For
7.51per hour, per hour, and
per hour
a.For
For
b.P( within $1.00per hour of )
For
For
c.P(greater than μ by $1.50 per hour or more)
For
7.52 hours, hours, and
hour
a.For
For
b.For
7.53 and
a.For
For
=.3830+.4279=.8109
b.For
7.54, and
a.For
For
P(81008500) = P(–1.67≤ z ≤ .83) = P(–1.67 ≤ z ≤ 0) + P(0 ≤ z ≤.83)
=.4525 + .2967 = .7492
b.P(within $200 of μ) =
For
For
=.3944+.3944=.7888
c.P(greater than μ by $300 or more)
For
7.55,, and
a.For
For
=.4972–.3508=.1464
b.
For
For
c.
For
7.56,, and
a.For
b.For
For
= .2704+.4868=.7572
c.
For
For
d.
For
7.57 inches, inches, and
inch
a.For
b.For
For
=.3944 + .4599=.8543
c. within .6 inches of )
For
For
d.is lower than by .5 inches or more)
For
7.58 hours, hours, and
hours
We are to find
For
For
7.59 inches, inch, and
inch
For
For
=.0062+.0062=.0124
7.60 and
7.61 and
7.62Number with characteristic in population
Number with characteristic in sample or 63
7.63Number in population with characteristic
Number in sample with characteristic
7.64a.The mean of is:
b.The standard deviation of is:
c.The sampling distribution of is approximately normal if np and nq are both greater than 5.
7.65Sampling error =
7.66Sampling error = – p = .33 – .29 =.04
7.67The estimator of p is the sample proportion.
The sample proportion is an unbiased estimator ofp, since the mean of is equal to p.
7.68The sample proportion is a consistent estimator of p, since decreases as the sample size is increased.
7.69, hence decreases asn increases.
7.70
a., and
b., and
7.71
a., and
b., and
7.72, and
a.n / N = 800 / 4000 = .20 > .05
b.,
7.73, and
a.
b.,
7.74A sample is considered large enough to apply the central limit theorem if np and nq are both greater than 5.
7.75a. and
Since and , the central limit theorem applies.
b.; since , the central limit theorem does not apply.
c. and
Since and , the central limit theorem applies.
d.; since , the central limit theorem does not apply.
7.76a. and
Since and , the central limit theorem applies.
b. and
Since and , the central limit theorem applies.
c.; since , the central limit theorem does not apply.
d.; since , the central limit theorem does not apply.
7.77a.The proportion of these TV sets that are good is
b.Total number of samples of size 5 is:
c & d.Let: G =good TV set and D = defective TV set
Let the six TV sets be denoted as: 1 = G, 2 = G, 3 = D, 4 = D, 5 = G, and 6 = G. The six possible samples, their sample proportions, and the sampling errors are given in the table below.
Sample / TV sets / / Sampling error1, 2, 3, 4, 5 / G, G, D, D, G / 3/5=.60 / .60 – .667 = –.067
1, 2, 3, 4, 6 / G, G, D, D, G / 3/5=.60 / .60 – .667 = .067
1, 2, 3, 5, 6 / G, G, D, G, G / 4/5=.80 / .80 – .667 = .133
1, 2, 4, 5, 6 / G, G, D, G, G / 4/5=.80 / .80 – .667 = .133
1, 3, 4, 5, 6 / G, D, D, G, G / 3/5=.60 / .60–.667=–.067
2, 3, 4, 5, 6 / G, D, D, G, G / 3/5=.60 / .60–.667=–.067
/ / Relative Frequency
.60 / 4 / 4/6=.667
.80 / 2 / 2/6=.333
.60 / .667
.80 / .333
7.78a.Two of the five fires of this seasonwere caused by arson. Hence, the proportion of firescaused by arson is:
- The total number of samples of size 3 that can be selected is:
c & d.Let the five fires of the season be denoted as:
A = arson, B = accident, C = accident, D = arson, and E = accident
The following tables list all the possible samples of size 3, the sample proportions, the sampling errors and the sampling distribution of the sample proportion.
Sample / Employees / / Sampling errorA, B, C / arson, accident, accident / 1/3 = .333 / .333 – .40 = –.067
A, B, D / arson, accident, arson / 2/3 = .667 / .667 – .40 = .267
A, B, E / arson, accident, accident / 1/3 = .333 / .333 – .40 = –.067
A, C, D / arson, accident, arson / 2/3 = .667 / .667 – .40 = .267
A, C, E / arson, accident, accident / 1/3 = .333 / .333– .40 = –.067
A, D, E / arson, arson, accident / 2/3 = .667 / .667 – .40 = .267
B, C, D / accident, accident, arson / 1/3 = .333 / .333– .40 = –.067
B, C, E / accident, accident, accident / 0/3 = .000 / .000 – .40 = –.400
B, D, E / accident, arson, accident / 1/3 = .333 / .333– .40 = –.067
C, D, E / accident, arson, accident / 1/3 = .333 / .333– .40 = –.067
/ / Relative Frequency
.000 / 1 / 1/10 = .10
.333 / 6 / 6/10 = .60
.667 / 3 / 3/10 = .30
.000 / .10
.333 / .60
.667 / .30
7.79, ,
and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
7.80
and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
7.81, ,
and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
7.82,
and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
7.83
Thus, 95.44% of the sample proportions will be within 2 standard deviations of the population proportion.
7.84
Thus, 99.74% of the sample proportions will be within 3 standard deviations of the population proportion.
7.85
a.
b.
c.
d.
7.86
a.
b.
c.
d.
7.87, , and
a.For
For
= .1808 + .3997 = .5805
b.For
7.88, , and
- For
For
=.4292 – .1700=.2592
b.For
7.89, and
a.For
For
=.4222+.4834=.9056
b.For
7.90, , and
a.For
For
=.3686+.2995=.6681
- For
7.91, , and
For
7.92, , and
a.The required probability is:
For
For
=.3944+.3944=.7888
b.We are to find the probability: .
For
c.
For
7.93 hours, hours, and
750 hours and hours
The sampling distribution of is normal because the population is normally distributed.
7.94minutes,minutes,
24 minutes and minutes
The sampling distribution of is approximately normal because
7.95 hours, hours, and
hours
a.For
b.For
For
=.4884–.3186=.1698
c. within 15 hours of )
For
For
d. is lower than by 20 hours or more)
For
7.96minutes, minutes,
minute
a.For
b.For
For
=.4236+.4979=.9215
c. within 1minute of )
For
For
=.4236+.4236=.8472
d. is greater than by 2 minutes or more)
For
7.97 hours, hours, and
hours
a.For
b.For
For
= .2704+.3962=.6666
c.
For
For
=.4332+.4332=.8664
d.
For
7.98 ounces, ounce, and
For
For
=.0062 + .0062=.0124
7.99, and
, and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
7.100 and
, and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
7.101, and
a.i.For
ii.For
For
=.4956–.4049=.0907
b.
For
For
c.
For
7.102 and
a.is within .05 ofp) =
For
For
b. is not within .05 ofp) = 1 – is within .05 ofp) = 1–.7372 (from part a) = .2628
c. is greater than p by .08 or more) =
For
d. is less than p by .06 or more) =
For
7.103 and
The required probability is:
For
For
7.104Given and
The corresponding z values are approximately z = 1.04 and z = –.52 respectively.
First we use to find .
We have and
Subtracting, gives so
Since , we have so .
7.105 ppm
We want
or
, so
Ten measurements are necessary.
7.106a.so we can use the normal approximation to
thebinomial distribution.
b.
so
, hence
The reporter should take a sample of 66 voters or more.
7.107a. and we assume that
The shape of the sampling distribution is approximately normal.
b.
so
, hence
The sample should include at least 754 voters.
7.108 feet, feet, and
feet
P(total length of three throws exceeds 885) = P(mean length of three throws exceeds 885/3)
=
For
7.109 pounds, pounds, and
Since is approximately normally distributed.
P (sum of 35 weights exceeds 6000 pounds) = P(mean weightexceeds 6000/35) =
For
Self – Review Test for Chapter Seven
1. b2. b3. a4. a5. b6. b
7. c8. a9. a10. a11. c12. a
13.According to the central limit theorem, for a large sample size, the sampling distribution of the sample mean is approximately normal irrespective of the shape of the population distribution. The mean and standard deviation of the sampling distribution of the sample mean are:and .
The sample size is usually considered to be large if
From the same theorem, the sampling distribution of is approximately normal for large samples. In the case of proportion, the sample is large if and
14. pounds and pounds
a.pounds and pounds
b.pounds and pounds
In both cases the sampling distribution of is approximately normal because the population has an approximate normal distribution.
15. minutes and minutes
a.minutes and minute
Since the population has an unknown distribution and , we can draw no conclusion about the shape of the sampling distribution of .
b.minutes and minutes
Since , the sampling distribution of is approximately normal.
16.seconds, seconds, and
seconds
a.For
For
=.4977– .2611=.2366
b.For
For
c. greater than by 1 second or more)
For
d.For
For
= .4830 + .4207 = .9037
e.
17. ounces, ounce, and
a.i.For
For
= .4868 – .3665 = .1203
ii.For
iii.For
b.
For
For
c.
For
18.
a., and
and nq = 80(.93) = 74.4
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
b., and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
c., and
and
Since np and nq are both greater than 5, the sampling distribution of is approximately normal.
19. and
- i.For
ii.For
For
= .1368+.4608=.5976
iii.For
iv.For
For
= .4608–.2580= .2028
b.
For
For
= .4207 +.4207 = .8414
c.
For
d.
For
For
= .3554 –.3554 = .7108
e.
For