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Solutions - Sampling Distributions
7.6 Exit poll and n:
a) The interval of values within the sample proportion will almost certainly fall
within three standard errors of the mean: 0.53 to 0.59.
b) Based on the interval calculated in (a), it would be unusual to observe a sample
proportion of 0.60.
0.565  0.559
c)
= 0.63. Since 0.565 falls 0.63 standard deviations above the mean,
.0095
it would not be unusual to observe a sample proportion of 0.565.
7.8 Relative frequency of heads:
a) sample proportion probability
0
0.5
1
0.5
b) sample proportion probability
0
0.25
½
0.50
1
0.25
c) sample proportion probability
0
0.125
1/3
0.375
2/3
0.375
1
0.125
d) The distribution begins to take a bell shape.
7.10 Effect of n on sample proportion:
a) (i) standard error =
p1  p 
=
n
.51  .5
= 0.05
100
(ii) standard error =
p1  p 
=
n
.51  .5
= 0.0158
1000
b) The sample proportion is likely to fall within three standard errors of the mean. In
(i), with a mean of 0.50 and a standard error of 0.05, this would be between 0.35
and 0.65. In (ii), with a mean of 0.50 and a standard error of 0.0158, this would be
between 0.45 and 0.55.
c) When the sample size is larger, the standard error is smaller (a reflection of the fact
that a larger representative sample is likely to be more accurate than a smaller
representative sample). Three standard errors from a larger sample, therefore, will
be smaller than three standard errors from a smaller sample. The interval will be
smaller, an indication of a more precise estimate of the population proportion.
7.14 Education of the self-employed:
a) The random variable X is years of education.
b) Mean =13.6; standard error =

=3.0/10= 0.30. The mean of the sampling
n
distribution is 13.6, the same as the mean of the population. The standard error is
smaller than the standard deviation of the population. It reflects the variability
among all possible samples of a given size taken from this population.
c) Mean =13.6; standard error = /nσ=3.0/400= 0.15. As n increases, the mean of the
sampling distribution stays the same, but the standard error gets smaller.
7.16 Playing roulette:
a) X
Probability
0
1/38 = 0.973684
35
37/38 = 0.026316
b) The mean would be the same as the population mean of 0.921. The standard error

=5.603/ 5040 = 0.079.
n
c) z = (1 – 0.921)/0.079 = 1. $1 is one z-score, or one standard deviation, above the
mean. The central limit theorem tells us that 0.68 of observations fall within one
standard deviation of the mean. That indicates that 0.32 of observations fall
beyond one standard deviation from the mean. Half of this (0.16) would be below
one standard deviation below the mean and half (0.16) would be above one
standard deviation above the mean. Thus, the probability that with this amount of
roulette playing, your mean winnings are at least $1, is 0.16.
would be
7.18 Income of farm workers:
a) Standard error =

n
=160/10= 16
b) It is almost certain that the sample mean will fall within three standard errors of
the mean. Three standard errors, or (3)(16), is 48.
c) z = (520-500)/16 = 20/16 = 1.25
Table A tells us that 0.8944 falls below the z-score of 1.25, and 0.1056 falls below
the z-score of -1.25. Subtracting 0.1056 from 0.8944 gives us 0.79, the proportion
that falls within $20 of the mean.
7.23 Household size:
a) The random variable, X = number of people in a household, is quantitative.
b) The center of the population distribution is the mean of the population of 2.6. The
spread is the standard deviation of the population, 1.5.
c) The center of the data distribution is 2.4. The spread is the standard deviation of
1.4.
d) The center of the sampling distribution of the sample mean is 2.6. The spread, or
standard error, is

n
=1.5/225= 0.1
7.28 Predicting the Florida senate race:
For n = 2775, the mean of the sampling distribution would be 0.50 and the standard
p1  p 
0.51  0.5
error =
=
0095. We would expect that all samples from this
n
2775
distribution would fall within three standard errors of the mean for a range of 0.47 to
0.53. The sample Nelson proportion of 0.61 is above this range, so we would have
been willing to predict Nelson as the winner
7.34 How common is bulimia?:
The sampling distribution of the sample proportion would have a mean of 0.01 and
0.011  0.01
standard deviation
=0.0026 if p=0.01. Since 30/1500=0.02 is (0.021500
0.01)/0.0026=3.85 standard errors above the assumed population proportion, the
sample provides strong evidence that the population proportion actually exceeds 0.01.