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Chapter 7 Full Solutions
Section 7.1
Check Your Understanding, page 446
1. The population is all M&M Milk Chocolate Candies produced by the factory in Hackettstown, NJ. The
parameter is p = the proportion of all M&M Milk Chocolate Candies produced by the factory in Hackettstown, NJ
that are orange. The parameter is claimed to be p = 0.25. The sample is the 50 M&M Milk Chocolate Candies
selected. The statistic is the proportion of the sample of 50 M&M’s that are orange, p̂ .
2. The graph below shows the population distribution.
3. Answers will vary. The graph below shows a possible distribution of sample data. For this sample, there are
11
pˆ = 0.22.
11 orange M&M’S so =
50
4. The middle graph is the approximate sampling distribution of pˆ . The statistic measures the proportion of
orange candies in samples of 50 M&M’S. Assuming that the company is correct, 25% of the M&M’S are orange,
so the center of the distribution of p̂ should be at approximately 0.25. The first graph shows the distribution of
the colors for one sample, rather than the distribution of p̂ from many samples, and the third graph is centered at
0.125, rather than 0.25.
Chapter 7: Sampling Distributions
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Check Your Understanding, page 452:
1. The median does not appear to be an unbiased estimator of the population median. The mean of the simulated
sampling distribution of the sample median (73.5) is not equal to the median of the population (75).
2. Increasing the sample size from 10 to 20 will decrease the variability of the sampling distribution. Larger
samples provide more precise estimates, because larger samples include more information about the population
distribution.
3. The sampling distribution of the sample median is skewed to the left and single peaked.
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Section 7.1 Exercises
7.1
Population: all people who signed a card saying that they intend to quit smoking. Parameter: p = the proportion
of the population who had not smoked over the past 6 months. Sample: the 1000 people who were selected at
random. Statistic: p̂ = the proportion in the sample who had not smoked over the past 6 months = 0.21.
7.2
Population: all U.S. adults. Parameter: p = the proportion of all U. S. adults who were unemployed in October
2016. Sample: the 60,000 randomly selected U.S. adults. Statistic: p̂ = the proportion in the sample who were
unemployed in October 2016 = 0.049.
7.3
Population: all dental practices in California. Parameter: the interquartile range of the price to fill a cavity for
all dental practices in California. Sample: the 10 randomly selected dental practices which provided the price
they charge to fill a cavity. Statistic: IQR = the interquartile range of the price to fill a cavity for the 10 selected
dental practices = $74.
7.4
Population: all points in the turkey. Parameter: minimum temperature in all points of the turkey. Sample: four
randomly chosen locations in the turkey. Statistic: minimum temperature in the sample of four locations= 170 F.
7.5
Population: all bottles of Arizona Iced Tea produced that day. Parameter: µ = the average number of ounces per
bottle in all bottles of Arizona Iced Tea produced that day. Sample: the 50 bottles of tea selected at random from
the day’s production. Statistic: x = the average number of ounces of tea contained in the 50 bottles = 19.6
ounces.
7.6
Population: all ball bearings in the production run. Parameter: µ = the average diameter of all ball bearings in
the production run. Sample: the 100 bearings selected at random from the run. Statistic: x = the average
diameter of the 100 ball bearings selected= 2.5009 cm.
Chapter 7: Sampling Distributions
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7.7
The 10 possible SRSs of size n = 2 and their sample means are:
Sample #1: Abigail(10), Bobby(5)
x = 7.5
Sample #2: Abigail(10), Carlos(10)
x = 10
Sample #3: Abigail(10), DeAnna(7)
x = 8.5
Sample #4: Abigail(10), Emily(9)
x = 9.5
Sample #5: Bobby(5), Carlos(10)
x = 7.5
Sample #6: Bobby(5), DeAnna(7)
x=6
Sample #7: Bobby(5), Emily(9)
x=7
Sample #8: Carlos(10), DeAnna(7)
x = 8.5
Sample #9: Carlos(10), Emily(9)
x = 9.5
Sample #10: DeAnna(7), Emily(9)
x=8
A dotplot of the sampling distribution of the sample mean is given below.
7.8
The 10 possible SRSs of size n = 3 and the minimum quiz score for each sample is:
Sample #1: Abigail(10), Bobby(5), Carlos(10)
Min = 5
Sample #2: Abigail(10), Bobby(5), DeAnna(7)
Min = 5
Sample #3: Abigail(10), Bobby(5), Emily(9)
Min = 5
Sample #4: Abigail(10), Carlos(10), DeAnna(7)
Min = 7
Sample #5: Abigail(10), Carlos(10), Emily(9)
Min = 9
Sample #6: Abigail(10), DeAnna(7), Emily(9)
Min = 7
Sample #7: Bobby(5), Carlos(10), DeAnna(7)
Min = 5
Sample #8: Bobby(5), Carlos(10), Emily(9)
Min = 5
Sample #9: Bobby(5), DeAnna(7), Emily(9)
Min = 5
Sample #10: Carlos(10), DeAnna(7), Emily(9)
Min = 7
A dotplot of the sampling distribution of the sample minimum is given below.
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7.9
The 10 possible SRSs of size n = 2 and the proportion of females in each sample is:
Sample #1: Abigail(F), Bobby(M)
p̂ = 0.50
p̂ = 0.50
Sample #2: Abigail(F), Carlos(M)
p̂ = 1
Sample #3: Abigail(F), DeAnna(F)
p̂ = 1
Sample #4: Abigail(F), Emily(F)
p̂ = 0
Sample #5: Bobby(M), Carlos(M)
p̂ = 0.50
Sample #6: Bobby(M), DeAnna(F)
Sample #7: Bobby(M), Emily(F)
Sample #8: Carlos(M), DeAnna(F)
Sample #9: Carlos(M), Emily(F)
Sample #10: DeAnna(F), Emily(F)
p̂ = 0.50
p̂ = 0.50
p̂ = 0.50
p̂ = 1
A dotplot of the sampling distribution of the sample proportion is given below.
7.10
The 10 possible SRSs of size n = 3 and the median score of each sample is:
Sample #1: Abigail(10), Bobby(5), Carlos(10)
median = 10
Sample #2: Abigail(10), Bobby(5), DeAnna(7)
median = 7
Sample #3: Abigail(10), Bobby(5), Emily(9)
median = 9
Sample #4: Abigail(10), Carlos(10), DeAnna(7)
median = 10
Sample #5: Abigail(10), Carlos(10), Emily(9)
median = 10
Sample #6: Abigail(10), DeAnna(7), Emily(9)
median = 9
Sample #7: Bobby(5), Carlos(10), DeAnna(7)
median = 7
Sample #8: Bobby(5), Carlos(10), Emily(9)
median = 9
Sample #9: Bobby(5), DeAnna(7), Emily(9)
median = 7
Sample #10: Carlos(10), DeAnna(7), Emily(9)
median = 9
A dotplot of the sampling distribution of the sample median is given below.
Chapter 7: Sampling Distributions
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7.11
(a) A graph of the population distribution is shown below.
(b) Answers will vary. An example bar graph is given.
7.12
(a) A graph of the population distribution is shown below.
(b) Answers will vary. An example dotplot is given below.
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7.13
45
= 0.45 , which is less than 0.60.
100
(b) It is possible that 60% of the students did their homework, and the students got a p̂ less than 60% because of
sampling variability. It is also possible that the sample proportion is less than 60% because less than 60% of all
the students did their homework.
(c) In one simulated SRS of 100 students, there were 73 students who did their assigned homework last week.
(d) Because there were no values of p̂ less than or equal to 0.45 in the simulation, it would be very surprising to
get a sample proportion of 0.45 or less in an SRS of size 100 from a population in which p = 0.60.
(e) Because it would be very surprising to get a sample proportion of p̂ = 0.45 or less in an SRS of size 100
when p = 0.60, we should be skeptical of the newspaper’s claim. Sampling variability is not a plausible
explanation for the difference between the observed proportion ( p̂ = 0.45) and the proportion claimed by the
newspaper (p = 0.60).
pˆ
(a) The sample proportion is=
7.14
(a) The sample mean was x = 64.7 inches, which is greater than 64 inches.
(b) It is possible that the average height of all 16-year-old females at this school is 64 inches, but the students got
a sample mean x greater than 64 inches because of sampling variability. It is also possible that the sample mean
is greater than 64 inches because the average height of all 16-year-old females at this school is greater than 64
inches.
(c) In one simulated SRS of 20 students, the average height was x = 62.5 inches.
(d) Because about 10% of the values of x were 64.7 or greater, it wouldn’t be surprising to get a sample mean of
64.7 or larger in an SRS of size 20 from a Normal population with µ = 64 and σ = 2.5.
(e) Because it isn’t surprising to get a sample mean of x = 64.7 or greater in an SRS of size 20 from a Normal
population with µ = 64 and σ = 2.5, we do not have convincing evidence that the population mean height at the
school is different than µ = 64. Sampling variability is a plausible explanation for the difference between the
sample mean ( x = 64.7) and the mean claimed by the National Center for Health Statistics ( µ = 64).
7.15
Because 78/250 or 31.2% of the values of p̂ are less than or equal to 0.57 in the simulation, it would not be
surprising to get a sample proportion of p̂ = 0.57 or less in an SRS of size 100 from a population in which p =
0.60. A sample proportion of p̂ = 0.57 does not provide convincing evidence that the population proportion of
students who did all their assigned homework is less than p̂ = 60%. Sampling variability is a plausible
explanation for the difference between the observed proportion ( p̂ = 0.57) and the proportion claimed by the
newspaper (p = 0.60).
7.16
Because none of the values of x were 65.8 or greater, it would be surprising to get a sample mean of x = 65.8 or
larger in an SRS of size 20 from a Normal population with µ = 64 and σ = 2.5. Therefore, a sample mean of x
= 65.8 does provide convincing evidence that the population mean height at the school is greater than µ = 64.
Sampling variability is not a plausible explanation for the difference between the sample mean ( x = 65.8) and the
mean claimed by the National Center for Health Statistics ( µ = 64).
Chapter 7: Sampling Distributions
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7.17
A sample standard deviation of 5ºF is quite large compared with what we would expect to happen by chance
alone, because none of the 300 simulated SRSs had a standard deviation that large. A sample standard deviation
of 5ºF provides convincing evidence that the manufacturer’s claim is false and that the thermostat actually has
more variability than claimed.
7.18
A sample minimum of 40 F is quite low compared with what we would expect to happen by chance alone,
because only 2 of the 300 simulated SRSs had a minimum that small. A sample minimum of 40ºF provides
convincing evidence that the manufacturer’s claim is false.
7.19
The 6 possible SRSs of size n = 2 and the sample proportion for each sample is:
Sample #1: Red, White
p̂ = 0.5
Sample #2: Red, Silver
p̂ = 0.5
Sample #3: Red, Red
p̂ = 1
Sample #4: White, Silver
p̂ = 0
Sample #5: White, Red
p̂ = 0.5
Sample #6: Silver, Red
p̂ = 0.5
A dotplot of the sampling distribution of the sample proportion is given below.
The population proportion of red cars is p = 2/4 = 0.5. The sample proportion is an unbiased estimator of the
population proportion. The mean of the sampling distribution of the sample proportion of red cars is equal to the
population proportion.
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7.20
The 6 possible SRSs of size n = 2 and the sample minimum for each sample is:
Sample #1: Red(1), White(5)
minimum = 1
Sample #2: Red(1), Silver(8)
minimum = 1
Sample #3: Red(1), Red(20)
minimum = 1
Sample #4: White(5), Silver(5)
minimum = 5
Sample #5: White(5), Red(20)
minimum = 5
Sample #6: Silver(5), Red(20)
minimum = 5
A dotplot of the sampling distribution of the sample minimum is given below.
The minimum age of the population of 4 cars is 1 year. The sample minimum is not an unbiased estimator of the
population minimum. The mean of the sampling distribution of the sample minimum of car age is 3, which is not
equal to the population minimum age.
7.21
The 4 possible SRSs of size n = 3 and the sample proportion for each sample is:
Sample #1: Red, White, Silver
p̂ = 1/3
Sample #2: Red, White, Red
p̂ = 2/3
Sample #3: Red, Silver, Red
p̂ = 2/3
Sample #4: White, Silver, Red
p̂ = 1/3
A dotplot of the sampling distribution of the sample proportion is given below.
The variability in this sampling distribution is less than the variability in the sampling distribution from Exercise
19. Increasing the sample size decreases the variability. The standard deviation of the sampling distribution based
upon SRSs of size n = 2 (in Exercise 19) is 0.289 and the standard deviation of the sampling distribution based
upon SRSs of size n = 3 (in this Exercise) is 0.167.
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7.22
The 4 possible SRSs of size n = 3 and the sample minimum for each sample is:
Sample #1: Red(1), White(5), Silver(8)
minimum = 1
Sample #2: Red(1), White(5), Red(20)
minimum = 1
Sample #3: Red(1), Silver(8), Red(20)
minimum = 1
Sample #4: White(5), Silver(8), Red(20)
minimum = 5
A dotplot of the sampling distribution of the sample minimum is given below.
The variability in this sampling distribution, as measured by the standard deviation, is less than the variability in
the sampling distribution from Exercise 20. The standard deviation of the sampling distribution based upon SRSs
of size n = 2 (in Exercise 20) is 2 and the standard deviation of the sampling distribution based upon SRSs of size
n = 3 (in this Exercise) is 1.732.
7.23
If we chose many SRSs and calculated the sample mean x for each sample, the distribution of x would be
centered at the value of µ . In other words, when we use x to estimate µ , we will not consistently underestimate
µ or consistently overestimate µ .
7.24
If we chose many random samples and calculated the sample proportion p̂ for each sample, the distribution of p̂
would be centered at the value of p. In other words, when we use p̂ to estimate p, we will not consistently
underestimate p or consistently overestimate p.
7.25
(a) Statistics ii and iii are unbiased estimators because the means of their sampling distributions appear to be equal
to the corresponding population parameters.
(b) Statistic ii does the best job at estimating the parameter. It is unbiased and has very little variability.
7.26
7.27
7.28
7.29
7.30
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c.
d.
a.
b.
c.
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7.31
(a) We are looking for the percentage of values that are 2.5 standard deviations or farther below the mean in a
Normal distribution. In other words, we are looking for P(z ≤ –2.5).
1. Draw a Normal distribution.
−2.5
Standardized BMD value
2. Perform Calculations.
(i) z = –2.5.
Using Table A: P(z ≤ –2.5) = 0.0062.
Using technology: P(z ≤ –2.5) = normalcdf(lower: –1000, upper: –2.50, mean: 0, SD: 1) = 0.0062.
Less than 1% of healthy young adults have osteoporosis.
(b) Let X be the BMD for women aged 70 to 79 on the standard scale. Then X follows a Normal distribution with
a mean of -2 and a standard deviation of 1 and we want to find P(X ≤ –2.5).
1. Draw a Normal distribution.
−2.5
Standardized BDM value
2. Perform Calculations.
−2.5 − ( −2 )
(i) z =
= −0.5 .
1
Using Table A: P(X ≤ –2.5) = P(z ≤ –0.5) = 0.3085.
Using technology: P(z ≤ –0.5) = normalcdf(lower: –1000, upper: –0.5, mean: 0, SD: 1) = 0.3085.
(ii) Using technology: P(X ≤ –2.5) = normalcdf(lower: –1000, upper: –2.5, mean: –2, SD: 1) = 0.3085.
About 30.85% of women aged 70–79 have osteoporosis.
Chapter 7: Sampling Distributions
11
7.32
(a) The linear model is appropriate for these data because there is no leftover pattern in the residual plot.
(b) The equation for the least squares regression line =
is: yˆ 1.4146 + 0.4399 x where ŷ is the predicted average
number of offspring per female and x is the index of the abundance of pine cones.
(c) r2 = 57.2% of the variability in the average number of offspring is accounted for by the least-squares
regression line with x = cone index.
s = 0.600309. The actual number of offspring per female is typically about 0.600309 away from the amount
predicted by the least-squares regression line with x = cone index.
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