Download Answers - Topic 17

Workshop Statistics: Discovery with Data,
Second Edition
Topic 17: Sampling Distributions II: Means
Activity 17-1: Coin Ages (cont.)
(a) observational units: pennies; variable: age, quantitative
(b) parameters; mean: ; standard deviation: 
(c) No, this distribution is skewed to the right.
Students' answers to (d)-(l) may differ since the data are chosen randomly. These are
meant to be sample answers.
(d) 13, 24, 17, 0, 0
(e) 10.8
(f)
Sample no.
1
2
3
4
5
Sample mean 10.8 13 5.8 12.8 17.8
(g) No, this is an example of sampling variability. This is a quantitative variable, rather
than a categorical one.
(h) mean of x values: 12.04; standard deviation of x-bar values: 4.33
(i) This mean is reasonably close to the population mean. This standard deviation is less
than the population standard deviation.
(j)
(k) This distribution appears to be centered near the population mean of 12.264. The
values are less spread out than the population distribution and the five sample means of
size 5.
(l) yes
Activity 17-2: Coin Ages (cont.)
Students' answers to (a)-(m) may differ since the data is chosen randomly. These are
meant to be sample answers.
(a)
30
20
10
Std. Dev = 9.72
Mean = 12.6
N = 100.00
0
0.0
5.0
10.0
15.0
20.0
25.0
30.0
35.0
40.0
MEAN99
Yes, it resembles the distribution of ages in the population.
(b) mean: 12.8; standard deviation: 9.228; Yes, they are reasonably close to their
population counterparts. The distribution is skewed to the right.
(c)
16
14
12
10
8
6
4
Std. Dev = 4.00
2
Mean = 12.5
N = 100.00
0
2.0
6.0
4.0
10.0
8.0
14.0
12.0
18.0
16.0
22.0
20.0
MEAN99
This distribution does not have as wide a spread. The shape is much less skewed.
(d) mean: 12.426; standard deviation: 4.072; The mean is pretty close, but the standard
deviation is lower.
(e)
16
14
12
10
8
6
4
Std. Dev = 1.98
2
Mean = 12.09
N = 100.00
0
0
.5
17
0
.5
16
0
.5
15
0
.5
14
0
.5
13
0
.5
12
0
.5
11
0
.5
10
50
9.
50
8.
50
7.
MEAN99
The spread is even more narrow and mound shaped now, as variability decreases.
(f) mean: 12.098; standard deviation: 1.715; The mean is still close, but the standard
deviation is even lower now.
(g)
20
10
Std. Dev = 1.34
Mean = 12.38
N = 100.00
0
9.00
10.00
9.50
11.00
10.50
12.00
11.50
13.00
12.50
14.00
13.50
15.00
14.50
16.00
15.50
16.50
MEAN99
Again, the distribution narrows and the variability decreases. The distribution has
become more normal.
(h) mean: 12.02; standard deviation: 1.371; The mean is still close, but the standard
deviation is slightly lower than before.
(i) Although answers may vary, the correct answer is yes.
(j)
12
10
8
6
4
2
Std. Dev = .03
Mean = .491
N = 100.00
0
75
.5
63
.5
50
.5
38
.5
25
.5
13
.5
00
.5
88
.4
75
.4
63
.4
50
.4
38
.4
25
.4
MEAN99
This distribution is roughly normal.
(k), (l, k)
(l)
Mean
Std. dev.
Shape
Population 1
 = .5
= .289
uniform
Sample Means of size n = 50 from 1
.49937
.03637
Population 2
 = .5
 = .354
Sample means of size n = 50 from 2
.50264
.04686
u-shaped
20
10
Std. Dev = .04
Mean = .479
N = 100.00
0
.413
.438
.425
.463
.450
.488
.475
.513
.500
.538
.525
.563
.550
.588
.575
.600
MEAN99
This distribution is roughly normal.
(m) Penny ages: 1.359; 1: .041; 2: .050
Activity 17-3: Christmas Shopping
(a) It is a statistic because the 922 adults are meant as a sample of the population of all
American adults who plan to buy Christmas gifts that year.
(b)




population of interest: American adults who expected to buy Christmas gifts in
1999
sample selected: 922 American adults polled on November 18-21
parameter of interest: money expected (mean amount) to be spent on Christmas
gifts
statistic calculated: sample mean was $857
(c)  is not necessarily equal to $857, though it is possible. All values listed for are
possible, with the different sample mean arising by chance alone.
(d) The central limit theorem says that the sample means would vary like a normal
distribution with mean $850 and standard deviation 250/sqrt(922)=8.23 if samples of size
n = 922 were taken over and over. This does not depend on the shape of the distribution
of expected expenditures in the population since the sample size is large.
(e)
(f) No, since this is a normal distribution, $857 is near the center of the distribution, well
within one standard deviation of the mean.
(g, d) The sample mean would vary over a normal distribution with mean $800 and
standard deviation 8.23 if samples of size n = 922 were taken over and over.
(g, e) The picture would be the same, just shifted to center around 800:

(g, f) Now $857 is way in the tail of the distribution. This would be a very unlikely
sample mean if the population mean was equal to $800.
(h) The CLT says that the standard deviation of the sampling distribution of the sample
mean should be 250/ 922 = 8.23. 
(i)


interval for  : ($840.54, $873.46)