Download Solutions 7(8, 12, 20), and

Solutions 7(8, 12, 20), and
8(4, 10, 16, 18, 40, 46, 54, 84)
Exercise 7.8
Random sample n measurements selected from population mean = 60, variance = 100
Sampling distribution
Mean Stdev
a. n=10
60
3.16
b. n =25
60
2
c. n=50
60
1.41
d. n=75
60
1.15
e. n=100
60
1
f n=500
60
0.45
g. n=1,000
60
0.32
Exercise 7.12
mean 293ppt, stdev 847ppt
a. n=50
b. The graph is a normal density, as
the result of CLT.
c. P(xbar exceeds 550ppt)
mean
293 ppt
z-score
Area
Probability 0.5-0.4842
stdev
119.78
2.15
0.4842
= 0.0158
= 1.5%
Exercise 7.20
mean 9.8sec, stdev .55 sec
n=50
a. mean for sampling distribution
Stdev for sampling distribution
b. p(xbar >9.70)
9.8 sec.
0.078 sec.
z-score
-1.28
Area
0.3997
Probability
0.8997
= 90%
c. xbar < 9.55 seconds
z-score
-3.21
Area
0.5
Probability
0.00 Almost 0%
The probability is almost zero that you would see a value less than 9.55 seconds.
Note that, in this case the z-score is greater than 3 standard deviations therefore,
it’s almost impossible to get such a small xbar.
d. With a smaller sample size the sampling distribution would become more
spread out (will be shorter and flatter). Therefore any inference about the population mean (mu)
based on the sample mean is very unreliable. (larger variation implies lower quality
which implies that the result is less reliable).
f. With a larger sample size the sampling distribution would
become less spread out (will be sharper, thinner, all possible values will be closer
to the average we’ve obtained from this sample). As a result, we have an estimate for
MU which is of high quality (less variation).
Exercise 8.4
Alpha= 0.05, alpha/2 =0.025
Stdev
95% confidence interval
from Table 5, z = 1.96:
a. n=35, xbqr = 26, variance =228.2
15.10629 5.004721 20.995279 31.00472
b. n=70, xbar = 24.1, variance = 198.4
14.085453 3.299726 20.800274 27.39973
c. n=105, xbar 24.2, variance 216.9
14.727525 2.817028 21.382972 27.01703
Exercise 8.10
xbar = 19.4, s =10.1, n =105
90% confidence interval
1.6214092
17.778591 21.02141
To say “the true mean number of VP's for a random sample of U.S firms would be
between 17 and 21 with a confidence interval of 90%” is not, statistically
speaking, a correct statement. Here is a correct one: We have at least 90%
confidence that this interval CONTAINS the true mean (i.e., the population mean).
Learn that, what is random is not the population mean, but the interval
we have constructed. Since this interval is based on a specific random sample.
The question is that are we sure that this interval captures the true mean?
Who knows, but we are sure that if you repeat the sampling 100 times and
construct a 90% confidence interval for MU based of each of these sample means,
then we are sure that at least 90 out of the 100 confidence interval we
have constructed contain the true mean.
I read the Definition 8.2 very slowly and carefully.
Exercise 8.16
Interval estimation procedure may not be applicable when the sample size is small.
Since the Central (which means, the average, the most widely used population parameter)
Limit (which means as sample size gets larger and larger)
Theorem (to make it sounds academic), does not apply
we can not assume that the sampling distribution is normal,
therefore cannot construct confidence interval as a measure
of accuracy for your estimate. The sample standard deviation
could be computed as usual but it is not a “good”
estimation for to the unknown population standard deviation.
Exercise 8.18
Mean
Stdev
a. 90% C.I. (using t-table, d.f. =4),
t=2.132
b. 99% confidence interval
Notice that the more confidence you
want, you must lower your expectation
(you get wider interval).
Confidence and Precision are inversely
related.
Exercise 8.40
a. 90% confidence interval
b. 95% confidence interval
c. 99% confidence interval
7
4
2
5
7
5
2.98
4
1
9
0
4
18
2.121
7.02
0.63
9.37
-8
3.0333333
1.7416467
-10.86501 -5.13499
-11.41363 -4.58637
-12.49345 -3.50655
Exercise 8.46
a. 90% confidence interval
0.38
0.0174
0.1319091
0.1630096 0.59699
b. Can the manufacturing process be established locally?
No since the C.I. does not contain ZERO (which means no difference).
So the difference is significant. The voltage reading at the new location
is smaller than at the old one.
Exercise 8.54
48
50
47
50
63
65
55
Mean
Stdev
t .05 , 6df
90% confidence interval
54
56
50
55
64
65
61
2.544836
1.943
-5.726032 -1.98825
-6
-6
-3
-5
-1
0
-6
-3.857143
Exercise 8.84
Calculate 90% confidence interval
a. s=2.5, n=50
90% CI
306.25
4.5367144
0.05
67.5048
8.80935
0.95
34.7642 df =50
b. s=.02, n=15
0.0056 23.6848
6.57063 df =14
90% CI
0.0002364 0.000852
c. s=31.6, n=22
20969.76 32.6705
11.5913 df = 21
90% CI
641.85611 1809.095
d. s=1.5, n=5
9 9.48773 0.710721 df=4
90% CI
0.9485936 12.6632
The conditions (NOT assumptions!) that the population from which the sample is selected is normally distributed,
and the sample is random must be satisfied for the C.I. to be valid.