Transcript Hwk 1 Soln

Hw 1 Prob 1
A sample of 20 cigarettes is tested to determine nicotine content and the average
value observed was 1.2 mg. Compute a 99 percent two-sided confidence interval
for the mean nicotine content of a cigarette if it is known that the standard
deviation of a cigarette’s nicotine content is σ = 0.2 mg.
1    P{ X  Z / 2

1   C . I . given by
X  z / 2

n
 1.2  2.575
0.2
20
 (1.08   1.32)

n
   X  Z / 2

n
}
Hw 1 Prob 2
•
In Problem 1, suppose that the population variance is not known in advance
of the experiment. If the sample variance is 0.04, compute a 99% two-sided
confidence interval for the mean nicotine content.
In problem 1, we knew  = 0.2. Now we must estimate  with the sample
standard deviation s=.2.
1    P{ X  t / 2 ,n1
s
s
   X  t / 2 ,n1
}
n
n

1   C . I . given by
s
X  t / 2 ,n1
n
 1.2  2.861
0.2
20
 (1.07   1.33)
Note: this is just slightly larger than
C.I. in problem one because we have
a fairly large sample. For large n, the
difference between the normal and t
distributions starts to disappear.
Hw 1 Prob 3
•
The daily dissolved oxygen concentration for a water stream has been
recorded over 30 days. If the sample average of the 30 values is 2.5 mg/liter
and the sample standard deviation is 2.12 mg/liter, determine a value which,
with 90 percent confidence, exceeds the mean daily concentration.
Since n=30, we could choose to use a t-distribution with 29 degrees of
freedom or a standard normal distribution. The difference will be small. I will
use the standard normal. For a standard normal a confidence interval is given
by
X  z / 2
s
n
 1.2 1.645
02.12
30
 (1.86    3.14)
Note that the way the problem, we are more likely interested only in an upper
limit. In this case, we could conduct a one sided confidence interval. In this
case, we have
X  z
s
n
 2.5 1.282
 (   2.996)
02.12
30
Hw 1 Prob 3
•
The daily dissolved oxygen concentration for a water stream has been
recorded over 30 days. If the sample average of the 30 values is 2.5 mg/liter
and the sample standard deviation is 2.12 mg/liter, determine a value which,
with 90 percent confidence, exceeds the mean daily concentration.
Since n=30, we could choose to use a t-distribution with 29 degrees of
freedom or a standard normal distribution. The difference will be small. I will
use the standard normal. For a standard normal a confidence interval is given
by
X  z / 2
s
n
 2.5 1.282
02.12
30
 (   2.996)
If we do a one sided C.I. using the t distribution with 29 degrees of freedom,
we get
s
X  t ,n1
n
 2.5 1.311
 (   3.007 )
02.12
30
Hw 1 Prob 4
•
Suppose that when sampling from a normal population having an unknown
mean µ and unknown variance σ2, we wish to determine a sample size n so as
to guarantee that the resulting 100(1 – α) percent confidence interval for µ
will be of size no greater than A, for given values α and A. Explain how we can
approximately do this by a double sampling scheme that first takes a
subsample of size 30 and then chooses the total sample size by using the
results of the first subsample.
1   C . I . given by
s
X  t / 2 ,n1
n
which means the precision is
E  t / 2 ,n1
s
t
s
or n    / 2 ,n1 
n
 E 
2
Since we do not know s until we collect data, we cannot know the t value to
look up. Iterative solution
1.
2.
3.
4.
5.
Estimate a sample size n
Collect data and calculate s
If we have achieved the desired precision, stop. If not, calculate a sample
size based on formula above.
Collect additional data and calculate new s.
Go to step 3.