### Notes2

```One- and Two-Sample Estimation Problems
Classical Methods of Estimation
A point estimate of some population parameter  is a
single value of a statistic.
Definition 9.1 A statistic is said to be an unbiased
estimator of the parameter  if
 = E() = .
Definition 9.2 If we consider all possible unbiased
estimators of some parameter , the one with the
smallest variance is called the most efficient estimator
of .
The interval L <  < U, computed from the selected
sample, is called a (1- )100% confidence interval, the
fraction 1- is called the confidence coefficient or the
degree of confidence, and the endpointsL and U,
are called the lower and upper confidence limits.
“
Ideally we prefer a short interval with a high degree of confidence”
Single Sample: Estimating the Mean
Confidence Interval of ;  Known. If x is the mean of
a random sample of size n from a population with
known variance 2, a (1 - )100% confidence interval
for  is given by


x - z/2 ------ <  < x + z/2 --------,
n
n
where z/2 is the z-value leaving an area of /2 to the
right.
4/252) An electrical firm manufactures light bulbs that have a
length of life that is approximately normally distributed with a
standard deviation of 40 hours. If a sample of 30 bulbs has an
average life of 780 hours, find a 96% confidence interval for the
population mean of all bulbs produced by this firm.
Soln:
Given: n = 30,  = 40 hours,x = 780 hours
96% confidence interval
From Table A.3, z/2 = z.02 = 2.05
__
__
x - z/2 /n <  < x + z/2 /n
__
__
780 – 2.05(40)/30 <  < 780 + 2.05(40)/30
765 <  < 795
Theorem 9.1 Ifx is used as an estimate of , we can
then be (1 - )100% confident that the error will not
exceed
_
z/2/n.
Error
_
x - z/2n
x

_
x + z/2n
6/252) The heights of a random sample of 50 college students showed
a mean of 174.5 centimeters and a standard deviation of 6.9
centimeters. Construct a 98% confidence interval for the mean height
of all college students. What can we assert with 98% confidence about
the possible size of our error if we estimate the mean height of all
college students to be 174.5 centimeters?
Soln:
Given: n = 50,  = 6.9 cm,x = 174.5 cm
a) 98% confidence interval
From Table A.3, z/2 = z.01 = 2.33
__
__
x - z/2 /n <  < x + z/2 /n
__
__
174.5 – 2.33(6.9)/50 <  < 174.5 + 2.33(6.9)/50
172.23 <  < 176.77
__
b) Error  2.33(6.9)/50
 2.27 cm
Theorem 9.2 Ifx is used as an estimate of , we can
be (1-)100% confident that the error will not exceed
a specified amount e when the sample size is
n = (z/2/e)2
8/253) How large a sample is needed in Exercise 4 if we wish to be
96% confident that our sample mean will be within 10 hours of the
true mean?
Soln:
Given: z/2 = 2.05,  = 40 hours, e = 10 hours
n = (z/2 / e)2
n = (2.05(40) / 10)2
n  68 samples
10/253) An efficiency expert wishes to determine the average time
that it takes to drill three holes in a certain metal clamp. How large
a sample will he need to be 95% confident that his sample mean
will be within 15 seconds of the true mean? Assume that it is known
from previous studies that  = 40 seconds.
Soln:
Given: z/2 = z.025 = 1.96,  = 40 seconds, e = 15 seconds
n = (z/2 / e)2
n = (1.96(40) / 15)2
n  28 samples
Confidence Interval for ;  Unknown. Ifx and s are the
mean and standard deviation of a random sample from a
normal population with unknown variance 2, a (1 –
)100% confidence interval for  is given by
_
_
x - t/2 s/n <  < x + t/2 s/n,
were t/2 is the t-value with  = n –1 degrees of freedom,
leaving an area of /2 to the right.
13/253) A machine is producing metal pieces that are cylindrical in
shape. A sample of pieces is taken and the diameters are 1.01, 0.97,
1.03, 1.04, 0.99, 0.98, 0.99, 1.01, and 1.03 centimeters. Find a 99%
confidence interval for the mean diameter of pieces from this machine,
assuming an approximate normal distribution.
Soln:
1.01 + 0.97 + 1.03 + 1.04 + 0.99 + 0.98 + 0.99 + 1.01 + 1.03
x = -------------------------------------------------------------------------9
x = 1.006
9
s2 =  (xi -x)2/(n-1) = 0.0048 / 8 : Therefore, s = 0.024
i=1
From Table A.4, t/2 = t .005 = 3.355
__
__
x - t/2 s/n <  < x + t/2 s/n,
__
__
1.006 – 3.355(.024)/9 <  < 1.006 + 3.355(.024)/9
16/253) A random sample of 12 graduates of a certain secretarial
school typed an average of 79.3 words per minute with a standard
deviation of 7.8 words per minute. Assuming a normal distribution
for the number of words typed per minute, find a 95% confidence
interval for the average number of words typed by all graduates of
this school.
Soln:
Given: n = 12, x = 79.3 words/minute, s = 7.8 words/minute
@ 95% confidence interval
From Table A.4, t/2 = = t .025 = 2.201
__
__
x - t/2 s/n <  < x + t/2 s/n,
__
__
79.3 – 2.201(7.8)/12 <  < 79.3 + 2.201(7.8)/12
74.344 <  < 84.256
Standard Error of a Point Estimate
_
The standard error ofX is /n. Simply put, the
standard error of an estimator is its standard deviation.
For the case ofX, the computed confidence limit
_
x  z/2  /n
is written as
x  z/2 s.e.(x), where s.e. is the standard error.
In the case where  is unknown and sampling is from a
normal distribution, s replaces  and the estimated
standard error s/n is involved. Thus the confidence
limits on  are given by
_
x  t/2 s/n = x  t/2s.e.(x).
Tolerance limits. For a normal distribution of
measurements with unknown mean  and unknown
standard deviation , tolerance limits are given byx 
ks, where k is determined so that one can assert with
100(1 - )% confidence that the given limits contain at
least the proportion 1 -  of the measurements.
18/253) The following measurements were recorded for the drying time, in hours,
of a certain brand of latex paint:
3.4
2.5
4.8
2.9
3.6
2.8
3.3
5.6
3.7
2.8
4.4
4.0
5.2
3.0
4.8
Assuming that the measurements represent a random sample from a normal
population, find the 99% tolerance limits that will contain 95% of the drying times.
Soln:
15
x =  xi / n = 3.787
i=1
s2
15
=  (xi -x)2/(n-1) : Therefore, s = 0.971
i=1
@ n = 15, (1- ) = 0.99, (1- ) = 0.95
From Table A.7, k = 3.507
x  ks
3.787  3.507(0.971)
3.787  3.405
Therefore, the tolerance limit is from 0.382 to 7.192 hours.
Two Samples: Estimating the Difference Between Two
Means
Confidence Interval For 1 - 2; 21 and 22 Known. Ifx1
andx2 are the means of independent random samples of
size n1 and n2 from populations with known variances 21
and 22, respectively, a (1 - )100% confidence interval for
1 - 2 is given by
_________
__________
(x1-x2)-z/221/n1+22/n2 < 1-2 < (x1-x2)+z/221/n1+22/n2,
where z/2 is the z-value leaving an area of /2 to the
right.
2/263) Two kinds of thread are being compared for strength. Fifty pieces of each
type of thread are tested under similar conditions. Brand A has an average tensile
strength of 78.3 kilograms with a standard deviation of 5.6 kilograms, while brand
B had an average tensile strength of 87.2 kilograms with a standard deviation of
6.3 kilograms. Construct a 95% confidence interval for the difference of the
population means.
Soln:
Given: Brand A: nA = 50, A = 5.6 kg,xA = 78.3 kg
Brand B: nB = 50, B = 6.3 kg,xB = 87.2 kg
@ 95% confidence interval
From Table A.3, z/2 = z.025 = 1.96
___________
___________
(xB-xA) - z/22A/nA+2B/nB <B-A< (xB-xA) + z/22A/nA+2B/nB,
________________
(87.2–78.3)–1.96(5.6)2/50 + (6.3)2/50 < B-A <
________________
(87.2–78.3)+1.96(5.6)2/50 + (6.3)2/50,
6.56 < B - A < 11.24
3/263) A study was made to determine if a certain metal treatment has any effect
on the amount of metal removed in a pickling operation. A random sample of 100
pieces was immersed in a bath of 24 hours without treatment, yielding an average
of 12.2 millimeters of metal removed and a sample standard deviation of 1.1
millimeters. A second sample of 200 pieces was exposed to the treatment followed
by the 24-hour immersion in the bath, resulting in an average removal of 9.1
milliliters of metal with a sample standard deviation of 0.9 millimeter. Compute a
98% confidence interval estimate for the difference between the population means.
Does the treatment appear to reduce the mean amount of metal removed?
Soln:
Sample1
n1 = 100
x1 = 12.2 ml
1 = 1.1 ml
@ 98% confidence interval,
From Table A.3, z/2 = z.01 = 2.33
Sample 2
n2 = 200
x2 = 9.1 ml
2 = 0.9 ml
(x1-x2)
_______________
- z/221/n1 + 22/n2 < 1 - 2 <
_____________
(x1-x2) + z/221/n1 + 22/n2,
__________________
(12.2–9.1)–2.33 (1.1)2/100 + (0.9)2/200 < 1-2 <
__________________
( 12.2–9.1)+2.33(1.1)2/100+(0.9)2/200
2.804 < 1-2 < 3.396
Confidence Interval for 1 - 2; 21 = 22 but Unknown.
Ifx1 andx2 are the means of independent random
samples of size n1 and n2, respectively, from approximate
normal populations with unknown but equal variances, a
(1 - )100% confidence interval for 1 - 2 is given by
_______
________
(x1-x2) - t/2sp1/n1+1/n2 < 1-2 < (x1 -x2) + t/2 sp1/n1+1/n2,
where sp is the pooled estimate of the population
standard deviation, given as
(n1 – 1)s21 + (n2 –1)s22
s2p = --------------------------------n1 + n2 - 2
and t/2 is the t-value with  = n1 + n2 –2 degrees of
freedom, leaving an area of /2 to the right.
8/264) An experiment reported in Popular Science, in 1981,
compared fuel economies for two types of similarly equipped
diesel mini-trucks. Let us suppose that 12 Volkswagen and 10
Toyota trucks are used in 90-kilometer per hour steady-speed tests.
If the 12 Volkswagen trucks average 16 kilometers per liter with a
standard deviation of 1.0 kilometer per liter and the10 Toyota
trucks average 11 kilometers per liter with a standard deviation of
0.8 kilometer per liter, construct a 90% confidence interval for the
difference between the average kilometers per liter of these two
mini-trucks. Assume that the distances per liter for each truck
model are approximately normally distributed with equal
variances.
Soln:
Volkswagen
n1 = 12
x1 = 16 km/l
s1 = 1.0 km/l
Toyota
n2 = 10
x2 = 11 km/l
s2 = 0.8 km/l
@ 90% confidence interval for the difference 2 - 1
From Table A.4 , @  = 20, t/2 = t.05 = 1.725
(12 –1)12 + (10-1)(0.8)2
s2p = --------------------------------- = 0.838
12 + 10 - 2
sp = 0.915
________
(16 - 11) – 1.725(0.915)1/12+1/10 < 2 - 1 <
_________
(16 - 11) + 1.725(0.915)1/12+1/10,
4.324 < 2 - 1 < 5.676
Confidence Interval for 1 - 2; 21  22 and Unknown. Ifx1 and
s21, andx2 and s22, are the means and variances of small
independent samples of size n1 and n2, respectively, from
approximate normal distributions with unknown and unequal
variances, an approximate (1 - )100% confidence interval for 1 2 is given by
__________
__________
(x1 -x2) - t/2s21/n1+s22/n2 < 1-2 < (x1 -x2) + t/2s21/n1+s22/n2,
where t/2 is the t-value with
(s21/n1 + s22/n2)2
v = --------------------------------------------------[(s21/n1)2/(n1 – 1)] + [(s22/n2)2/(n2 – 1)]
degrees of freedom, leaving an area of /2 to the right.
Paired Observations
Confidence Interval for D = 1 - 2 for Paired Observations. If d
and sd are the mean and standard deviation of the normally
distributed differences of n random pairs of measurements, a
(1 - )100% confidence interval for D = 1 - 2 is
_
_
d - t/2 (sd / n) <  < d + t/2 (sd / n),
where t/2 is the t-value with  = n –1 degrees of freedom, leaving
an area of /2 to the right,
and
_________________
|
n
n
| n  d2i – (  di )2
i=1
i=1
|
sd =
| ------------------------
n(n-1)
Single Sample: Estimating a Proportion
Large-Sample Confidence Interval for p. If p is the proportion of
successes in a random sample of size n, andq = 1 -p, an
approximate (1 - )100% confidence interval for the binomial
parameter p is given by
_______
_______
p - z/2pq / n < p <p + z/2pq /n ,
where z/2 is the z-value leaving an areas of /2 to the right.
Theorem 9.3 Ifp is used as an estimate of p, we can be
(1- )100% confident that the error will not exceed
______
z/2pq / n
Theorem 9.4 Ifp is used as an estimate of p, we can be
(1 - )100% confident that the error will be less than a
specified amount e when the sample size is
approximately
n = z2/2pq / e2
Theorem 9.5 If p is used as an estimate of p, we can be
at least (1 - )100% confident that the error will not
exceed a specified amount e when the sample size is
n = z2/2 / 4e2
Two Samples: Estimating the Difference B/n Two Proportions
Large-Sample Confidence Interval for p1-p2. Ifp1 andp2 are the
proportion of successes in random samples of size n1 and n2,
respectively,q1 = 1-p1 and q2 = 1-p2, an approximate
(1- )100% confidence interval for the difference of two binomial
parameters p1-p2, is given by
________________
(p1 - p2) - z/2p1q1/n1 +p2q2/n2 < p1 – p2 <
__________________
(p1 -p2) + z/2p1q1/n1 + p2q2 / n2,
where z/2 is the z-value leaving an area of /2 to the right.
Single Sample: Estimating the Variance
Confidence Interval for 2. If s2 is the variance of a random sample
of size n from a normal population, a (1 - )100% confidence
interval for 2 is given by
(n –1)s2/ 2/2 < 2 < (n - 1)s2/ 21-/2,
where 2/2 and 21-/2 are 2-values with  = n –1 degrees of
freedom, leaving areas of /2 and 1- /2, respectively, to the right.
Two Samples: Estimating the Ratio of Two Variances
Confidence Interval for 21/22. If s21 and s22 are the variances of
independent samples of size n1 and n2, respectively, from normal
populations, then a (1 - )100% confidence interval for 21/22 is
(s21/s22)(1/f/2(1, 2)) < 21/22 < (s21/s22)f/2(1, 2),
where f/2(1, 2) is an f-value with 1 = n1 –1 and 2 = n2 –1
degrees of freedom leaving an area of /2 to the right, and f/2(1,
2) is a similar f-value with 2 = n2 – 1 and 1 = n1 –1 degrees of
freedom.
```