6.3. Confidence intervals for the mean of population that is
normally distributed: population variance known
Let
be
a random sample of n
observation from a normal population with unknown
and
known variance
.
Let
be
the sample mean. Then
confidence
interval for the population mean with known variances is given by
,
where is the number for which and the random variable Z has a standard normal distribution.
Example:
Given
a random sample of 36 observations from a normal population for which
is unknown and
,
the sample mean is found to be
.
Construct a 95% confidence interval for .
Solution:
and 
From
we obtain that
So,
is
a 95% confidence interval for
.
It means, if sample of 36 observations are drawn repeatedly and independently from the population, then over a very large number of repeated trials, 95% of these intervals will contain the value of the true population mean.
Exercises
1. Find for each of the following confidence levels
a) 88%; b) 94%; c) 96%; d) 99%
2.
The standard deviation for a population is
.
A sample of
36 observations selected from this population gave a mean equal to 74.8.
a) Make a 90 % confidence interval for .
b) Construct a 95 % confidence interval for .
c) Determine a 99 % confidence interval for .
3.
The standard deviation for a population is
.
A sample of 121 observations selected from this population gave a
mean equal to 84.5.
a) Make a 99 % confidence interval for
b) Construct a 95 % confidence interval for
c) Determine a 90 % confidence interval for
4.
The standard deviation for a population is
.
A random sample selected from this population gave a mean equal to
78.90.
a) Make a 99 % confidence interval for assuming n =36
b) Make a 99 % confidence interval for assuming n =81
c) Make a 99 % confidence interval for assuming n =100
Explain your results.
5.
Given that a sample of size 16 from a normal distribution yielded
.
The population variance is known to be 64. Find
a) 90 % confidence interval for
b) 95 % confidence interval for
c) 99 % confidence interval for .
Answers
1. a) 1.555; b) 1.88; c) 2.055; d) 2.575; 2. a) (71.35; 78.26);
b) (70.68; 78.92); c) (69.39; 80.21); 3. a) 82.56 to 86.44; b) 83.03 to 85.97; c) 83.26 to 85.74; 4. a) 76.19 to 81.61; b) 77.09 to 80.71; c) 77.27 to 80.53; 5. a) (21.71, 28.29); b) (21.08, 28.92); c) (19.85, 30.15).
6.4. Confidence intervals for the mean of population that is
normally distributed: large sample size
Let
be
a random sample of n
observation from a normal population with unknown
and
unknown variance
.
Let
be
the sample mean. If the sample size n
is large
,
then according to the central limit theorem, for a large sample the
sampling distribution of the sample mean
is (approximately) normal irrespective of the shape of the population
from which the sample is drawn. Therefore, when the sample size is 30
or larger, we will use the normal distribution to construct a
confidence interval for
.
If
the variance of the population is unknown, it should be estimated by
the sample variance,
,
and
replaced by
in
confidence interval formula for the case when population variance is
known.
Remark:
For large sample sizes, is usually close to the true value of .
An approximate confidence interval for the population mean with unknown variance is given by
where is based on a sample of at least thirty observations, is the number for which and the random variable Z has a standard normal distribution.
Example:
A
sample of 64 observations from a large population yielded the sample
values,
and
.
Find an approximate 99 % confidence interval for
.
Solution:
First we find the standard deviation of . Because is not known, we will use as an estimator of .
.
Then
and 
Substituting all the values in the formula, the 99 % confidence interval
for is
An approximate 99 % confidence interval for is (166.4, 177.6).
Example:
Radiation
measurements on a sample of 69 microwave ovens produced
and
.
Determine a 94 % confidence interval for the mean radiation.
Solution:
Again
since
we
can use
as
an estimator of
.
Then confidence interval for the population mean with unknown variance is given by
and 
Substituting all the values in the formula we obtain
.
Thus, we can state with 94 % confidence that average radiation measure of microwave ovens is between (0.21, 0.139).
Exercises
1.
Determine a 90 % confidence interval for
if
,
,
and
.
2.
Determine a 98 % confidence interval for
if
,
,
and
.
3.
A sample of size 50 from a population yielded the sample values
,
and
Find a 95 per cent confidence interval for
.
4.
For a sample data set,
,
and
a) Construct a 95 % confidence interval for assuming n = 50.
b) Construct a 90 % confidence interval for assuming n = 50. Is the width of the 90 % confidence interval smaller than the width of the 95 % confidence interval calculated in part a? If yes, why it is so?
c) Find a 95% confidence interval for assuming n = 100. Is the width of the 95 % confidence interval for with n = 100 smaller than the width of the 95 % confidence interval for with n = 50 calculated in part a?
If so, why?
5.
a) A sample of 100 observations taken from a population produced a sample mean equal to 55.32 and a standard deviation equal to 8.4. Make a 90 % confidence interval for .
b) Another sample of 100 observations taken from the same population produced a sample mean equal to 57.40 and a standard deviation equal
to 7.5. Make a 90 % confidence interval for .
c) A third sample of 100 observations taken from the same population produced a sample mean equal to 56.25 and a standard deviation equal
to 7.9. Make a 90 % confidence interval for .
d) The true population mean for this population is 55.80. How many of the confidence intervals constructed in a-c cover this population mean and how many do not?
6. The mean annual salaries of managers at the certain company is $80 722 for males and $65 258 for females. These mean salaries are based on samples of 400 male and 200 female managers. Assume that the standard deviation of the annual salaries of male managers is $11 500 and standard deviation of the female managers is $8 400.
a) Construct a 95 % confidence interval for the mean annual salary of male managers.
b) Construct a 95 % confidence interval for the mean annual salary of female managers.
7. From a random sample of 70 high school seniors, the sample mean and standard deviation of the math scores are found to be 96 and 17 respectively. Determine a 96% confidence interval for the mean math score of all seniors in the school.
8. With a random sample of size n =144, someone proposes
to be a confidence interval for . What then is the level of confidence?
Answers
1. (84.63, 88.38); 2. (0.854, 0.876); 3. (182.2, 197.8); 4. a) (14.53, 17.47);
b) (14.76, 17.24); c) (14.96, 17.04); 5. a) 53.94 to 56.70; b) 56.17 to 58.63; c) 54.95 to 57.55; 6. a) ($79 595, $81 849); b) ($64 093, $66 422);
7. 91.83 to 100.17; 8. 0.8502 or about 85 %.