Download 7.4 Estimating a Population Mean

7.4 Estimating a Population Mean –
Standard Deviation Not Known:
Objectives:
1. Use the Student t distribution.
2. Construct a confidence interval for population mean when standard deviation is not known.
Overview:
This section presents methods for estimating a population mean when the population standard deviation
is not known. With σ unknown, we use the Student t distribution assuming that the relevant
requirements are satisfied.
Sample Mean as a Point Estimate:
1. For all populations, the sample mean x is an unbiased estimator of the population mean µ,
meaning that the distribution of sample means tends to center about the value of the population
mean µ.
2. The sample mean x is the best point estimate of the population mean µ.
Student t Distribution:
If the distribution of a population is essentially normal, and the standard deviation is not known, then the
Student t distribution is used to estimate the population mean. The distribution
t=
x−µ
s
n
is a Student t distribution for all samples of size n. It is often referred to as a t distribution and is used
to find critical values denoted by tα/2. Table A-3 is used to find t values.
Because we do not know the value of the population standard deviation, we use the sample stamndard
deviation instead. This introduces another source of unreliability, especially with small samples. In
order to maintain a desired confidence level such as 95%, we compensate for this additional unreliability
by making the confidence interval wider. The Student t distribution satisfies this requirement.
Degrees of Freedom:
The number of degrees of freedom for a collection of sample data is the number of sample values that
can vary after certain restrictions have been imposed on all data values. The degree of freedom is often
abbreviated df.
Degrees of freedom = n – 1 will be used in this section.
Example: If the sum of n = 5 data values is 100, then 4 of these values may take on any value. The last
data value MUST make the sum 100 and is therefore not have the freedom to be any value. Therefore
the degree of freedom is: n - 1 = 5 - 1 = 4.
Margin of Error:
When data from a simple random sample are used to estimate a population mean, the margin of error,
denoted by E, is the maximum likely difference (with probability 1 – α, such as 0.95) between the
observed mean and the true value of the population mean.
The margin of error E is also called the maximum error of the estimate and can be found by multiplying
the critical value and the standard deviation divided by the square root of the sample size:
E = tα 2 ⋅
s
n
where ta/2 has n – 1 degrees of freedom.
Constructing a Confidence Interval for Estimating a Population Mean
(with σ Unknown):
Notation:
µ = population mean
s = sample standard deviation
x = sample mean
n = number of sample values
E = margin of error
tα/2 = z score separating an area of a/2 in the right tail of the standard normal distribution
Requirements:
1. The sample is a simple random sample.
2. Either the sample is from a normally distributed population or n > 30.
Confidence Interval:
The confidence interval may be written as:
x−E<µ<x+E
where
E = tα 2 ⋅
s
n
The confidence interval is often expressed in the following equivalent forms:
x±E
or
( x − E, x + E)
Procedure for Constructing Confidence Intervals:
1.
2.
3.
4.
Verify that requirements are met.
Using n-1 df, find the critical t value for the confidence level.
Find the margin of error.
Determine confidence interval.
Example: A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was
225 milligrams with s = 15.7 milligrams. Construct a 95% confidence interval for the true mean
cholesterol content of all such eggs.
Solution: Assume requirements are met. The t-score corresponding to a 95% confidence level and 11
degrees of freedom is 2.201.
The margin of error is:
s
n
 15.7 
E = 2.201

12


E = 9.975
E = tα 2 ⋅
The confidence interval x − E < µ < x + E is:
225 − 9.975 < µ < 225 + 9.975
215.025 < µ < 234.975
215 < µ < 235
We can be 95% confident that the actual mean will fall between 215 and 235.
Example: Thirty randomly selected students took the calculus final. If the sample mean was 83 and the
standard deviation was 13.5, construct a 99% confidence interval for the mean score of all students.
Solution: Assume requirements are met. The t-score corresponding to a 99% confidence level and 29
degrees of freedom is 2.756.
The margin of error is:
E = tα 2 ⋅
s
n
 13.5 
E = 2.756

 30 
E = 6.792
The confidence interval x − E < µ < x + E is:
83 − 6.792 < µ < 83 + 6.792
76.208 < µ < 89.792
76 < µ < 90
We can be 99% confident that the actual mean will fall between 76 and 90.
Example: A sociologist develops a test to measure attitudes towards public transportation, and 27
randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is
21.4. Construct the 95% confidence interval for the mean score of all such subjects.
Solution: Assume requirements are met. The t-score corresponding to a 95% confidence level and 26
degrees of freedom is 2.056.
The margin of error is:
E = tα 2 ⋅
s
n
 21.4 
E = 2.056

 27 
E = 8.467
The confidence interval x − E < µ < x + E is:
76.2 − 8.467 < µ < 76.2 + 8.467
67.733 < µ < 84.667
68 < µ < 85
We can be 95% confident that the actual mean will fall between 68 and 85.
Example: A savings and loan association needs information concerning the checking account balances
of its local customers. A random sample of 14 accounts was checked and yielded a mean balance of
$664.14 and a standard deviation of $297.29. Find a 98% confidence interval for the true mean checking
account balance for local customers.
Solution: Assume requirements are met. The t-score corresponding to a 98% confidence level and 13
degrees of freedom is 2.65.
The margin of error is:
s
n
 297.29 
E = 2.65

14


E = 210.553
E = tα 2 ⋅
The confidence interval x − E < µ < x + E is:
664.14 − 210.553 < µ < 664.14 + 210.553
453.587 < µ < 874.693
454 < µ < 875
We can be 95% confident that the actual mean will fall between 454 and 875.
Important Properties of the Student t Distribution:
1. The Student t distribution is different for different sample sizes (see the following graph,
for the cases n = 3 and n = 12).
2. The Student t distribution has the same general symmetric bell shape as the standard normal
distribution but it reflects the greater variability (with wider distributions) that is expected with
small samples.
3. The Student t distribution has a mean of t = 0 (just as the standard normal distribution has a mean
of z = 0).
4. The standard deviation of the Student t distribution varies with the sample size and is greater
than 1 (unlike the standard normal distribution, which has a s = 1).
5. As the sample size n gets larger, the Student t distribution gets closer to the normal distribution.
Finding the Point Estimate and E from a Confidence Interval:
Sometimes we want to better understand a confidence interval that might have been obtained from a
journal article or generated by computer software. If we already know the confidence interval limits,
then the sample point estimate and E can be found as follows:
Point estimate of p:
x=
(upper confidence limit) + (lower confidence limit)
2
Margin of Error:
E=
(upper confidence limit) - (lower confidence limit)
2