Transcript 6.1 Confidence Intervals for the Mean (Large Samples) Statistics Spring Semester Mrs. Spitz Objectives/Assignment     How to find a point estimate and a maximum error of estimate How.

6.1 Confidence Intervals for
the Mean (Large Samples)
Statistics
Spring Semester
Mrs. Spitz
Objectives/Assignment




How to find a point estimate and a maximum
error of estimate
How to construct and interpret confidence
intervals for the population mean
How to determine the required minimum
sample size when estimating 
Assignment: pp. 259-261 #1-40 all
Estimating Population Parameters

In this chapter, you will learn an important technique
of statistical inference—to use sample statistics to
estimate the value of an unknown population
parameter. In this particular section, you will learn
how to use sample data to make an estimate of the
population parameter  when the sample size is at
least 30 or the standard deviation  is known. To
make such an inference, begin by fnding a point
estimate.
DEFINITION

A point estimate is a single value estimate for
a population parameter. The most unbiased
point estimate of the population mean  is the
sample mean x .
Ex. 1: Finding a Point Estimate

Market researchers use the number of sentences per advertisement as a
measure of readability for magazine advertisements. The following
represents a random sample of the number of sentences found in 54
advertisements. Find a point estimate of the population mean .
x
671
x 

 12 .4
n
54
So, your point estimate for the mean length of all
magazine advertisements is 12.4 sentences.
You try it: Finding a Point Estimate

Another random
sample of the number
of sentences found in
30 magazine
advertisements is listed
below. Use this sample
to find another point
estimate of the
population mean .


Find the sample mean.
Estimate the mean sentence
length of the population.
Number of Sentences
16
99
17
9
18
6
14
13
11
11
12
17
17
5
18
12
9
20
6
5
14
11
7
18
11
6
12
4
12
13
xx 443

xx 


14
.
767
nn
30
More about point estimates

In Example 1, the probability that the
population mean is exactly 12.4 is virtually
zero. So, instead of estimating  to be exactly
12.4, you can increase your accuracy by
estimating that it lies within an interval. This
is called “making an interval estimate.”
More about interval estimates

Although you can assume that the point estimate in Example 1 is not
equal to the actual population mean, it is probably close to it. To form an
interval estimate, use the point estimate as the center of the interval, then
add and subtract a margin of error. For instance, if the margin of error is
2.1, then an interval estimate would be given by 12.4 ± 2.1 or 10.3 <  <
14.5. The point estimate and interval estimate are as follows.
Before finding an interval estimate, you should first
determine how confident you need to be that your interval
contains the population mean .
Study Tip:

In this course, you will usually use 90%, 95%, and
99% levels of confidence. The following z-scores
correspond to these levels of confidence.
Level of Confidence
zc
90%
1.645
95%
1.96
99%
2.575
DEFINITION

The level of confidence, c, is the probability
that the interval estimate contains the
population parameter.
Critical Values

You know from the Central Limit Theorem
that when n ≥ 30, the sampling distribution of
sample means is a normal distribution. The
level of confidence, c, is the area under the
standard normal curve between the critical
values, -zc and zc.
Critical Values

You can see from the graph that c is the percent of area under
the normal curve between -zc and zc. The area remaining is 1
– c, so the area in each tail is ½ (1 – c). For instance, if c =
90%, then 5% of the area lies to the left of -zc = -1.645 and
5% lies to the right of zc = 1.645
Error of Estimate

The distance between the point estimate and the actual
parameter value is called the error of estimate. When
estimating , the error of estimate is the distance
|x|

In most cases, of course,  is unknown and x varies from
sample to sample. However, you an calculate a maximum
value for the error if you know the level of confidence and the
sampling distribution.
DEFINITION

Given a level of confidence, the maximum error of estimate
(or error tolerance), E, is the greatest possible distance
between the point estimate and the value of the parameter it is
estimating.
E  zc c 


n
When n ≥ 30, the sample standard deviation s can be used in
place of .
Ex. 2: Finding the Maximum Error of
Estimate

Use the data in Example 1 and a 95%
confidence interval to find the maximum error
of estimate for the number of sentences in a
magazine advertisement.
Ex. 2: Finding the Maximum Error of
Estimate--Solution

The z-score that corresponds to a 95%
confidence interval is 1.96. This implies that
95% of the area under the curve falls within
1.96 standard deviations of the mean. You
don’t know the populations standard deviation
. But because n ≥ 30, you can use s in place
of .
Ex. 2: Finding the Maximum Error of
Estimate--Solution
( x  x ) 2
s
n 1
1333.2
s
 5.0
53
Using the values zc =
1.96, s ≈5.0, and n = 54.
E  zc
E  1.3
s
5.0
 1.96 
n
54
You are 95% confident
that the maximum error
of estimate is about 1.3
sentences per magazine
advertisement.
Confidence Intervals for the Population Mean

Using a point estimate and a maximum error
of estimate, you can construct an interval
estimate of a population parameter such as .
This interval estimate is called a confidence
interval.
Ex. 3: Constructing a Confidence
Interval


Construct a 95% confidence interval for the
mean number of sentences in a magazine
advertisement.
Solution: In examples 1 and 2, you found that
x = 12.4 and E = 1.3. The confidence inteval
is as follows:
Insight:

A larger sample size tends to give you a
narrower confidence interval—for the same
level of confidence.
Ex. 4: Constructing a Confidence
Interval using Technology


Use a graphing calculator to construct a 99%
confidence interval for the mean number of
sentences in a magazine advertisement, using the
sample in Example 1.
Solution: To use a calculator to solve the problem,
enter the data and find that the sample standard
deviation is s≈ 5.0. Then, use the confidence
interval command to calculate the confidence
interval (ZInterval for the TI-83 or TI-84). The
display should look like the one on the next slide.
http://calculator.maconstate.edu/confidence_int_z/index.html#
So, a 99% confidence interval for  is (10.7, 14.2). With
99% confidence, you can say that the mean number of
sentences is between 10.7 and 14.2.
More about intervals

In Ex. 4, and Try it Yourself 4, the same
sample data was used to construct intervals
with different levels of confidence. Notice
that as the level of confidence increases, the
width of the confidence interval also
increases. In other words, using the same
sample data, the greater the level of
confidence, the wider the interval.
In Ex. 5, notice that if  is known, then the
sample size can be less than 30.

A college admissions director wishes to estimate the
mean age of all students currently enrolled. In a
random sample of 20 students, the mean age is found
to be 22.9 years. From past studies, the standard
deviation is known to be 1.5 years. Construct a 90%
confidence interval of the population mean age.
E  zc
E  .55
s
1.5
 1.645
n
20
In Ex. 5, notice that if  is known, then the
sample size can be less than 30.

The 90% confidence interval is as follows:
So, with 90% confidence, you can say that the mean age of the
students is between 22.35 and 23.45 years.
Interpreting the results


After constructing a confidence interval, it is important that
you interpret the results correctly. Consider the 90%
confidence interval constructed in ex. 5. Because  already
exists, it is either in the interval or not. It is not correct to say,
“There is a 90% probability that the actual mean is in the
interval (22.51, 23.29.”
The correct way to interpret your confidence interval is,
“There is a 90% probability that the confidence interval you
described contains .” This means also, of course, that there
is a 10% probability that your confidence interval WILL NOT
contain .
Sample Size

As the level of confidence increases, the confidence
interval widens. As the confidence interval widens,
the precision of the estimate decreases. One way to
improve the precision of an estimate without
decreasing the level of confidence is to increase the
sample size. But how large a sample size is needed
to guarantee a certain level of confidence for a given
maximum error of estimate?
Ex. 6: Determining a Minimum Sample Size
You want to estimate a mean number of sentences in a magazine
advertisement. How many magazine advertisements must be
included in the sample if you want to be 95% confident that the
sample mean is within one sentence of the population mean?
Ex. 6: Determining a Minimum Sample Size
Using c = 0.95, zc = 1.96, s ≈5.0 (from example 2), and E
= 1, you can solve for the minimum sample size, n.
 zcs 
 1.96  5.0 
n
 
  96.04
1


 E 
2
2
When necessary, round up to obtain a whole number. So,
you should include at least 97 magazine advertisements in
your sample. (You already have 54, so you need 43 more.)