Chapter 8: Confidence Intervals

8.1 Confidence Interval for a Mean, Standard Deviation Known

Introduction

Assume we have a sample from a previous class, with sample size 36, where the mean age of

students in the class was 23.1 years old. Using this sample, can we make a statement about the

population mean of all Cabrillo College students? Assume in this situation that the population

standard deviation of the age of all Cabrillo students is known to be 4.5 years.

Definitions

Point estimate:

Confidence Interval:

Margin of error:

Hence, the margin of error in this example is

Margin of error =

Figure 8.1 demonstrates this situation for a 95% confidence interval.

For this situation, a 95% confidence interval for µ is _______________________.

Confidence Level:

Critical Values

The value 1.96 is called the critical value. The critical value is a z-score that separates the

desired area from the remaining area under the Standard Normal Curve.

Example

a)Find the critical value 𝑧𝛼 2⁄ for a confidence level of 88%.

b)Assume we have a sample of size 43, a sample mean of 44.5, and the population standard

deviation is known to be 1.3. Construct an 88% confidence interval in this situation.

Example

Find the confidence level associated with a critical value of 1.04.

Assumptions

The method we have described for constructing a confidence interval requires the following

assumptions.

1.

2.

Procedure for Constructing a Confidence Interval for µ When σ is Known

Check to be sure the assumptions are satisfied. If they are, then proceed with the following

steps.

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

OR USE A CALCULATOR…

Rounding off the final result

If you are given the value of �̅�, round the final result of the confidence interval to the same

number of decimal places as �̅�. If you are given data, then round �̅� and the final result of the

confidence interval to one more decimal place than is given in the data. Do not round off the

calculations you make along the way for this may affect the accuracy of your final answer.

Note that 95% is the most common confidence level - if a question at any point asks you to

construct a confidence interval and DOES NOT give you a confidence level, assume 95%

confidence is desired.

Example

Find a 95% confidence interval for the starting salary of college graduates who have taken a

statistics class, given a simple random sample of 28 such college graduates who have a mean

salary of $45,678 and the population standard deviation is known to be $9900. Assume this

population is normally distributed.

Solution

We check the assumptions. We find that

More confidence means a bigger margin of error

Example

In a sample of 150 households, the sample mean number of personal computers was 1.32.

Assume the population standard deviation is 0.41 . Construct a confidence interval for the

mean number of personal computers per household at the

a)90% confidence level

b)98% confidence level

We check the assumptions. We find that we have a problem…

Let’s make the proper assumption…(so just assume for now that we are good to go)

Example

The volume of beverage in 6 12oz bottles was measured. The volumes found were:

12.3 12.3 12.2 12.4 12.2 12.5

Assume that the volumes of beverage in 12 oz bottles is normally distributed with a population

standard deviation of 0.35 oz. Construct a 97% confidence interval for the mean volume of

beverage in 12 oz. bottles. What does this suggest about 12 oz. bottles?

Solution

The confidence level measures the success rate of the method used to construct the interval

We cannot be sure whether we got one of the confidence intervals that covers µ, or whether we

were unlucky enough to get one of the unsuccessful ones. What we do know is that our

confidence interval was constructed by a method that succeeds 95% (or whatever the confidence

level is) of the time. The confidence level describes the success rate of the method used to

construct a confidence interval, not the success of any particular interval. See picture page 349.

More Confidence Interval Fun:

Example

Suppose you are told that a 95% confidence interval estimate for the mean weight of a peanut is

(0.35,0.47).

a)What is the point estimate for the mean weight of a peanut?

b)What is the margin of error for this confidence interval?

c)Could (0.30,0.52) be a 90% confidence interval or a 99% confidence interval for the mean

weight of a peanut?

d)If the Planters company makes the claim that the mean weight of their peanuts is 0.46 grams,

would that be a problem? What if they claim that the mean weight of their peanuts is 0.50

grams?

Finding the Necessary Sample Size

We have seen that we can make the margin of error smaller if we reduce _________________.

We can also reduce the margin of error by increasing the _________________. The formula for

the margin of error is

m =

We can manipulate the formula using algebra to express the sample size n in terms of the margin

of error m.

With this formula, if we know how small we want the margin of error to be, we can compute the

sample size needed to achieve the desired margin of error.

If the value of n given by the formula is not a whole number, then round it up to the

nearest whole number.

Example

Find the sample size necessary to estimate the mean number of minutes per day college students

watch TV to within 15 minutes with 96% confidence. Assume that the population standard

deviation is known to be 112.2 minutes.

Solution

𝑧𝛼 2⁄ =

σ =

m =

SO n =

We must sample______________college students to obtain a 96% confidence interval with a

margin of error of 15 minutes.

Factors that limit sample size

Since larger sample size result in narrower confidence intervals, it is natural to wonder why we

don’t always collect a large sample when we want to construct a confidence interval. Can you

think of some reason why large samples are not always feasible in practice?

Distinguish Between Confidence and Probability

Example

A survey organization drew a simple random sample of 625 households from a city of 100,000

households. The sample mean number of people per household was 2.30, and a 95% confidence

interval for the mean number of people in the city was (2.16,2.44). True or false:

a)We are 95% confident that the mean number of people in the 625 households is between 2.16

and 2.44.

b)We are 95% confident that the mean number of people per household in the city is between

2.16 and 2.44.

c)The probability is 0.95 that the mean number of people per household in the city is between

2.16 and 2.44.

d)95% of the households in the sample contain between 2.16 and 2.44 people.

Do You Know

How to construct and interpret confidence intervals for a population mean when the

population standard deviation is known?

How to find critical values for confidence intervals?

How to describe the relationship between the confidence level and the margin of error?

How to find the sample size necessary to obtain a confidence interval of a given width?

The difference between confidence and probability?

8.2 Confidence Interval for Mean, Standard Deviation Unknown

Introduction

In Section 8.1, we showed how to construct a confidence interval for the mean µ of a normal

population when the population standard deviation σ is known. The critical value is __________

because the quantity

�̅� − 𝜇

𝜎 √𝑛⁄

has a normal distribution.

In practice, it is more common that σ is unknown. When we don’t know the value of σ, we

replace it with the sample standard deviation s. However, we cannot then use 𝑧𝛼 2⁄ as the critical

value, because the quantity

�̅� − 𝜇

𝑠 √𝑛⁄

does not have a normal distribution. The distribution of this quantity is called the

degrees of freedom:

Number of degrees of freedom =

Figure 8.9 presents t distributions for several different degrees of freedom, along with a standard

normal distribution for comparison.

Summary

The Student’s t distribution has the following properties:

Finding the critical value

To find the critical value for a confidence interval with the given level, 1 − 𝛼 be the confidence

level expressed as a decimal.

𝑡𝛼 2⁄ =

Example

A simple random sample of size 12 is drawn from a normal population. Find the critical value

𝑡𝛼 2⁄ for a 96% confidence interval.

Solution

The sample size is n = 10, so the number of degrees of freedom is _________________.

Example 8.11 Finding a critical value

Find the critical value for a 99% confidence interval for a sample of size 58.

Solution

Assumptions for Constructing a Confidence Interval for µ When σ Is Unknown

1.

2.

When the sample size is small (n ≤ 30), we must check to determine whether the sample comes

from a population that is approximately normal. A simple method is to draw a dotplot or boxplot

of the sample. If there are no outliers, and if the sample is not strongly skewed, then it is

reasonable to construct a confidence interval using the t statistic.

Constructing a Confidence Interval for µ

The ingredients for a confidence interval for a population mean µ when σ is unknown are

The margin of error is ____________________.

When the assumptions for the Student’s t distribution are met, we can use the following

procedure for constructing a confidence interval for a population mean.

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

OR USE A CALCULATOR

Example

Construct a 99% confidence interval for elbow to fingertip length of men, given a simple random

sample of size 32, with sample mean 14.50 inches, sample standard deviation 0.70 inches,

a)by Steps 1-5 above

b)by calculator

Solution

We check the assumptions. We have

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

Example

The EPA air quality standard for lead is 1.5 microgram per cubic meter. On 6 different days

following 9/11/01, measured amounts of lead in the air around New York City were:

5.40 1.10 0.42 0.73 0.48 1.10

Construct a 95% confidence interval for the mean amount of lead in the air. Let’s just do this

blindly, and then talk about the problems with this question.

Example

Would you use a z or t-distribution, or neither, to construct a confidence interval for the mean in

the following situations?

a)95% confidence level, population appears normal, s = 32.5

b)98% confidence level, n = 45, population is skewed, 3.2

c)90% confidence level, n = 9, 4.5 , population skewed

Note: use the same formula as in the last section to do sample size determination, but replace

population standard deviation with sample standard deviation.

Do You Know

The properties of the Student’s t distribution?

How to construct confidence intervals for a population mean when the population

standard deviation is unknown?

8.3 Confidence Interval for a Proportion

Introduction

Example

Suppose we take a simple random sample of 80 Cabrillo students and find that 60% of those

sampled think the cafeteria should remain open for breakfast after 10am. Could we use this

information to find a confidence interval to estimate the population proportion of students who

want the cafeteria to remain open?

This is an example of a population whose items fall into two categories.

1.

2.

We are interested in the population proportion of those who believe the cafeteria should remain

open.

Notation

______________ is the population proportion of individuals who are in a specified

category.

______________ is the number of individuals in the sample who are in the specified

category.

______________ is the sample size.

______________ is the sample proportion of individuals who are in the specified

category. It is equal to ______________.

To construct a confidence interval, we need:

The point estimate for the population proportion is

The margin of error is

The confidence interval is

Assumptions for Constructing a Confidence Interval for p

1.

2.

3.

4.

Procedure for Constructing a Confidence Interval for p

Check to be sure the assumptions are satisfied. If they are then proceed with the following steps.

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

OR AS ALWAYS, USE A CALCULATOR…

When constructing confidence intervals for a proportion, we will round the final result to

three decimal places. Only round the final result and not the calculations along the way.

Example

Suppose we take a simple random sample of 80 Cabrillo students and find that 60% of those

sampled think the cafeteria should remain open for breakfast after 10am

(a) Construct a 90% confidence interval for the proportion of Cabrillo students who think the

cafeteria should remain open for breakfast after 10am.

(b)Taher Foods claims that less than 50% of Cabrillo students think the cafeteria should remain

open past 10am for breakfast. Does the confidence interval contradict this claim?

Solution

(a) We check the assumptions. We have

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

OR BY CALCULATOR

(b)

Example

In a Gallup poll, a simple random sample of 491 adults were asked whether they are in favor of

the death penalty for a person convicted of murder and 319 said yes.

a)Find a95% confidence interval for the proportion of adults in favor of the death penalty in this

situation.

b)What is the point estimate for this proportion?

c)What is the margin of error in this confidence interval?

d)Can we conclude that the majority of adults are in favor of the death penalty in this situation?

Finding the Necessary Sample Size

If we wish to make the margin of error of a confidence interval smaller while keeping the

confidence level the same, we can do this by making the ______________ larger. Sometimes

we have a specific value m that we would like the margin of error to attain, and we wish to

compute a sample size n that is likely to give us a margin of error of that size.

Recall the formula of the margin of error:

Manipulating this formula algebraically, we can solve for n:

Therefore, sample size n depends on

We don’t know ahead of time what �̂� is going to be. There are two ways to determine the value

of �̂�.

1.

2.

Summary

Let m be the desired margin of error, and let be the critical value. The sample size n needed so

that a confidence interval for a proportion will have margin of error approximately equal to m is

n = if a value for �̂� is available

n = if no value of �̂� is available

If the value of n is not a whole number, round it up to the nearest whole number.

Note:The margin of error does not depend on the population size, only on sample size. So

we could find the margin of error for a proportion about Cabrillo students or about

California residents with the same sample size.

Example

Determine the sample size needed to estimate the population proportion with a margin of error of

0.038, a 95% confidence level, where no previous studies have been performed.

Solution

Example

60 Minutes recently had a share of 20, meaning 20% of TV’s in use at the time it aired were

turned to 60 Minutes.

a)Based on a sample size of 4000 randomly selected TV’s, construct a 97% confidence interval

estimate for the proportion of all TV sets turned to 60 Minutes when it aired.

b)What sample size would be required to estimate the percentage of TV’s tuned to 60 Minutes if

we wanted 99% confidence and a margin of error of 0.5%? (Here, we can assume that a sample

proportion is 20%, since that is the information given to us in the question).

c)At the same time 60 Minutes aired, “Exposed: Pro Wrestling” was on another network. It got a

share of 11. Can we conclude 60 Minutes had a greater proportion of viewers?

Do You Know

How to construct a confidence interval for a population proportion?

How to find the sample size necessary to obtain a confidence interval for a population

proportion of a given width?

How to construct confidence intervals with small samples?

8.4 Confidence Interval for a Standard Deviation

Introduction

When the population is normal, it is possible to construct confidence intervals for the standard

deviation or variance. These confidence intervals are based on a distribution known as the

________________________ distribution, denoted ____________________.

There are many different chi-square distributions, each with a different number of degrees of

freedom. Figure 8.13 below presents chi-square distributions for several different degrees of

freedom.

There are two important points to notice about the chi-square distribution.

1.

2.

A confidence interval for a standard deviations and variances consists of only a couple of things

which is different than confidence intervals for means and proportions.

1.

2.

The point estimate is _______________ for a population variance, 𝜎2. The critical values for a

level 100(1 – α)% confidence interval are the values that contain the middle 100(1 – α)% of the

area under the curve between them. The notation for the critical values tells how much area is to

the ________________ of the critical value. For a level 1 – α confidence interval, the critical

values are denoted

1.

2.

See Figure 8.14.

Note:You can find critical values for chi-square by table in certain situations. But the

calculator will always work…

Example

Find the critical values of 2 :

a)that contain an area of 0.05 in each tail, if n = 10

b)for a 98% confidence interval with n = 53

.

Confidence Intervals for the Variance and Standard Deviation

Let 𝑠2 be the sample variance from a simple random sample of size n from a normal population.

A level 100(1 – α)% confidence interval for the population variance 𝜎2 is

A level 100(1 – α)% confidence interval for the population standard deviation σ is

The critical values are taken from a chi-square distribution with ____________ degrees of

freedom.

Procedure for Constructing Confidence Intervals for the Variance and Standard Deviation

of a Normal Distribution

Note: You can only do this with a population that is normally distributed. n > 30 is not good

enough with 2 .

Step 1:

Step 2:

Step 3:

Step 4:

NOTE: There is no calculator shortcut for this. You can find the necessary values on your

calculator, but there is no one-step calculator program for this confidence interval as there was in

the previous 3 sections.

Example

The starting salaries of college graduates who have taken statistics are normally distributed. A

sample of 51 such graduates has a mean salary of $45,678 and a sample standard deviation of

$9900. Construct a 95% confidence interval for the population standard deviation σ.

Solution

Step 1:

Step 2:

Step 3:

Step 4:

Example

A package of 12 doughnuts is advertised to weigh 42 ounces. A quality control supervisor at the

doughnut company knows that he is in “good shape” if the doughnuts he is producing have a

mean weight of 3.50 ounces and a standard deviation of 0.06 ounces or less (not too large to lose

profit, and not too small for customers to realize that their doughnuts are not as large as what

they paid for). Assume doughnuts have normally distributed weights.

12 doughnuts are randomly selected from the production line and are found to have the following

weights:

3.43 3.37 3.58 3.50 3.68 3.61 3.42 3.52 3.66 3.50 3.36 3.42

Construct a 95% confidence interval for the population standard deviation of the doughnuts that

are being produced.

Do You Know

How to find critical values of the chi-square distribution?

How to construct confidence intervals for the variance and standard deviation of a normal

distribution?