Topic:Confidence Intervals

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Contents 1 Topic Highlights 2 Introduction and Motivation 3 Learning Activities 4 Learning Objectives 5 Topic Notes: The Basics of Confidence Intervals 5.1 What is a Confidence Interval Anyway? 5.1.1 Example 1 5.1.2 Example 2 5.2 What are the Common Confidence Intervals? 5.2.1 Example 3 5.2.2 Example 4 5.3 Summary 6 Topic Notes: Confidence Intervals for the Mean of a Normal Population with Known Standard Deviation 6.1 Example 5 7 Practice Problems 7.1 Practice Problem 1 7.2 Practice Problem 2 (ES)

Topic Highlights

(What you will learn)

• An introduction to the concept of confidence intervals
• How to apply confidence intervals to cases where population mean, , is estimated using sample mean,

Introduction and Motivation

(Why learn it)

Way back in the topic Introduction to Business Statistics, we looked at the decision-making process and the difference between descriptive and inferential statistics. Then recently, in the topic A First Look at the Central Limit Theorem, we looked more closely at inferential statistics - we looked at how managers can infer something about the population mean, , for example, from an estimate he or she obtains from a sample mean, .

In this topic, we are going to try to understand how the precision of such an estimate can be measured and represented. This is done using something called a confidence interval.

Learning Activities

(How the levels of understanding will be gained)

Learning activities for this topic

Type	Name	Direction
Reading	Section 7.1 of Bowerman et al. Section 7.3 of Kvanli et al.	Self-directed
In-class Worksheet	Worksheet - Confidence Intervals	Self-directed
Lecture and discussion	The Basics of Confidence Intervals (below on this page) Confidence Intervals for the Mean of a Normal Population (with Known Standard Deviation) (also below)	Instructor-directed
Personal activities	e.g. reflection and review Review the learning objectives (below)	Self-directed

Learning Objectives

(Levels of understanding to be gained)

Learning objectives for this topic

Level of Understanding	Objective(s)
Very best	Can I solve Practice Problem 2?
Highly satisfactory	Do I understand how the concept of a confidence interval applies to the average, as in Example 5? Can I solve Example 5? Can I solve Practice Problem 1?
Satisfactory	Can I do Example 1 and Example 2 on my own?
Maybe just enough to pass	Do I have the z-scores and confidence intervals from Example 3 and Example 4 on my equation sheet? Do I know how to use them?

Topic Notes: The Basics of Confidence Intervals

These notes are intended to facilitate an introductory discussion of the confidence interval concept.

What is a Confidence Interval Anyway?

Let's start our discussion with an example.

Example 1

You are told that a population is distributed normally with a mean, = 10, and a standard deviation, = 2.

In this case, 80% of the data fall between what two values?

Solution

From our discussions about normal curves, you should recognize the situation to be as follows: Where the area on each side of the mean = 0.80/2 = 0.40, and where you are asked to find xmin and xmax. You also know that this can be represented as follows using the standard normal curve: Using your normal curve Table A.4, you can find the values of z: zmax = 1.28 zmin = -1.28 And using the rearranged form of the z-score equation, we find that: xmax = + 1.28 = 10 + 1.28x2 = 12.56 xmin = -1.28 = 10 - 1.28x2 = 7.44 In other words, for a population distributed normally with  = 10 and  = 2: 80% of the data fall in the range between 7.44 and 12.56.

You just figured out your first confidence interval!

In the language of confidence intervals, you could say that for the situation described above:

• You are 80% confident that any point chosen at random will fall between 7.44 and 12.56
• Your 80% confidence interval is [7.44, 12.56]

As you've probably gathered, confidence intervals are expressed in percentages, such as the 80% confidence interval or the 95% confidence interval. The percentage values 80% and 95% are known as the confidence level.

Let's take a look at another example.

Example 2

What is the 80% confidence interval for a population for which = 36.20 and = 12.30?

Solution
We know from the last example that for an 80% confidence level, the values of z are: zmax = 1.28 zmin = -1.28 And using the rearranged form of the z-score equation again, we find that: xmax = + 1.28 = 36.20 + 1.28x12.30 = 20.456 xmin = -1.28 = 36.20 - 1.28x12.30 = 51.944 In other words, for a population distributed normally with  = 36.2 and  = 12.3: The 80% confidence interval is [20.456, 51.944]

What are the Common Confidence Intervals?

Example 3

What are the z-scores corresponding to the following confidence levels?

a) 90%

b) 95%

c) 97.5%

d) 98%

e) 99%

Solution
a) Look up 0.4500 in Table A.4 to get: z = 1.645 b) Look up 0.4750 to get: z = 1.96 c) Look up 0.4875 to get z = 2.24 d) Look up 0.4900 to get approximately z = 2.33 e) Look up 0.4950 to get z = 2.575

Example 4

What are the corresponding confidence intervals?

Solution
a) For 90% where z = 1.645, the interval is: [ - 1.645, + 1.645 ] b) For 95% where z = 1.96, the interval is: [ - 1.96, + 1.96 ] c) For 97.5% where z = 2.24, the interval is: [ - 2.24, + 2.24 ] d) For 98% where z = 2.33, the interval is: [ - 2.33, + 2.33 ] e) For 99% where z = 2.575, the interval is: [ - 2.575, + 2.575 ]

This is a good set of values to put on your equation sheet.

Kvanli et al. generalize the confidence interval with the following equation:

[ - Z_/2 , + Z_/2 ]

Where Z_/2 is the z-score corresponding to = 0.01 for a 99% confidence interval, and =0.05 for a 95% confidence interval, and so on.

Summary

The following table summarizes some of our findings here:

Confidence level	Associated alpha ()	Z	Times out of ...
90%	0.10	1.645	9 times out of 10
95%	0.05	1.96	19 times out of 20
97.5%	0.025	2.24	39 times out of 40
98%	0.02	2.33	49 times out of 50
99%	0.01	2.575	99 times out of 100

Topic Notes: Confidence Intervals for the Mean of a Normal Population with Known Standard Deviation

These topic notes are intended to facilitate a discussion of confidence intervals for the mean when the standard deviation is known.

A typical application of the confidence interval is for decribing the precision with which a sample mean, , approximates the population mean, . We're going to take a look at this here, but first let's recall what we learned in the topic A First Look at the Central Limit Theorem.

If we sample a population many times, as shown below:

and compute the sample mean for each set of sample, your plot of sample means might look like the one below:

The Central Limit Theorem tells us that the sample means themselves make up a random variable, , that has a normal distribution just like the ones we looked at in topics like Continuous and Normal Random Variables. The concept is shown below:

Okay, so what's the point?

The point is that because each is just an estimate of the true mean , you can use a confidence interval for to express the precision of that estimate. Let's see how this works...

Example 5

You know the standard deviation of a population, = 2, but you don't know its average, . So, in order to estimate the average, you take 36 samples and compute a sample mean of = 9.2. What is the 90% confidence interval for ?

Solution

First, let's just solve the problem: As we saw earlier, z = 1.645 for a confidence interval of 90% And we can compute: = 2/sqrt(36) = 0.3333 So the confidence interval is: [ - 1.645, + 1.645 ] = [ 9.2 - 1.645(0.3333), 9.2 + 1.645(0.3333) ] = [ 8.64, 9.75 ] Then, let's make sure you recognize what's going on here: We were told we knew , but not So we estimated by sampling and getting The Central Limit Theorem gives us the distribution of the means that can be used for the confidence interval: Also be sure you recognize that we solved for the values xmin and xmax corresponding to the shaded area in the following:

Practice Problems

Practice Problem 1

A random sample of 100 observations is obtained from a normally distributed population with a standard deviation of 10. What is a 95% confidence interval for the mean of the population if the sample mean is 40?

Solution
As we saw in Example 3, z = 1.96 for a confidence interval of 95% And we can compute: = 10/sqrt(100) = 1.00 So the confidence interval is: [ - 1.96, + 1.96 ] = [ 40 - 1.96(1.00), 40 + 1.96(1.00) ] = [ 38.04, 41.96 ]

Practice Problem 2 (ES)

You work in a factory in which the manufacturing equipment creates engine parts of a certain average length, , and standard deviation, = 0.100 mm. The equipment did this perfectly for years until Buddy Rogers (a not-too-swift colleague of yours) dropped a hammer on it last week. You're pretty sure that the standard deviation hasn't changed, but you're not sure about the mean. So, you sample 100 parts coming off the line and find that the sample mean is 9.200 mm.

a) What is the 98% confidence interval for your estimate of the sample mean?

b) Interpret your solution

Solution

a) As we saw in Example 3, z = 2.33 for a confidence interval of 98% And we can compute: = 0.100/sqrt(100) = 0.010 So the confidence interval is: [ - 2.33, + 2.33 ] = [ 9.200 - 2.33(0.010), 9.200 + 2.33(0.010) ] = [ 9.177, 9.223 ] b) In other words, given your measurement of 9.200 mm, you can be 98% confident that the average length of the engine parts coming off the line now falls between 9.177 and 9.223 mm.