Section 8.3 – Confidence Intervals for Population Means
The Problem of Unknown σ
 Because σ is usually unknown, we estimate it by the sample deviation s.
Standard Error
The standard error of the sample mean x is
s
. ***s is the standard deviation of the sample.
n
***When we know the value σ, we base confidence intervals and tests for µ on the z –statistic: z 
x 

.
n
**The z-statistic has a normal distribution.
***When we do not know σ, we substitute the standard error
s
. This statistic does not have a normal
n
distribution.
So what do we use?
The t-distribution, Degrees of Freedom
 New table: Table B : )
Draw an SRS of size n from a population that has the normal distribution with mean µ and standard
x 
deviation σ. The one-sample t statistic t 
has the t distribution with n – 1 degrees of freedom.
s
n
***This statistic has the same interpretation as any standardized statistic: it says how far the sample mean is
from the population mean in standard deviation units.
Different t-distributions:
There is a different t-distribution for each sample size n.
So what do we notice? As n increases, the t
distribution gets closer to the standard normal
distribution. Show picture on p. 512.
t-distribution
t* is a critical value with probability p to the right.
Tail probability is found by: (1 – C) / 2
Degrees of freedom = df = n - 1
Example, p. 513: Finding t* Critical Values
What critical value t* from Table B should be used in constructing a confidence interval for the population
mean in each of the following settings?
(a) A 95% confidence interval based on an SRS of size n = 12.
Tail probability = (1 – 0.95) / 2 = 0.025
df = 12 – 1 = 11
**You can also look at the Confidence level at the
bottom of the chart.
So t* = 2.201
(b) A 0=90% confidence interval from a random sample of 48 observations.
Df = 48 – 1 = 47 **NOT ON CHART!!!** We’ll use the more conservative number of 40.
So t* = 1.684
You can also find t* on your calculator. Under invT( and then put the tail probability and df in the parentheses.
Before constructing a confidence interval for a population mean, check the conditions:
1. Random: The data come from a well-designed random sample or randomized experiment.
1
 10%: If sampling without replacement, check 𝑛 ≤ 10 𝑁.
2. Normal/Large Sample: The population has a Normal distribution or the sample size is large (n ≥ 30). If
the population has unknown shape and n < 30, use a graph of the sample data to assess the Normality of
the population. Do not use t procedures if the graph shows strong skewness or outliers. (You can use
histograms, stem-and-leaf plots, boxplots, Normal probability plots – but you must sketch them on your
paper!!!)
If one of the conditions is violated, there is no point in constructing a confidence interval for µ.
One-sample t procedures
When the conditions are met, a C% confidence interval for the unknown mean µ is
x  t*
s
n
Where t* is the critical value for the tn - 1 distribution, with C% of the area between –t* and t*.
Example, p. 519: Auto Pollution
Environmentalists, government officials, and vehicle manufacturers are all interested in studying the auto
exhaust emissions produced by motor vehicles. The major pollutants in auto exhaust from gasoline engines are
hydrocarbons, carbon monoxide, and nitrogen oxides (NOX). Researchers collected data on the NOX levels (in
grams/mile) for a random sample of 40 light-duty engines of the same type. The mean NOX reading was 1.2675
and the standard deviation was 0.3332.
(a) Construct and interpret a 95% confidence interval for the mean amount of NOX emitted by light-duty
engines of this type.
State: We want to estimate the true mean amount µ of NOX emitted by all light-duty engines of this
type at a 95% confidence level.
Plan: One-sample t-interval for µ (using t instead of z because we do not know σ)
Conditions: Random: Random sample of 40 light-duty engines of the same type

1
10%: Sampling without replacement, so check: 40 ≤ 10(All light-duty
engines). Reasonable to assume there are at least 400 light-duty engines.
Normal/Large sample: By Central Limit Theorem: 40 ≥ 30, so we’re safe to use
the t – distribution.
Do:
𝑥̅ = 1.2675
x  t*
sx = 0.3332
n = 40
CL: 95%, with df = 40 – 1 = 39  Not on chart
We’ll use df = 30 for a conservative approach.
So t* = 2.042
s
0.3332
= 1.2675 ± 2.042
= 1.2675 ± 0.1076 = (1.1599,1.3751)
√40
n
Conclude: We are 95% confident that the interval from 1.1609 to 1.3741 grams/mile captures the true
mean level of nitrogen oxides emitted by this type of light-duty engine.
(b) The Environmental Protection Agency (EPA) sets a limit of 1.0 gram/mile for average NOX emissions.
Are you convinced that this type of engine violates the EPA limit? Use your interval from (a) to support
your answer.
The confidence interval in (a) tells us that any value from 1.1609 to 1.3741 grams/mile is a plausible
value of the mean NOX level µ for this type of engine. Because the entire interval exceeds 1.0, it
appears that this type of engine violates EPA limits.
Example, p. 520: Video Screen Tension
A manufacturer of high-resolution video terminals must control the tension on the mesh of fine wires that lies
behind the surface of the viewing screen. Too much tension will tear the mesh, and too little will allow
wrinkles. The tension is measured by an electrical device with output readings in millivolts (mV). Some
variation is inherent in the production process. Here are the tension readings from a random sample of 20
screens from a single day’s production.
269.5
264.7
297.0
307.7
269.6
310.0
283.3
343.3
304.8
328.1
280.4
342.6
233.5
338.8
257.4
340.1
317.5
374.6
327.4
336.1
Construct and interpret a 90% confidence interval for the mean tension µ of all the screens produced on this
day.
State: We want to estimate the true mean tension µ of all the video terminals produced on this day with
90% confidence.
Plan: One-sample t-interval for µ
Conditions: Random: Random sample of 20 screens from a single day’s production

1
10%: Sampling without replacement, so check: 20 ≤ 10(All video
terminals produced this day). We must assume there are at least 200 video
terminals produced this day.
Normal/Large sample: Because the sample is not large enough, we will need to
graph the data.
The dotplot and boxplot do not show strong skewness or any outliers. The Normal
probability plot looks roughly linear. No reason to doubt the Normality of the
population.
Do: Use calculator to find sample mean and standard deviation. 𝑥̅ = 306.32 mV and sx = 36.21 mV.
n = 20
CL: 90%, df = 20 – 1 = 19, so t*= 1.729
x  t*
s
36.21
= 306.32 ± 1.729
= 306.32 ± 14.00 = (292.32, 320.32)
√20
n
Conclude: We are 90% confident that the interval from 292.32 to 320.32 mV captures the true mean µ
tension in the entire batch of video terminals produced that day.
It is not enough to just plot the data in your calculator. You must sketch an appropriate graph to receive
credit.
Choosing a sample size for a desired margin of error when estimating µ
To determine the sample size n that will yield a C% confidence interval for a population mean with a specified
margin of error ME:
 Get a reasonable value for the population standard deviation σ from an earlier or pilot study.
 Find the critical value z* from a standard Normal curve for confidence level C.
 Set the expression for the margin of error to be less than or equal to ME and solve for n:
𝜎
𝑧∗
≤ 𝑀𝐸
√𝑛
Example, p. 524: How Many Monkeys?
Researchers would like to estimate the mean cholesterol level µ of a particular variety of monkey that is often
used in laboratory experiments. They would like their estimate to be within 1 milligrams per deciliter (mg/dl) of
the true value of µ at a 95% confidence level. A previous study involving this variety of monkey suggests that
the standard deviation of cholesterol level is about 5 mg/dl.
What is the minimum number of monkeys the researchers will need to get a satisfactory estimate?
σ = 5 mg/dl
CL: 95%, so z* = 1.96
1.96
5
√𝑛
ME = 1 mg/dl
≤1
1.96 · 5
≤ √𝑛
1
96.04 ≤ 𝑛
We will need at least 97 monkeys to estimate the mean cholesterol level µ to within 1 mg/dl.
HW: p. 527 #55, 59, 67, 71, 74
Due: Tuesday