Download Confidence Intervals for the Mean

5.5 Confidence Intervals for a
Population Mean ; t
distributions
•
•
•
t distributions
t confidence intervals for a
population mean 
Sample size required to
estimate 
The Importance of the Central
Limit Theorem

When we select simple random
samples of size n, the sample means
we find will vary from sample to sample.
We can model the distribution of these
sample means with a probability model
that is
 

N  ,

n

Since the sampling model for
x is the normal model, when
we standardize x we get the
standard normal z
x
z

n
Note that SD( x ) 

n
SD( x ) 

If  is unknown, we probably
n
don’t know  either.
The sample standard deviation s provides an estimate of
the population standard deviation 
For a sample of size n,
the sample standard deviation s is:
1
2
s
(
x

x
)

i
n 1
n − 1 is the “degrees of freedom.”
The value s/√n is called the standard error of x , denoted
SE(x).
s
SE ( x ) 
n
Standardize using s for 

Substitute s (sample standard deviation)
for 
x
x
z
s s s zs ss s



s
n
n
Note quite correct
Not knowing  means using z is no
longer correct
t-distributions
Suppose that a Simple Random Sample of size n is drawn
from a population whose distribution can be approximated by
a N(µ, σ) model. When  is known, the sampling model for
the mean x is N(, /√n), so
x
 n
is approximately Z~N(0,1).
When  is estimated from the sample standard deviation s,
x
s n
the sampling model for
follows a
t distribution with degrees of freedom n − 1.
x 
t
s n
is the 1-sample t statistic
Confidence Interval Estimates






CONFIDENCE
INTERVAL for 
s
x t
n
where:
t = Critical value from tdistribution with n-1
degrees of freedom
x = Sample mean
s = Sample standard
deviation
n = Sample size



For very small samples (n < 15),
the data should follow a Normal
model very closely.
For moderate sample sizes (n
between 15 and 40), t methods
will work well as long as the data
are unimodal and reasonably
symmetric.
For sample sizes larger than 40, t
methods are safe to use unless
the data are extremely skewed. If
outliers are present, analyses can
be performed twice, with the
outliers and without.
t distributions
Very similar to z~N(0, 1)
 Sometimes called Student’s t
distribution; Gossett, brewery employee
 Properties:
i) symmetric around 0 (like z)
ii)degrees of freedom 

if  > 1, E(t ) = 0
if  > 2,  =   - 2, which is always
bigger than 1.
Student’s t Distribution
x - x
z =
x
x - x
s
t =
, sx =
sx
n
Z
-3
-3
-2
-2
-1
-1
00
11
22
33
Student’s t Distribution
z=
x - x
x - x
t=
s
n

n
Z
t
-3
-3
-2
-2
-1
-1
00
11
22
33
Figure 11.3, Page 372
Student’s t Distribution
Degrees of Freedom
s =
x - x
t=
s
n
s2
n
s2 =
2
(X

X)
 i
i=1
Z
n -1
t1
-3
-3
-2
-2
-1
-1
00
11
22
33
Figure 11.3, Page 372
Student’s t Distribution
Degrees of Freedom
s =
x - x
t=
s
n
s2
n
s2 =
2
(X

X)
 i
i=1
Z
n -1
t1
t7
-3
-3
-2
-2
-1
-1
00
11
22
33
Figure 11.3, Page 372
t-Table: text- inside back cover;
Course Pack- inside back cover
and following p. 15, lect. unit 5

90% confidence interval; df = n-1 = 10
Degrees of Freedom
1
2
.
.
10
0.80
3.0777
1.8856
.
.
1.3722
0.90
6.314
2.9200
.
.
1.8125
0.95
0.98
12.706
4.3027
.
.
2.2281
31.821
6.9645
.
.
2.7638
.
.
.
.
.
.
.
.
.
.
100

1.2901
1.282
1.6604
1.6449
1.9840
1.9600
s
90% confidence interval : x  1.8125
11
2.3642
2.3263
0.99
63.657
9.9250
.
.
3.1693
.
.
2.6259
2.5758
Student’s t Distribution
P(t > 1.8125) = .05
P(t < -1.8125) = .05
.90
.05
-1.8125
0
.05
1.8125
t10
Comparing t and z Critical
Values
z = 1.645
z = 1.96
z = 2.33
z = 2.58
Conf.
level
90%
95%
98%
99%
n = 30
t = 1.6991
t = 2.0452
t = 2.4620
t = 2.7564

Example
– An investor is trying to estimate the return
on investment in companies that won
quality awards last year.
– A random sample of 41 such companies is
selected, and the return on investment is
recorded for each company. The data for
the 41 companies have
x  14.75 s  8.18
– Construct a 95% confidence interval for the
mean return.
s
x t
n
x  14.75 s  8.18
degrees of freedom  41  1  40
d. f .  n 1
from t - table, t  2.0211
s
8.18
x t
 14.75  2.0211
n
41
 14.75  2.61  12.14,17.36
We are 95% confident that the interval
(12.14, 17.36) contains the population mean
return on investment for companies that win
quality awards.
Example



Because cardiac deaths increase after heavy
snowfalls, a study was conducted to measure
the cardiac demands of shoveling snow by
hand
The maximum heart rates for 10 adult males
were recorded while shoveling snow. The
sample mean and sample standard deviation
were x 175, s 15
Find a 90% CI for the population mean max.
heart rate for those who shovel snow.
Solution
s
x t
n
d. f .  n 1
x  175, s  15 n  10
From the t - table, t 1.8331
15
175  1.8331
 175  8.70
10
 (166.30, 183.70)
We are 90% confident that the interval
(166.30, 183.70) contains the mean
maximum heart rate for snow shovelers
EXAMPLE: Consumer Protection
Agency


Selected random sample of 16 packages of a
product whose packages are marked as
weighing 1 pound.
From the 16 packages: x  1.10 pounds, s  .36 pound
a. find a 95% CI for the mean weight 
of the 1-pound packages
 b. should the company’s claim that the
mean weight  is 1 pound be
challenged ?

EXAMPLE
s
x t
n
d. f .  n 1
95% CI, n=16, df=15, x=1.10
s=.36
critical value of t is t  2.1315
s
x t
becomes
n
 .36 
1.10  (2.1315) 
  1.10  .19  .91, 1.29 
 16 
Since 1 pound is in the interval, the company's
claim appears reasonable.