Ch 10 Estimating and Constructing Confidence Intervals
10.1 Interval Estimation
Finding a value for Z 
Show the 95% confidence interval picture and how to find a two tailed Z  when
2
  .05 .
10.2 Confidence Intervals for the Mean with Known Population Variance
The population mean  is an unknown parameter.
We wish to estimate  based on a sample.
x is a statistic which estimates  . (remember  x   )
We call x a point estimate because its value is a point on the real line.
Unfortunately, if we sample from a continuous distribution, P( x   )  0
We are sure that our estimate is wrong.
Statisticians prefer interval estimates.
x  "something" or x  E
E (the Error Tolerance) depends on the amount of variability in the data and how certain
we went to be that we are correct.
The degree of certainty that we are correct is known as the Level of Confidence.
When  is known Z 
x

n
has approximately a standard normal distribution.
Show Picture
P  z  Z  z   1  
2
2





x
P   z 
 z   1  

2
2


n


  1
P  z 
 x    z 

2
2
n
n

  1
P x  z 
   x  z 

2
2
n
n

Therefore, E  Z (

n
)
Notice that increasing the level of confidence, decreases the probability of error.
However it also increases the width of the interval.
Notice as sample size increases the width of the interval decreases.
Notice the more variability in the population the wider the interval.
Example
A sample of 100 visa accounts were studied for the amount of unpaid balance.
n  100
x  $645
  $132
Construct a 95% confidence interval
  1.96132
  25.87
E  Z  



n
100 


xE
645  25.87
619.13,679.87 
We are 95% confident that the mean unpaid balance of visa accounts is between $619.13
and $670.87.
Construct a 99% confidence interval
  2.576132
  34.00
E  Z  



n
100




xE
645  34.00
611.00,679.00
We are 99% confident that the mean unpaid balance of visa accounts is between $611.00
and $679.00.
Choosing the Sample Size
In the design stages of statistical research, it is good to decide in advance the confidence
level you wish to use and to select the error tolerance you want for the project.
This will let us decide how big our sample needs to be.
Sample size for estimating 
 Z 
n

 E 
2
If the value of  is not known, we do preliminary sampling to approximate it.
Example
We wish to estimate the number of patient-visit hours per week physicians in solo
practice spent. How large a sample is needed if we want to be 99% confident that our
point estimate is within 1 hour of the population mean? Assume a standard deviation of
11.97 hours.
 Z 
 2.576  11.97 
n
 
  950.8  951
1
 E 


2
2
10.3 Student’s t Distribution
 is seldom known.
To avoid the error involved in replacing  by s , we will introduce a new random
variable called Student’s t variable. (t distribution)
If we sample from a normal distribution t 
freedom.
x
has a t distribution with n-1 degrees of
s
n
The assumption that we sample from a normal population is important for small n but not
for large n.
Properties of the t-distribution
1. continuous and symmetric about 0
2. more variable and slightly different shape than the standard normal
3. As n becomes large, the t distribution can be approximated by the standard normal
distribution (The bottom row of the t distribution is Z(alpha))
Go to the back cover of the book to view the t table.
With 10 degrees of freedom and a Confidence level of .95, what is the two tailed t value?
t.025,10  2.228
With 10 degrees of freedom.
P(T > 2.228) = .025
P(T < -2.228) = .025 draw pictures
or
P(|T| > 2.228) = .05
Notice the bottom row of the t table gives Z 
For large n, t ,n 1 is well approximated by Z 
10.4 Confidence Intervals for the Mean with Unknown Population Variance
All confidence intervals are two-sided probabilities with a total area of  .
s
For  unknown, E  t ( )
n
xE  xE
100% confidence interval.
 0
t 
Our confidence interval is simply the entire real number line.
We knew this before we took the sample
Point estimate is a 0% confidence interval.
 1
t 0
Example
Mileage of tires in 1000’s of miles n  10
42,36,46,43,41,35,43,45,40,39
Compute a 95% confidence interval for 
x  41
s  3.59
n = 10
alpha = .05
t (.05,9)  2.262
 s 
 3.59 
E  t 
  2.262
  2.568
 n
 10 
xE
41  2.568
38.432,43.568
We are 95% confident that the population mean mileage of tires is between 38,432 and
43,568 miles.
Example
A random sample of 20 apples yields x  9.2 oz. and s  1.1 oz.
Find a 99% confidence interval.
T = 2.861
 s 
 1.1 
E  t 
  2.861
  .704
 n
 20 
xE
9.2  .704
8.496,9.904
We are 99% confident that the population mean weight of apples is between 8.496 and
9.904 oz.
10.5 Confidence Intervals for Proportions (Large Samples)
Inference for population proportion p
p̂ is a point estimate of p
We cannot expect p̂ to be exactly equal to p
For large n, Z 
EZ
pˆ  p
pq
n
is approximately standard normal for large n.
pˆ qˆ
n
p̂  E
Example
A survey of 1,200 registered voters yields 540 who plan to vote for the republican
candidate.
p = proportion of all voters who plan to vote for the republican candidate
p̂ 
540
 0.45  45%
1200
Find a 95% confidence interval for p
p̂(1  p̂)
.45(.55)
 1.96
 .028
n
1200
(.45 - .028, .45 +.028)
(.422, .478)
EZ
We are 95% confident that the true proportion of voters who will vote for the republican
candidate is between 42.2% and 47.8%.