Download Confidence Interval for Mean

Inference for a Mean
when you have a “small” sample...
As long as you have a
“large” sample….
A confidence interval for a population mean is:
 s 
xZ 

 n
*
where the average, standard deviation, and n depend on
the sample. Z* depends on the confidence level.
As long as you have a
“large” sample….
A test statistic for a population mean is:
x
Z
s/ n
where the average, standard deviation, and n depend
on the sample.  is the value specified in the null.
Example
Random sample of 59 students spent an
average of $273.20 on Spring 1998 textbooks.
Sample standard deviation was $94.40.
 94.4 
273.20  1.96
  273.20  24.09
 59 
We can be 95% confident that the average amount spent
by all students was between $249.11 and $297.29.
Example
A sample of 59 students spent an average of
$273.20 on textbooks with a standard deviation
of $94.40. Do students spend less than $300 on
average?
273.2  300 

P( X  300) P Z 
  P( Z  2.18)  0.015
94.4 / 59 

There is enough evidence, at 0.05 level, to conclude that,
on average, students spend less than $300 on textbooks.
What happens if you can only
take a “small” sample?
• Random sample of 15 students slept an
average of 6.4 hours last night with standard
deviation of 1 hour.
• What is the average amount all students
slept last night?
• Is the average amount less than 7 hours?
If you have a “small” sample...
Replace the Z multiplier with a t multiplier
to get:
 s 
x t 

 n
*
where “t*” comes from Student t distribution,
and depends on the sample size through the
degrees of freedom “n-1”.
If you have a “small” sample...
Replace the Z statistic with the t statistic:
x
t
s/ n
Again, “t” follows the Student’s t distribution,
which depends on the sample size through the
degrees of freedom “n-1”.
Student’s t distribution versus
Normal Z distribution
T-distribution and Standard Normal Z distribution
0.4
Z distribution
density
0.3
0.2
T with 5 d.f.
0.1
0.0
-5
0
Value
5
Student t distribution
• Shaped like standard normal distribution
(symmetric around 0, bell-shaped).
• But, t depends on the degrees of freedom
“n-1”.
• And, more likely to get extreme t values
than extreme Z values.
Graphical Comparison of
t and Z Multipliers
5
4
T with 5 df
3
2
Z distribution
1
0
0.90
0.92
0.94
0.96
0.98
Cumulative Probability
1.00
Tabular Comparison of
t and Z Multipliers
Confidence t value with Z value
level
5 d.f
2.015
1.65
90%
2.571
1.96
95%
4.032
2.58
99%
For small samples, t value is larger than Z value.
So, t interval is longer than a Z interval, and for a
given test statistic the P-value is larger.
Back to our CI example!
Sample of 15 students slept an average of 6.4
hours last night with standard deviation of 1 hour.
Need t with n-1 = 15-1 = 14 d.f.
For 95% confidence, t14 = 2.145
 s 
 1 
x  t
  6.4  2.145
  6.4  0.55
 n
 15 
That is...
We can be 95% confident that average amount
slept last night by all students is between 5.85
and 6.95 hours.
Hmmm! Adults need 8 hours of sleep each
night.
Logical conclusion:
On average, students need more sleep.
(Just don’t get it in this class!)
T-Interval for Mean in Minitab
T Confidence Intervals
Variable
Comb
N
89
Mean
2.011
StDev
1.563
SE Mean
0.166
95.0 % CI
(1.682, 2.340)
We can be 95% confident that the average number of
times a “Stat-250-like” student combs his or her hair is
between 1.7 and 2.3 times a day.
T- interval in Minitab
•
•
•
•
•
•
Select Stat.
Select Basic Statistics.
Select 1-Sample t…
Select desired variable.
Specify desired confidence level.
Say OK.
And to our HT example!
New sample of 18 students slept an average of 6.01
hrs last night with standard deviation of 1.11 hrs. Do
students sleep less than 7 hours on average?
H0:  = 7 vs. HA:  < 7
6.01  7
t
 3.78
1.11 / 18
If the population mean is 7, how likely is it that a sample of 18
students would sleep an average as low as 6.01 hours?
Or, how likely is it that we’d get a t statistic as low as -3.78?
T-test for Mean in Minitab
T-Test of the Mean
Test of mu = 7.000 vs mu < 7.000
Variable
SleepHrs
N
18
Mean
6.011
StDev
1.113
SE Mean
0.262
T
-3.77
P
0.0008
If the population mean was 7, it is not likely (P-value = 0.0008)
that we’d get a sample mean as small as 6.011
Reject the null hypothesis. There is enough evidence to conclude
that students sleep on average less than 7 hours.
T- test in Minitab
•
•
•
•
•
•
•
Select Stat.
Select Basic Statistics.
Select 1-Sample t…
Select desired variable.
Specify the null mean in “Test mean” box.
Select the alternative hypothesis.
Say OK.
What happens as
sample gets larger?
T-distribution and Standard Normal Z distribution
0.4
Z distribution
density
0.3
T with 60 d.f.
0.2
0.1
0.0
-5
0
Value
5
Example
Random sample of 64 students spent an average of 3.8
hours on homework last night with a sample standard
deviation of 3.1 hours.
Z Confidence Intervals The assumed sigma = 3.10
Variable
Homework
N
Mean
64 3.797
StDev
3.100
95.0 % CI
(3.037,
4.556)
T Confidence Intervals
Variable N
Mean
Homework 64 3.797
StDev
3.100
95.0 % CI
(3.022,
4.571)
Example
Random sample of 139 students own an average of 12.7 pairs
of shoes with a sample standard deviation of 9.6 pairs.
Z-Test
Test of mu = 10.000 vs mu > 10.000
The assumed sigma = 9.63
Variable
N
Shoes
139
Mean
12.669
StDev
9.625
SE Mean
0.816
Z
3.27
P
0.0006
T
3.27
P
0.0007
T-Test of the Mean
Test of mu = 10.000 vs mu > 10.000
Variable
N
Shoes
139
Mean
12.669
StDev
9.625
SE Mean
0.816
One not-so-small problem!
• It is only OK to use the t interval for small
samples if your original measurements
are normally distributed.
Strategy
• If you have a large sample of, say, 30 or more
measurements, then don’t worry about
normality, and use a t-interval or do a t-test.
• If you have a small sample and your data are
normally distributed, then use a t-interval or
do a t-test.
• If you have a small sample and your data are
not normally distributed, then use
nonparametric hypothesis tests.