Transcript Pwpt 8.2

What if it is impossible or
impractical to use a large
sample?
Apply the Student’s t distribution.
Student’s t Variable
x
t
s
n
The shape of the t distribution
depends only only the sample
size, n, if the basic variable x
has a normal distribution.
When using the t distribution, we will
assume that the x distribution is normal.
Table 6 in Appendix II gives
values of the variable t
corresponding to the number
of degrees of freedom (d.f.)
Degrees of Freedom
d.f. = n – 1
where n = sample size
The t Distribution has a Shape
Similar to that of the the
Normal Distribution
A Normal
distribution
A “t”
distribution
Find the critical value tc for a
95% confidence interval if n = 8.
d.f.=7
c
'
''
d.f.
...
6
7
8
...0.90
...
0.95
...
0.98
...
0.99
...
...1.943
...1.895
...1.860
2.447
2.365
2.306
3.143
2.998
2.896
3.707
3.499
3.355
Confidence Interval for the
Mean of Small Samples (n < 30)
from Normal Populations
xE xE
x  Sample Mean
s
E  tc
n
c = confidence level (0 < c < 1)
tc = critical value for confidence level c, and
degrees of freedom = n - 1
where
The mean weight of eight fish
caught in a local lake is 15.7
ounces with a standard
deviation of 2.3 ounces.
Construct a 90% confidence
interval for the mean weight of
the population of fish in the lake.
Mean = 15.7 ounces
Standard deviation = 2.3
n = 8, so d.f. = nounces.
–1=7
•
• For c = 0.90, Table 6 in Appendix II gives
t0.90 = 1.895.
s
2.3
E  tc
 1.895
 1.54
n
8
Mean = 15.7 ounces
Standard deviation = 2.3 ounces.
E = 1.54
The 90% confidence interval is:
xE xE
15.7 - 1.54 <  < 15.7 + 1.54
14.16 <  < 17.24
The 90% Confidence Interval:
14.16 <  < 17.24
We are 90% sure that the true
mean weight of the fish in the
lake is between 14.16 and 17.24
ounces.