Download 9.5 T-Interval

Confidence intervals for m when s is
unknown
When s is unknown, we use the sample standard
deviation s to estimate s. In place of z-scores, we
must use the following to standardize the values:
x m
t 
s
n
The use of the value of s introduces extra
variability. Therefore the distribution
of t values has more variability than a
standard normal curve.
Important Properties of t
Distributions
1)
The t distribution is bell shaped and centered at
zero (just like the standard normal (z)
distribution).
2) Each t distribution is more spread out than the
standard normal distribution.
z curve
t curve for 2 df
0
3) The shape of a particular t-
distribution curve depends on df.
The number of degree of freedom is
defined as the number of observation
that can be chosen freely.
df = n-1
Example-1:
Find the value of t for 16 df and
0.05 area in the right tail of a tdistribution curve.
Central area captured:
Confidence level:
1
2
3
4
5
6
D
7
e
8
g
9
r
10
11
e
12
e
13
s
14
15
16
o
17
f
18
19
20
f
21
r
22
e
23
24
e
25
d
26
o
27
m
28
29
30
40
60
120
z critical values
0.80
0.90
0.95
0.98
0.99
0.998
0.999
80%
90%
95%
98%
99%
99.8%
99.9%
3.08
1.89
1.64
1.53
1.48
1.44
1.41
1.40
1.38
1.37
1.36
1.36
1.35
1.35
1.34
1.34
1.33
1.33
1.33
1.33
1.32
1.32
1.32
1.32
1.32
1.31
1.31
1.31
1.31
1.31
1.30
1.30
1.29
1.28
6.31
2.92
2.35
2.13
2.02
1.94
1.89
1.86
1.83
1.81
1.80
1.78
1.77
1.76
1.75
1.75
1.74
1.73
1.73
1.72
1.72
1.72
1.71
1.71
1.71
1.71
1.70
1.70
1.70
1.70
1.68
1.67
1.66
1.645
12.71
4.30
3.18
2.78
2.57
2.45
2.36
2.31
2.26
2.23
2.20
2.18
2.16
2.14
2.13
2.12
2.11
2.10
2.09
2.09
2.08
2.07
2.07
2.06
2.06
2.06
2.05
2.05
2.05
2.04
2.02
2.00
1.98
1.96
31.82
6.96
4.54
3.75
3.36
3.14
3.00
2.90
2.82
2.76
2.72
2.68
2.65
2.62
2.60
2.58
2.57
2.55
2.54
2.53
2.52
2.51
2.50
2.49
2.49
2.48
2.47
2.47
2.46
2.46
2.42
2.39
2.36
2.33
63.66
9.92
5.84
4.60
4.03
3.71
3.50
3.36
3.25
3.17
3.11
3.05
3.01
2.98
2.95
2.92
2.90
2.88
2.86
2.85
2.83
2.82
2.81
2.80
2.79
2.78
2.77
2.76
2.76
2.75
2.70
2.66
2.62
2.58
318.29
22.33
10.21
7.17
5.89
5.21
4.79
4.50
4.30
4.14
4.02
3.93
3.85
3.79
3.73
3.69
3.65
3.61
3.58
3.55
3.53
3.50
3.48
3.47
3.45
3.43
3.42
3.41
3.40
3.39
3.31
3.23
3.16
3.09
636.58
31.60
12.92
8.61
6.87
5.96
5.41
5.04
4.78
4.59
4.44
4.32
4.22
4.14
4.07
4.01
3.97
3.92
3.88
3.85
3.82
3.79
3.77
3.75
3.73
3.71
3.69
3.67
3.66
3.65
3.55
3.46
3.37
3.29
4) As the number of degrees of freedom increases,
the spread of the corresponding t distribution
decreases.
t curve for 8 df
t curve for 2 df
0
5) As the number of degrees of freedom increases,
the corresponding sequence of t distributions
approaches the standard normal distribution.
For what df would the t
distribution be
approximately the
same as a standard
normal distribution?
z curve
t curve for 2 df
t curve for 5 df
0
How to find t-critical using GDC?
1. Menu
2. Statistics (6)
3. Distributions (5)
4. Inverse t (6)
Note that Area here is “Area to the left
of the t-critical value”
Examples to Try
1. Find the value of t for a t-distribution
with 14 degree of freedom and 0.01 area
in the upper tail.
2. Find the value of t for a t-distribution
for confidence level 90% and n=19
Confidence intervals for m when s is
unknown
The general formula for a confidence interval for a
population mean m based on a sample of size n
when . . .
This confidence interval is appropriate for small n
ONLYsample
when mean
the population
distribution
is (at
1) x is the
from a random
sample,
least
approximately)
normal.or the
2) the population
distribution
is normal,
sample size n is large (n > 30), and
3) s, the population standard deviation, is unknown
s 

is
x  (t critical value)

Where the t critical value is based on df = n 
- 1. n 
Example-2
Ten randomly selected shut-ins were
each asked to list how many hours of
television they watched per week. The
results are
82
66
90
84
75
88
80
94
110
91
Find a 90% confidence interval
estimate for the true mean number
of hours of television watched per
week by shut-ins.
Calculating the sample mean and standard
deviation we have n = 10, x 
= 86,
86 s = 11.842
We find the critical t value of 1.833 by looking on
the t table in the row corresponding to df = 9, in
the column with bottom label 90%. Computing
the confidence interval for is
s
11.842
x  t*
 86  (1.833)
 86  6.86
n
10
(79.14, 92.86)
To calculate the confidence interval, we
had to make the assumption that the
distribution of weekly viewing times was
normally distributed. Consider the normal
plot of the 10 data points produced with
Minitab that is given on the next slide.
Notice that the normal plot looks reasonably
linear so it is reasonable to assume that the
number of hours of television watched per
week by shut-ins is normally distributed.
Normal Probability Plot
.999
.99
.95
Probability
Typically if the
p-value is more than 0.05
we assume that the
distribution is normal
.80
.50
.20
.05
.01
.001
70
80
90
100
110
Hours
Average: 86
Anderson-Darling Normality Test
A-Squared: 0.226
P-Value: 0.753
StDev: 11.8415
N: 10
Anderson-Darling Normality Test
A-Squared: 0.226
P-Value: 0.753
How to use GDC to find the
intervals????????
to be taught on Wednesday!
Test on ch-9
(to be discussed and decided)
Thursday 21st, January 2016
or
Monday 22nd, Feb 2016
or
Tuesday 23rd , Feb 2016 ??