Download 9.2 A Significance Test for a Mean

Section 9.2
Significance Test
for a Mean
Homework
Activity 9.2a, page 581
• Due tomorrow
• Use complete sentences to support your
answers.
• For each question, be sure to explain why
you made the choice you did.
Test Statistic
When we tested for the mean of a
sampling distribution, we used:
Recall, μ0 is hypothesized population mean
Population Mean
To test H0:   o , we need a test statistic.
However, we do not know  so what do we
do?
Test Statistic
We substitute s for  and t for z.
Test Statistic
We substitute s for  and t for z.
The test statistic becomes:
Test Statistic
The weights of newly minted U.S. pennies
are approximately normally distributed and
are targeted to have a mean weight of
3.11 g. The random sample of nine
pennies has a mean weight of 3.16 g with
a standard deviation of 0.065.
What is the value of a test statistic for a test
to determine whether the mean has
moved away from the target mean?
Test Statistic
3
.
16

3
.
11
t
0.065 / 9
Test Statistic
3
.
16

3
.
11
t
0.065 / 9
t  2.31
t-Distributions
Suppose you draw random samples of size
n from a normally distributed population
with mean μ and unknown standard
deviation σ.
t-Distributions
Suppose you draw random samples of size n from
a normally distributed population with mean 
and unknown standard deviation  .
The distribution of the
values of:
is called a t-distribution. There is a different
t-distribution for each degree of freedom.
t-Distributions
A t-distribution is:
• mound-shaped
• with mean 0 and
• a spread that depends on the value of
the degrees of freedom (df).
t-Distributions
A t-distribution is:
• mound-shaped
• with mean 0 and
• a spread that depends on the value of
the degrees of freedom (df).
The greater the df, the smaller the
spread.
t-Distributions
A t-distribution is mound-shaped, with mean
0 and a spread that depends on the value
of df. The greater the df, the smaller the
spread.
The spread of any t-distribution is greater
than that of the standard normal
distribution.
t-Distributions
This shows the t-distribution for df = 4
plotted on the same graph as the standard
normal distribution.
p. 584
t-Distributions
1) For the degrees of freedom to increase,
what has to happen?
t-Distributions
1) For the degrees of freedom to increase,
what has to happen? Sample size must
increase as df = n -1.
t-Distributions
1) For the degrees of freedom to increase,
what has to happen? Sample size must
increase as df = n -1.
2) As sample size, n, increases what
happens to s?
t-Distributions
1) For the degrees of freedom to increase,
what has to happen? Sample size must
increase as df = n -1.
2) As sample size, n, increases what
happens to s? s  
t-Distributions
1) For the degrees of freedom to increase,
what has to happen? Sample size must
increase as df = n -1.
2) As sample size, n, increases what
happens to s? s  
3) As s   , what happens to the
t-distribution?
t-Distributions
1) For the degrees of freedom to increase, what
has to happen? Sample size must increase as df
= n -1.
2) As sample size, n, increases what happens to
s? s  
3) As s   , what happens to the t-distribution?
t-distribution more closely approximates a
standard normal distribution
t-Distributions
Therefore, as the degrees of freedom
increase, the t-distribution more closely
approximates a standard normal
distribution
t-Distributions
This shows a t-distribution with df = 9.
t-Distributions
df = 4
df = 9.
P-value
If we know the t value, how do we determine
the associated P-value?
P-value
If we know the t value, how do we determine
the associated P-value?
P-value
Recall, if we had a z-distribution we would
use __________.
Since we have a t-distribution we need to
use _______?
P-value
Recall, if we had a z-distribution we would
use normalcdf.
Since we have a t-distribution we need to
use _______?
P-value
Recall, if we had a z-distribution we would
use normalcdf.
Since we have a t-distribution we need to
use tcdf.
2nd, DISTR, 6: tcdf
P-value
Find the P-value for the graph if t = 2.31 and
df = 8.
P-value
Syntax: tcdf(lower bound, upper bound, df)
Recall: There is a different t-distribution for
each degree of freedom so we have to
specify df here
P-value
Syntax: tcdf(lower bound, upper bound, df)
P-value = 2[ tcdf (-1EE99, - 2.31, 8)]
= 2(.0248445422)
= 0.0496890845
Components of a Significance Test
for a Mean
4 components:
Components of a Significance Test
for a Mean
4 components:
1) Name the test and check conditions
Components of a Significance Test
for a Mean
4 components:
1) Name the test and check conditions
2) State hypotheses
Components of a Significance Test
for a Mean
4 components:
1) Name the test and check conditions
2) State hypotheses
3) Compute test statistic, find Pvalue, and draw a sketch
Components of a Significance Test
for a Mean
4 components:
1) Name the test and check conditions
2) State hypotheses
3) Compute test statistic, find P-value, and
draw a sketch
4) Write conclusion linked to
computations and in context.
Name Test
Two-sided significance test for a mean
or
One-sided significance test for a mean
Check Conditions
1) Randomness
Check Conditions
1) Randomness
–Survey: random sample
Check Conditions
1) Randomness
–Survey: random sample
–Experiment: randomly assigned
treatments
Check Conditions
2)
Normality
Check Conditions
2) Normality
Sample must look like it’s reasonable to
assume that the sample came from a
normally distributed population
or
Sample size must be large enough that the
sampling distribution of the sample mean
is approximately normal (will discuss in
Section 9.3)
Check Conditions
3) Population
size
Check Conditions
3) Population size
Survey: population size is at least 10 times
as large as the sample
Check Conditions
3) Population size
Survey: population size is at least 10 times
as large as the sample
Experiment: not applicable
State Hypotheses
State Hypotheses
Null hypothesis is that the population
mean, μ, has a particular value μo.
Define μ in context.
H o:
  o
State Hypotheses
Null hypothesis is that the population
mean, μ, has a particular value μo.
Define μ in context.
H o:
  o
Example: Ho: μ = 3.85, where μ is the mean
GPA for acceptance at Harvard
State Hypotheses
Alternative hypothesis is one of these:
Ha:   o
Ha:  < o
H a:  >
o
Use specific value for o
Computations
Compute test statistic, find P-value, and
draw sketch.
Computations by Calculator
STAT
Computations by Calculator
STAT
Computations by Calculator
Conclusion
Write conclusion linked to your computations
and in the context of the problem.
Remember, if you are not given a level of
significance, you can assume  = 0.05
Page 596, P10
Find the test statistic, t, and the P-value.
T-test
Inpt: Data Stats
μ0: 72
List: L1
Freq: 1
μ: ≠ μ0 < μo > μo
Calculate
T-test
μ ≠ 72
t = - 2.931419509
p = .0262386012
x = 70.14285714
Sx = 1.67616342
n=7
t ≈ ± 2.9314
Page 596, P10
Page 597, P14
Page 597, P14
The test statistic, t, is approx. ± 2.9315
P-value = 2[ tcdf (-1EE99, - 2.9315, 6)]
= 0.0262358556
Page 597, P14
The test statistic, t, is approx. -2.9315
P-value = 2[ tcdf (-1EE99, - 2.9315, 6)]
= 0.0262358556
If the mean temperature at the desk is
actually 720, the probability that
temperatures taken on seven randomly
selected days would give a result as
extreme or more extreme than  2.9315
is about 0.0262
Page 597, P16
Page 597, P16
a) Ho: μ = 720, where μ is the mean
temperature at my desk
Page 597, P16
a) Ho: μ = 720, where μ is the mean
temperature at my desk
Ha: μ ≠ 720.
Page 597, P16
b) P-value is about 0.0262
10% level: reject Ho because P-value of
about 0.0262 is less than  = 0.10
5% level:
1% level:
Page 597, P16
b) P-value is about 0.0262
10% level: reject Ho because P-value of
about 0.0262 is less than  = 0.10
5% level: reject Ho because P-value of
about 0.0262 is less than  = 0.05
1% level:
Page 597, P16
b) P-value is about 0.0262
10% level: reject Ho because P-value of
about 0.0262 is less than  = 0.10
5% level: reject Ho because P-value of
about 0.0262 is less than  = 0.05
1% level: do not reject Ho because P-value
of about 0.0262 is greater than  = 0.01
Questions?