Transcript Reject H0 - BrainMass

Lesson Four: Student t
Distribution and
Comparing Samples
Hypotheses
Do You Remember the
Coefficient of Variation? We
looked at the three samples of
BRUS, comparing to the total.
We wondered if they were
significantly different from each
other.
Books R US Sample
Coefficient of Variation
WK 1-14
WK 15-28 WK 29-42
Standard
Deviation
8241
11502
7557
Mean
9084
19543
20574
Coefficient
of Var.
0.91
0.59
0.37
Mean=
16,400
S.D. =
10,324
C.V. =
0.63
Total all
Weeks
We asked: Why the Change in
Variation?
The first sample looks VERY different from
the other two. Let’s develop another
formula to compare the two.
One-Tailed Tests about a Population
Mean: Large-Sample Case (n > 30)
Test Statistic
Formula #4 
Known
x  0
z
s/ n
x  0
z
/ n
Rejection Rule
Formula #5  Unknown
Right-Tailed
Reject H0 if z > 
Left-Tailed
Reject H0 if z < 
Example: Two Tail Test
To be 95% confident, you have 5%
chance of error. Divide this between
both tails: 2.5%
on each tail
Reject H0
0.025
-1.96
Do Not Reject H0
0.95 of Area
Under the
Curve
0
Reject H0
0.025
1.96
z
Example: One Tail Test
To be 95% confident, you have 5%
chance of error. All 5% is in One Tail.
H1 > 
Reject H0
Do Not Reject H0
0.95 of Area
Under the
Curve
0.05
0
1.65
z
0.05
1.6
0.4505
Example: Books R US
Left tail hypothesis, Reject H0 if z < 
 = .05, find by .5000 - .4500
“Critical Value” of Z = -1.65
A value of 1.65 for a “Z Statistic” is found
by locating the value closest to .4500
(.4505 in this case, round up) and find 1.6
on the row heading, and 0.5 on the
column heading.
Applying the Formula to Calculate Z
x  0
z
 / n =(9,084 – 16,400)
10,324 / 14
Z = -7,316
2,760
Z = -2.65
Example: Books R US
Left tail hypothesis, Reject H0 if z < 
2.65 is less than (further to the left of)
-1.65
Reject H0
Do Not Reject H0
-2.65 -1.65
0
z
Problem Statements
Tells
Tells
Tells
Tells
Tells
what is going on
when it is happening
who is impacted by it
where the problem occurs
how the problem occurs
A problem statement is a question about possible
relationship between the manipulated and
responding variables in a situation that implies
something to do or try
The Five-Step Process for
Hypothesis Testing (Thinking Stages)
State the null and alternative hypotheses
Find the level of significance



.01 = scientific research
.05 = consumer and product research
.10 = political polling
Hypothesis Test
Develop the null and alternative hypotheses.
“Ho: The sample mean of the first sample > 
H1: The sample mean of the first sample < ”
Specify the level of significance (.05)
Select the test statistic: (Z statistic).
Determine the critical value for Ho: -1.65
Collect sample data, compute test statistic.
Computed value = -2.65, smaller than
-1.65
Decision: Reject Ho, the first sample is
statistically, significantly smaller than .
Hypotheses
In our last lesson, we were dealing with
the following:
• H0 = no effect, chance differences x = 
• H1 = effect or difference exists x = 
This is for a two – tail test.
We’ve set a critical value of 1.96, but let’s
say that the   .05. This is the
critical value of the p-value.
Hypotheses
For a one tail test, we might want to see if
something is GREATER THAN the mean, or LESS
THAN the mean.
H0 = no effect, chance differences x > 
H1 = is an effect, it is likely that x <  our Books
R US data.
Let’s combine this with what we just did with the
first sample. Remember that we had Z = -2.65
On the Z table, this gives us 0.4960
Figure 6-16: The P -value of a z statistic can be approximated by noting which levels from
Table D it
falls between. Here, P lies between 0.20 and 0.25.
Reject H0
Do Not Reject H0
-2.65 -1.65
0.05
2.6
0.4960
P - Value
In order to REJECT the null, the p-value
must be less than the  level, in this
example, .05.
.5000 - .4960 = .004
The smaller the P, the stronger the
evidence that H0 is false
Now, we REJECT the null, ACCEPT H1
Why? Because the P value is smaller than
the critical value of .
What if all we have is a sample
standard deviation, and a sample
mean?
In that case, we use this formula:
 Unknown
x  0
z
s/ n
But with the BRUS data, n < 30 in our
sample, so we must use the t Statistic.
Point and Interval Estimates
If the population standard
deviation is unknown and the
sample is less than 30 we use the
t distribution.
s
Formula #6 X  t
n
Problem Using Formula #6
In the second 14 weeks of the Books R
Use data, the mean total sales were
$19,543. The standard deviation was
$11,502. At the .05 level of significance,
what was the confidence interval? Did the
value $16,400 fall inside our outside the
confidence interval?
One-Tailed Tests about a Population Mean:
Small-Sample Case (n < 30)
Test Statistic
Formula 7
 Known
x  0
t
/ n
Formula 8
 Unknown
x  0
t
s/ n
This test statistic has a t distribution with n - 1 degrees
of freedom, or “DF”.
Rejection Rule
Right-Tailed
Reject H0 if t >t

Left-Tailed
Reject H0 if t <-t

So, Do we Choose the Z or
the t Statistic?
Remember our three sets of weeks? There were
1 in each set.
Since there are fewer than 30 observations in a
sample, we’ll use the t test.
Use this formula:
x  0
t
s/ n
This test statistic has a t distribution with n - 1
degrees of freedom, or “DF”.
For weeks 1 – 14, the X was $9,084, s was
$8,241 o was $16,400, and. n = 14
So, Do we Choose the Z or
the t Statistic? We Use t.
t
=
t
=
$9,084 - $16,400
( $8,241 / 14 )
-3.32
We need to find the
“critical value” of t:
-2.160 See next slide
- $7,316
=
2,202.5
x  0
t
s/ n
We can use
a 1 – tail
test
We can use
a 2 – tail
test
df
DF = n – 1
DF = 13
STUDENT’S
T
DISTRIBUTION
13
2.160
Hypothesis Test
Develop the null and alternative hypotheses.
“Ho: The sample mean of the first sample = 
H1: The sample mean of the first sample = ”
Specify the level of significance (.05)
Select the test statistic: (t statistic).
Determine the critical value for Ho: 2.160
Collect sample data, compute test statistic.
Calculated t = -3.32, further to left of
-2.160
Decision: Reject Ho, the first sample is
statistically, significantly different from .
Using a P – Value Calculator
For the P-Value of a t statistic, go to
http://www.danielsoper.com/statcalc
Choose the Student t Distribution. In this case,
it is 0.005531 for a two tail test, which is <
the critical value of .05, so we reject the null.
Use the same website to calculate the P-Value
for a Z statistic.
In many cases, modern computer programs will
print the p-Value, so it is important to be able
to understand its meaning.
Summary of Formulas
Z = (X – )

CV = s
X
 Known
x  0
z
/ n
 UnKnown
x  0
z
s/ n
 Known
x  0
t
/ n
 UnKnown
x  0
t
s/ n
Confidence Range
For a t Statistic
X t
s
n
Comparing Two Samples
Apply hypothesis testing to
different populations and samples
in business research situations.
Comparing Two Samples
We want to apply hypothesis testing to
different populations & samples in bus.
Research situations.
Examples of when do 2 independent samples
when sample size is 30 or greater.
Ex. When do 2 independent samples when
sample size is less than 30.
Hypothesis Testing
Single Samples (<30)
We compared the results of a single sample
to a population value
We determined whether the proposed
population value was reasonable
We used the ‘Steps in Hypothesis Testing”
(handout) to answer our research question
about our sample
One-tailed vs. Two-tailed
Hypothesis Testing
Population Means: Large Samples
Is there a difference in the mean amount to
residential real estate sold by male agents and
female agents in south Florida?
Let’s select random samples from 2 populations. We
wish to investigate if these populations have the same
mean
Want to determine whether the samples are from the
same or equal populations
If the 2 populations are the same, we would expect
the difference between the 2 sample means to be zero
2 assumptions needed:


Both samples are at least 30
The samples are from independent populations
Population Means: Large Samples
Formulas
Example
A financial analyst wants to compare the turnover
rates, in percent, for shares of oil-related stocks
versus other stocks, such as GE and IBM. She selected
32 oil-related stocks and 49 other stocks. The mean
turnover rate of oil-related stocks is 31.4 percent and
the standard deviation 5.1 percent. For the other
stocks, the mean rate was computed to be 34.9
percent and the standard deviation 6.7 percent. Is
there a significant difference in the turnover rates of
the two types of stock? Use the .01 significance level.
Hypothesis Testing
Population Means: Small Samples
Is the mean salary of nurses larger than that of
school teachers?
The sample size is less than 30
‘Small sample test of means’
The 2 sample variances are pooled to estimate
population variance; weighted mean
The weights are the degrees of freedom that each
sample provides
Assumptions:



1. The sampled populations follow the normal distribution
2. The two samples are from independent populations
3. The standard deviations of the two populations are equal
Population Means: Small Samples
Formulas
Example
A recent study compared the time spent together by
single- and dual-earner couples. According to the
records kept by the wives during the study, the mean
amount of time spent together watching television
among the single-earner couples was 61 minutes per
day, with a standard deviation of 15.5 minutes. For
the dual-earner couples, the mean number of minutes
spent watching television was 48.4 minutes, with a
standard deviation of 18.1 minutes. At the .01
significance level, can we conclude that the singleearner couples on average spend more time watching
television together? There were 15 single-earner and
12 dual-earner couples studied.
Application to Lemonade
Stand Results