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```10- 1
Chapter
Ten
McGraw-Hill/Irwin
10- 2
Chapter Ten
One-Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
ONE
Define a hypothesis and hypothesis testing.
TWO
Describe the five step hypothesis testing procedure.
THREE
Distinguish between a one-tailed and a two-tailed test of
hypothesis.
FOUR
Conduct a test of hypothesis about a population mean.
10- 3
Chapter Ten continued
One-Sample Tests of Hypothesis
GOALS
When you have completed this chapter, you will be able to:
FIVE
Conduct a test of hypothesis about a population proportion.
SIX
Define Type I and Type II errors.
SEVEN
Compute the probability of a Type II error.
10- 4
A statement
value of a
population
parameter
developed for
the purpose of
testing.
The mean monthly
income for systems
analysts is \$6,325.
What is a
Hypothesis?
Twenty percent of all
customers at Bovine’s Chop
House return for another
meal within a month.
What is a Hypothesis?
10- 5
Hypothesis testing
Based on
sample
evidence and
probability
theory
Used to determine
whether the hypothesis is
a reasonable statement
and should not be
rejected, or is
unreasonable and should
be rejected
What is Hypothesis Testing?
10- 6
Step 1: State null and alternate hypotheses
Step 2: Select a level of significance
Step 3: Identify the test statistic
Step 4: Formulate a decision rule
Step 5: Take a sample, arrive at a decision
Do not reject null
Reject null and accept alternate
Hypothesis Testing
10- 7
Step One: State the null and alternate hypotheses
Null Hypothesis H0
value of a population
parameter
Alternative Hypothesis H1:
A statement that is
accepted if the sample data
provide evidence that the
null hypothesis is false
10- 8
Step One: State the null and alternate
hypotheses
Three
possibilities
regarding
means
H 0: m = 0
H 1: m = 0
H 0: m < 0
H 1: m > 0
H0: m > 0
H1: m < 0
The null
hypothesis
always contains
equality.
10- 9
Type I Error
Level of Significance
Rejecting the null
hypothesis when it
is actually true (a).
The probability of
rejecting the null
hypothesis when it is
actually true; the level of
risk in so doing.
Type II Error
Accepting the null
hypothesis when it
is actually false (b).
Step Two: Select a Level of
Significance.
10- 10
Step Two: Select a Level of Significance.
Null
Hypothesis
Ho is true
Ho is false
Researcher
Accepts
Rejects
Ho
Ho
Correct
Type I
error
decision
(a)
Type II
Correct
Error
decision
(b)
Risk
table
10- 11
Test statistic
A value, determined
from sample
information, used to
determine whether
or not to reject the
null hypothesis.
Examples: z, t, F, c2
z Distribution as a
test statistic
X m
z
/ n
The z value is based on the
sampling distribution of X,
which is normally
distributed when the sample
is reasonably large (recall
Central Limit Theorem).
Step Three: Select the test statistic.
10- 12
Step Four: Formulate the decision rule.
Critical value: The dividing point between the region
where the null hypothesis is rejected and the region
where it is not rejected.
Sampling Distribution
Of the Statistic z, a
Right-Tailed Test, .05
Level of Significance
Region of
Do not
rejection
reject
[Probability =.95]
0
[Probability=.05]
1.65
Critical value
10- 13
Decision Rule
Reject the null
hypothesis and
accept the
alternate
hypothesis if
Computed -z < Critical -z
or
Computed z > Critical z
Decision Rule
10- 14
p-Value
The probability, assuming that the null
hypothesis is true, of finding a value of
the test statistic at least as extreme as the
computed value for the test
Decision Rule
If the p-Value is larger
If the p-Value is
than or equal to the
smaller than the
significance level, a, H0
significance level, a,
is not rejected.
H0 is rejected.
Calculated from the
probability distribution
function or by computer
Using the p-Value in Hypothesis
Testing
10- 15
Interpreting p-values
>.05
p
>.001
p
.10
SOME evidence Ho is not true
>.01
p
STRONG evidence
Ho is not true
.05
.01
VERY STRONG
evidence Ho is not true
10- 16
Step Five: Make a decision.
Movie
10- 17
One-Tailed Tests of Significance
The
alternate
hypothesis,
H1, states a
direction
H1: The mean
speed of trucks
traveling on I95 in Georgia
is less than 60
miles per hour.
(µ<60)
H1: The mean yearly
commissions earned by
full-time realtors is more
than \$35,000. (µ>\$35,000)
H1: Less than 20
percent of the
customers pay
cash for their
gasoline purchase.
(p<.20)
One-Tailed Tests of Significance
10- 18
Sampling Distribution
Of the Statistic z, a
Right-Tailed Test, .05
Level of Significance
Region of
Do not
rejection
reject
[Probability =.95]
0
[Probability=.05]
1.65
Critical value
One-Tailed Test of Significance
.
10- 19
Two-Tailed Tests of Significance
No direction is specified in the alternate hypothesis H1.
H1: The mean
amount spent by
customers at the
Wal-mart in
Georgetown is
not equal to \$25.
(µ ne \$25).
H1: The mean
price for a
gallon of
gasoline is not
equal to \$1.54.
(µ ne \$1.54).
Two-Tailed Tests of Significance
10- 20
Two-Tailed Tests of Significance
Regions of
Nonrejection
and Rejection
for a TwoTailed Test,
.05 Level of
Significance
Region of
Region of
Do not
rejection
rejection
reject
[Probability=.025]
[Probability =.95]
-1.96
Critical value
0
[Probability=.025]
1.96
Critical value
10- 21
Test for the population
mean from a large sample
with population standard
deviation known
X m
z
/ n
Testing for the Population Mean: Large
Sample, Population Standard Deviation
Known
10- 22
The processors of Fries’ Catsup
indicate on the label that the
bottle contains 16 ounces of
catsup. The standard deviation
of the process is 0.5 ounces. A
sample of 36 bottles from last
hour’s production revealed a
mean weight of 16.12 ounces per
bottle. At the .05 significance
level is the process out of
control? That is, can we
conclude that the mean amount
per bottle is different from 16
ounces?
Example 1
10- 23
Step 5
Make a decision and
interpret the results.
Step 4
State the decision rule.
Reject H0 if z > 1.96
or z < -1.96
or if p < .05.
Step 3
Identify the test statistic. Because
we know the population standard
deviation, the test statistic is z.
Step 1
State the null and the
alternative hypotheses
H0: m = 16
H1: m  16
Step 2
Select the significance level.
The significance level is .05.
EXAMPLE 1
10- 24
Step 5: Make a
decision and
interpret the results.
z
X m


n
oComputed
16 .12  16 .00
0.5
z of 1.44
< Critical z of 1.96,
op of .1499 > a of .05,
Do not reject the null
hypothesis.
36
 1.44
The p(z > 1.44)
is .1499 for a
two-tailed test.
We cannot
conclude the
mean is different
from 16 ounces.
Example 1
10- 25
Testing for the
Population Mean:
Large Sample,
Population Standard
Deviation Unknown
Here  is unknown,
so we estimate it
with the sample
standard deviation s.
As long as the
sample size n > 30, z
can be approximated
using
X m
z
s/ n
Testing for the Population Mean: Large Sample,
Population Standard Deviation Unknown
10- 26
Roder’s Discount Store
chain issues its own
credit card. Lisa, the
credit manager, wants to
find out if the mean
monthly unpaid balance
is more than \$400. The
level of significance is set
at .05. A random check
of 172 unpaid balances
revealed the sample
mean to be \$407 and the
sample standard
deviation to be \$38.
Should Lisa conclude
that the population
mean is greater than
\$400, or is it reasonable
to assume that the
difference of \$7 (\$407\$400) is due to chance?
Example 2
10- 27
Step 5
Make a decision
and interpret the
results.
Step 4
H0 is rejected if
z > 1.65
or if p < .05.
Step 3
Because the sample is large
we can use the z
distribution as the test
statistic.
Step 1
H0: µ < \$400
H1: µ > \$400
Step 2
The significance
level is .05.
Example 2
Step 5
Make a decision
and interpret the
results.
10- 28
z
oComputed
z of 2.42
> Critical z of 1.65,
op of .0078 < a of .05.
Reject H0.
Lisa can conclude that
the mean unpaid balance
is greater than \$400.
X m
s
n

\$407  \$400
\$38
 2.42
172
The p(z > 2.42) is
.0078 for a onetailed test.
10- 29
Testing for a
Population Mean:
Small Sample,
Population
Standard Deviation
Unknown
The test statistic
is the t
distribution.
t
X m
s/ n
The critical value of t is
determined by its degrees of
freedom equal to n-1.
Testing for a Population Mean: Small Sample,
Population Standard Deviation Unknown
The current rate for producing
5 amp fuses at Neary Electric
Co. is 250 per hour. A new
machine has been purchased
and installed that, according
to the supplier, will increase
the production rate. The
production hours are normally
distributed. A sample of 10
randomly selected hours from
last month revealed that the
mean hourly production on
the new machine was 256
units, with a sample standard
deviation of 6 per hour.
10- 30
At the .05 significance
level can Neary conclude
that the new machine is
faster?
Example 3
10- 31
The null hypothesis is rejected if t > 1.833 or, using the
p-value, the null hypothesis is rejected if p < .05.
Step 4
State the decision rule.
There are 10 – 1 = 9
degrees of freedom.
Step 1
State the null and
alternate hypotheses.
H0: µ < 250
H1: µ > 250
Step 3
Find a test statistic. Use
the t distribution since 
is not known and n < 30.
Step 2
Select the level of
significance. It is .05.
10- 32
Step 5
Make a decision
and interpret the
results.
t
X m
s
n

256  250
6 10
 3.162
The p(t >3.162) is
.0058 for a onetailed test.
oComputed
t of 3.162
>Critical t of 1.833
op of .0058 < a of .05
Reject Ho
The mean number
of amps produced is
more than 250 per
hour.
Example 3
10- 33
Proportion
The sample
proportion is p
and p is the
population
proportion.
The fraction or percentage
that indicates the part of the
population or sample having
a particular trait of interest.
Number of successesin the sample
p
Number sampled
Test Statistic for
Testing a Single
Population Proportion
z
p p
p (1  p )
n
10- 34
In the past, 15% of the
mail order solicitations for
a certain charity resulted in
a financial contribution. A
new solicitation letter that
has been drafted is sent to
a sample of 200 people
and 45 responded with a
contribution. At the .05
significance level can it be
concluded that the new
letter is more effective?
Example 4
10- 35
Step 3
Find a test statistic.
The z distribution
is the test statistic.
Step 4
State the decision rule.
The null hypothesis is
rejected if z is greater
than 1.65 or if p < .05.
Step 5
Make a decision and interpret the results.
Step 1
State the null and the
alternate hypothesis.
H0: p < .15
H1: p > .15
Step 2
Select the level of
significance. It is .05.
Example 4
10- 36
Step 5: Make a decision and interpret the results.
45
 .15
p p
z
 200
 2.97
p (1  p )
.15 (1  .15 )
n
200
p( z > 2.97)
= .0015.
Because the computed z of 2.97 > critical z of
1.65, the p of .0015 < a of .05, the null
hypothesis is rejected. More than 15 percent
responding with a pledge. The new letter is
more effective.
Example 4
```