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Transcript
```Hypothesis Testing
 Developing Null and Alternative Hypotheses
 Type I and Type II Errors
 One-Tailed Tests About a Population Mean:
Large-Sample Case
 Two-Tailed Tests About a Population Mean:
Large-Sample Case
 Tests About a Population Mean:
Small-Sample Case
Developing Null and Alternative Hypotheses
 Hypothesis testing can be used to determine whether
a statement about the value of a population parameter
should or should not be rejected.
 The null hypothesis, denoted by H0 , is a tentative
 The alternative hypothesis, denoted by Ha, is the
opposite of what is stated in the null hypothesis.
 The alternative hypothesis is what the test is
attempting to establish.
Developing Null and Alternative
Hypotheses
• Testing Research Hypotheses
•
Hypothesis testing is proof by contradiction.
•
The research hypothesis should be expressed as
the alternative hypothesis.
•
The conclusion that the research hypothesis is true
comes from sample data that contradict the null
hypothesis.
Developing Null and Alternative
Hypotheses
• Testing the Validity of a Claim
•
Manufacturers’ claims are usually given the benefit
of the doubt and stated as the null hypothesis.
•
The conclusion that the claim is false comes from
sample data that contradict the null hypothesis.
Developing Null and Alternative
Hypotheses
• Testing in Decision-Making Situations
•
A decision maker might have to choose between
two courses of action, one associated with the null
hypothesis and another associated with the
alternative hypothesis.
•
Example: Accepting a shipment of goods from a
supplier or returning the shipment of goods to the
supplier
Summary of Forms for Null and Alternative

The equality part of the hypotheses always appears
in the null hypothesis.
 In general, a hypothesis test about the value of a
population mean  must take one of the following
three forms (where 0 is the hypothesized value of
the population mean).
H 0 :   0
H a :   0
H 0 :   0
H a :   0
H 0 :   0
H a :   0
One-tailed
(lower-tail)
One-tailed
(upper-tail)
Two-tailed
Example: Metro EMS
• Null and Alternative Hypotheses
A major west coast city provides
one of the most comprehensive
emergency medical services in
the world.
Operating in a multiple
hospital system with
approximately 20 mobile medical
units, the service goal is to respond to medical
emergencies with a mean time of 12 minutes or
less.
Example: Metro EMS

Null and Alternative Hypotheses
The director of medical services
wants to formulate a hypothesis
test that could use a sample of
emergency response times to
determine whether or not the
service goal of 12 minutes or less
is being achieved.
Null and Alternative Hypotheses
H0:  
The emergency service is meeting
the response goal; no follow-up
action is necessary.
Ha: 
The emergency service is not
meeting the response goal;
appropriate follow-up action is
necessary.
where:  = mean response time for the population
of medical emergency requests
Type I and Type II Errors
 Because hypothesis tests are based on sample data,
we must allow for the possibility of errors.


A Type I error is rejecting H0 when it is true.
The person conducting the hypothesis test specifies
the maximum allowable probability of making a
Type I error, denoted by  and called the level of
significance.
Type I and Type II Errors

A Type II error is accepting H0 when it is false.

It is difficult to control for the probability of making
a Type II error, denoted by .

Statisticians avoid the risk of making a Type II
error by using “do not reject H0” and not “accept H0”.
Type I and Type II Errors
Population Condition
Conclusion
H0 True
( < 12)
H0 False
( > 12)
Accept H0
(Conclude  < 12)
Correct
Decision
Type II Error
Type I Error
Correct
Decision
Reject H0
(Conclude  > 12)
Using the Test Statistic
 The test statistic z has a standard normal probability
distribution.
 We can use the standard normal probability
distribution table to find the z-value with an area
of  in the lower (or upper) tail of the distribution.
 The value of the test statistic that established the
boundary of the rejection region is called the
critical value for the test.

The rejection rule is:
• Lower tail: Reject H0 if z < z.
• Upper tail: Reject H0 if z > z.
Using the p-Value
 The p-value is the probability of obtaining a sample
result that is at least as unlikely as what is observed.
 If the p-value is less than the level of significance ,
the value of the test statistic is in the rejection region.
 Reject H0 if the p-value <  .
Steps of Hypothesis Testing
1. Determine the null and alternative hypotheses.
2. Specify the level of significance .
3. Select the test statistic that will be used to test the
hypothesis.
Using the Test Statistic
4. Use to determine the critical value for the test
statistic and state the rejection rule for H0.
5. Collect the sample data and compute the value
of the test statistic.
6. Use the value of the test statistic and the rejection
rule to determine whether to reject H0.
Steps of Hypothesis Testing
Using the p-Value
4. Collect the sample data and compute the value of
the test statistic.
5. Use the value of the test statistic to compute the
p-value.
6. Reject H0 if p-value < .
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)

Hypotheses
H 0 :   
or
Ha: 

Test Statistic
 Known
z

x  0
/ n
H0:   
Ha: 
 Unknown
z
x  0
s/ n
Rejection Rule
Reject H0 if |z| > z
One-Tailed Test about a Population Mean:
Large-Sample Case (n > 30)
Let  = .05
Sampling
distribution
of z  x   0
/ n
Reject H0
Do Not Reject H0

z
0
z = 1.645
One-Tailed Test about a Population Mean:
Large-Sample Case (n > 30)
Let  = .10
Sampling
distribution
of z  x   0
/ n
Reject H0

Do Not Reject H0
z
z = 1.28
0
Example: Metro EMS
Null and Alternative Hypotheses
The response times for a random
sample of 40 medical emergencies
were tabulated. The sample mean
is 13.25 minutes and the sample
standard deviation is 3.2
minutes.
The director of medical services
wants to perform a hypothesis test, with a
.05 level of significance, to determine whether or not the
service goal of 12 minutes or less is being achieved.

One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
1. Determine the hypotheses.
H0:  
Ha: 
2. Specify the level of significance.  = .05
3. Select the test statistic.
z
x  0
s/ n
( is not known)
4. State the rejection rule.
Reject H0 if z > 1.645
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
5. Compute the value of the test statistic.
x   13.25  12
z

 2.47
s / n 3.2 / 40
6. Determine whether to reject H0.
Because 2.47 > 1.645, we reject H0.
We are 95% confident that Metro EMS is
not meeting the response goal of 12 minutes.
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the pValue
4. Compute the value of the test statistic.
x   13.25  12
z

 2.47
s / n 3.2 / 40
5. Compute the p–value.
For z = 2.47, cumulative probability = .9932.
p–value = 1  .9932 = .0068
6. Determine whether to reject H0.
Because p–value = .0068 <  = .05, we reject H0.
One-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
• Using the p-value
 = .05
p-value

z
0
z =
1.645
z=
2.47
Using Excel to Conduct a
One-Tailed Hypothesis Test

Formula Worksheet
A
Response
1
Time
2
19.5
3
15.2
4
11.0
5
12.8
6
12.4
7
20.3
8
9.6
9
10.9
10
16.2
11
13.4
12
19.7
B
C
Sample Size 40
Sample Mean =AVERAGE(A2:A41)
Sample Std. Dev. =STDEV(A2:A41)
Lev. of Signif. 0.05
Critical Value =NORMSINV(1-C5)
Hypoth. Value
Standard Error
Test Statistic
p -Value
Conclusion
12
=C3/SQRT(C1)
=(C2-C8)/C9
=1-NORMSDIST(C10)
=IF(C11<C5,"Reject","Do Not Reject")
Note: Rows 13-41 are not shown.
Using Excel to Conduct a
One-Tailed Hypothesis Test

Value Worksheet
A
Response
1
Time
2
19.5
3
15.2
4
11.0
5
12.8
6
12.4
7
20.3
8
9.6
9
10.9
10
16.2
11
13.4
12
19.7
B
C
Sample Size 40
Sample Mean 13.25
Sample Std. Dev. 3.20
Lev. of Signif. 0.05
Critical Value 1.645
Hypoth. Value
Standard Error
Test Statistic
p -Value
Conclusion
12
0.5060
2.471
0.0067
Reject
Note: Rows 13-41 are not shown.
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)

Hypotheses
H 0 :   0
H a :   0

Test Statistic
 Known
x  0
z
/ n

 Unknown
z
x  0
s/ n
Rejection Rule
Reject H0 if |z| > z
Example: Glow Toothpaste
• Two-Tailed Tests about a Population Mean:
Large n
The production line for Glow toothpaste
is designed to fill tubes with a mean weight
of 6 oz. Periodically, a sample of 30 tubes
will be selected in order to check the
filling process.
Quality assurance procedures call for
the continuation of the filling process if the
sample results are consistent with the assumption that
the mean filling weight for the population of toothpaste
tubes is 6 oz.; otherwise the process will be adjusted.
Example: Glow Toothpaste
Two-Tailed Tests about a Population Mean: Large n
Assume that a sample of 30 toothpaste
tubes provides a sample mean of 6.1 oz.
and standard deviation of 0.2 oz.
Perform a hypothesis test, at the .05
level of significance, to help determine
whether the filling process should
continue operating or be stopped and
corrected.

Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
1. Determine the hypotheses.
H0:  
Ha:   6
(two-tailed test)
2. Specify the level of significance.  = .05
3. Select the test statistic.
4. State the rejection rule.
x  0
s/ n
( is not known)
z
Reject H0 if |z| > 1.96
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
Sampling
distribution
x  0
of z 
/ n
Reject H0
Reject H0
Do Not Reject H0


-1.96
0
1.96
z
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the Test Statistic
5. Compute the value of the test statistic.
x  0
6.1  6
z

 2.74
s / n .2 / 30
6. Determine whether to reject H0.
Because 2.74 > 1.96, we reject H0.
We are 95% confident that the mean filling
weight of the toothpaste tubes is not 6 oz.
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
• Using the p-Value
Suppose we define the p-value for a two-tailed test
as double the area found in the tail of the distribution.
With z = 2.74, the cumulative standard normal
probability table shows there is a 1.0 - .9969 = .0031
probability of a z–score greater than 2.74 in the upper
tail of the distribution.
Considering the same probability of a z-score less
than –2.74 in the lower tail of the distribution, we have
p-value = 2(.0031) = .0062.
The p-value .0062 is less than  = .05, so H0 is rejected.
Two-Tailed Tests about a Population Mean:
Large-Sample Case (n > 30)
 Using the p-Value
1/2
p-value
= .0031
1/2
p-value
= .0031
 
 


z
z = -2.74
-z/2 = -1.96
0
z/2 = 1.96
z = 2.74
Using Excel to Conduct a
Two-Tailed Hypothesis Test

Formula Worksheet
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
C
Weight
Sample Size 30
6.04
Sample Mean =AVERAGE(A2:A31)
5.99
Sample Std. Dev. =STDEV(A2:A31)
5.92
6.03
Lev. of Signif. 0.05
6.01
Crit. Value (lower) =NORMSINV(C5/2)
5.95 Crit. Value (upper) =NORMSINV(1-C5/2)
6.09
6.07
Hypoth. Value 6
6.07
Standard Error =C3/SQRT(C1)
5.97
Test Statistic =(C2-C9)/C10
5.96
p -Value =2*NORMSDIST(C11)
6.08
Conclusion =IF(C12<C5,"Reject","Do Not Reject")
Note: Rows 14-31 are not shown.
Using Excel to Conduct a
Two-Tailed Hypothesis Test

Value Worksheet
1
2
3
4
5
6
7
8
9
10
11
12
13
A
B
Weight
Sample Size 30
6.04
Sample Mean 6.1
5.99
Sample Std. Dev. 0.2
5.92
6.03
Lev. of Signif. 0.05
6.01
Crit. Value (lower) -1.960
5.95 Crit. Value (upper) 1.960
6.09
6.07
Hypoth. Value 6
6.07
Standard Error 0.0365
5.97
Test Statistic 2.739
5.96
p -Value 0.006
6.08
Conclusion Reject
Note: Rows 14-31 are not shown.
C
Confidence Interval Approach to a
Two-Tailed Test about a Population Mean
 Select a simple random sample from the population
and use the value of the sample mean x to develop
the confidence interval for the population mean .
(Confidence intervals are covered in Chapter 8.)
 If the confidence interval contains the hypothesized
value 0, do not reject H0. Otherwise, reject H0.
Confidence Interval Approach to a
Two-Tailed Test about a Population Mean
The 95% confidence interval for  is
x  z / 2

n
 6.1  1. 96(. 2
30 )  6.1. 0716
or 6.0284 to 6.1716
Because the hypothesized value for the
population mean, 0 = 6, is not in this interval,
the hypothesis-testing conclusion is that the
null hypothesis, H0:  = 6, can be rejected.
Small-Sample Case (n < 30)
• Test Statistic
 Known
x  0
z
/ n
 Unknown
t
x  0
s/ n
This test statistic has a t distribution
with n - 1 degrees of freedom.
Small-Sample Case (n < 30)

Rejection Rule
H0:   
Reject H0 if t > t
H0:   
Reject H0 if t < -t
H0: 
Reject H0 if |t| > t
p -Values and the t Distribution
 The format of the t distribution table provided in most
statistics textbooks does not have sufficient detail
to determine the exact p-value for a hypothesis test.
 However, we can still use the t distribution table to
identify a range for the p-value.
 An advantage of computer software packages is that
the computer output will provide the p-value for the
t distribution.
Example: Highway Patrol
• One-Tailed Test about a Population Mean:
Small n
A State Highway Patrol periodically samples
vehicle speeds at various locations
The sample of vehicle speeds
is used to test the hypothesis
H0:  < 65
The locations where H0 is rejected are deemed the best
Example: Highway Patrol

One-Tailed Test about a Population Mean: Small n
At Location F, a sample of 16 vehicles shows a
mean speed of 68.2 mph with a
standard deviation of
3.8 mph. Use  = .05 to
test the hypothesis.
One-Tailed Test about a Population Mean:
Small-Sample Case (n < 30)
 Using the Test Statistic
1. Determine the hypotheses.
H0:  < 65
Ha:  > 65
2. Specify the level of significance.
3. Select the test statistic.
t
 = .05
x  0
s/ n
( is not known)
4. State the rejection rule.
Reject H0 if t > 1.753
(d.f. = 16-1 = 15)
One-Tailed Test about a Population Mean:
Small-Sample Case (n < 30)
Reject H0
Do Not Reject H0
0

1.753
(Critical value)
t
One-Tailed Test about a Population Mean:
Small-Sample Case (n < 30)
 Using the Test Statistic
5. Compute the value of the test statistic.
t
x  0 68.2  65

 3.37
s / n 3.8/ 16
6. Determine whether to reject H0.
Because 3.37 > 1.753, we reject H0.
We are at least 95% confident that the mean speed
of vehicles at Location F is greater than 65 mph.
Location F is a good candidate for a radar trap.
Using Excel to Conduct a One-Tailed
Hypothesis Test: Small-Sample Case

Formula Worksheet
A
B
C
Vehicle
1 Speed
Sample Size 16
2
69.6
Sample Mean =AVERAGE(A2:A17)
3
73.5
Sample Std. Dev. =STDEV(A2:A17)
4
74.1
5
64.4
Lev. of Signif. 0.05
6
66.3
Critical Value =TINV(2*C5,C1-1)
7
68.7
8
69.0
Hypoth. Value 65
9
65.2
Standard Error =C3/SQRT(C1)
10
71.1
Test Statistic =(C2-C8)/C9
11
70.8
p -Value =TDIST(C10, C1-1,1)
12
64.6
Conclusion =IF(C11<C5,"Reject","Do Not Reject")
Note: Rows 13-17 are not shown.
Using Excel to Conduct a One-Tailed
Hypothesis Test: Small-Sample Case

Value Worksheet
A
B
Vehicle
1 Speed
Sample Size 16
2
68.2
Sample Mean 68.20
3
77.0
Sample Std. Dev. 3.80
4
71.0
5
64.2
Lev. of Signif. 0.05
6
66.8
Critical Value 1.753
7
68.3
8
65.9
Hypoth. Value 65
9
63.9
Standard Error 0.9490
10
71.1
Test Statistic 3.372
11
71.6
p -Value 0.0021
12
60.7
Conclusion Reject
Note: Rows 13-17 are not shown.
C
```
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