7 hypothesis testing

Hypothesis Testing
 Developing Null and Alternative Hypotheses
 Type I and Type II Errors
 Population Mean: s Known
 Population Mean: s Unknown
 Population Proportion
 Hypothesis Testing and Decision Making
 Calculating the Probability of Type II Errors
 Determining the Sample Size for
a Hypothesis Test About a Population mean

Hypothesis Testing
 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
assumption about a population parameter.
 The alternative hypothesis, denoted by Ha, is the
opposite of what is stated in the null hypothesis.
 The hypothesis testing procedure uses data from a
sample to test the two competing statements
indicated by H0 and Ha.

Developing Null and Alternative Hypotheses
• It is not always obvious how the null and alternative
hypotheses should be formulated.
• Care must be taken to structure the hypotheses
appropriately so that the test conclusion provides
the information the researcher wants.
• The context of the situation is very important in
determining how the hypotheses should be stated.
• In some cases it is easier to identify the alternative
hypothesis first. In other cases the null is easier.
• Correct hypothesis formulation will take practice.

Developing Null and Alternative
Hypotheses
 Alternative Hypothesis as a Research Hypothesis
• Many applications of hypothesis testing involve
an attempt to gather evidence in support of a
research hypothesis.
• In such cases, it is often best to begin with the
alternative hypothesis and make it the conclusion
that the researcher hopes to support.
• The conclusion that the research hypothesis is true
is made if the sample data provide sufficient
evidence to show that the null hypothesis can be
rejected.

 Alternative Hypothesis as a Research Hypothesis
Developing Null and Alternative
Hypotheses
• Example:
A new teaching method is developed that is
believed to be better than the current method.
• Alternative Hypothesis:
The new teaching method is better.
• Null Hypothesis:
The new method is no better than the old method.

• Example:
A new sales force bonus plan is developed in an
attempt to increase sales.
The new bonus plan increase sales.
The new bonus plan does not increase sales.

• Example:
A new drug is developed with the goal of lowering
blood pressure more than the existing drug.
The new drug lowers blood pressure more than
the existing drug.
The new drug does not lower blood pressure more
than the existing drug.

Null Hypothesis as an Assumption to be
Challenged
We might begin with a belief or assumption that
a statement about the value of a population
parameter is true.
We then using a hypothesis test to challenge the
assumption and determine if there is statistical
evidence to conclude that the assumption is
incorrect.
In these situations, it is helpful to develop the null
hypothesis first.

Null Hypothesis as an Assumption to be
Challenged
• Example:
The label on a soft drink bottle states that it
contains 67.6 fluid ounces.
The label is correct. m > 67.6 ounces.
The label is incorrect. m < 67.6 ounces.

One-tailed
(lower-tail)
One-tailed
(upper-tail)
Two-tailed
0 0:H m m
0:aH m m
0 0:H m m
0:aH m m
0 0:H m m
0:aH m m
Summary of Forms for Null and Alternative
Hypotheses about a Population Mean
 The equality part of the hypotheses always appears
in the null hypothesis.
 In general, a hypothesis test about the value of a
population mean m must take one of the following
three forms (where m0 is the hypothesized value of
the population mean).

Null and Alternative Hypotheses
 Example: Metro EMS
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.
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
The emergency service is meeting
the response goal; no follow-up
action is necessary.
The emergency service is not
meeting the response goal;
appropriate follow-up action is
necessary.
H0: m 
Ha: m
where: m = mean response time for the population
of medical emergency requests

Type I Error
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 probability of making a Type I error when the
null hypothesis is true as an equality is called the
level of significance.
Applications of hypothesis testing that only control
the Type I error are often called significance tests.

Type II Error
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.
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
Correct
Decision
Type II Error
Correct
Decision
Type I Error
Reject H0
(Conclude m > 12)
Accept H0
(Conclude m < 12)
H0 True
(m < 12)
H0 False
(m > 12)Conclusion
Population Condition

p-Value Approach to
One-Tailed Hypothesis Testing
Reject H0 if the p-value < .
The p-value is the probability, computed using the
test statistic, that measures the support (or lack of
support) provided by the sample for the null
hypothesis.
If the p-value is less than or equal to the level of
significance , the value of the test statistic is in the
rejection region.

Suggested Guidelines for Interpreting p-Values
Less than .01
Overwhelming evidence to conclude Ha is true.
Between .01 and .05
Strong evidence to conclude Ha is true.
Between .05 and .10
Weak evidence to conclude Ha is true.
Greater than .10
Insufficient evidence to conclude Ha is true.

 p-Value Approach
p-value
7
0-z =
-1.28
 = .10
z
z =
-1.46
Lower-Tailed Test About a Population Mean:
s Known
Sampling
distribution
of z
x
n

m
s
0
/
p-Value <  ,
so reject H0.

p-Value

0 z =
1.75
 = .04
z
z =
2.29
Upper-Tailed Test About a Population Mean:
s Known
Sampling
distribution
of z
x
n

m
s
0
/
p-Value <  ,
so reject H0.

Critical Value Approach to
One-Tailed Hypothesis Testing
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

 
0z = 1.28
Reject H0
Do Not Reject H0
z
Sampling
distribution
of z
x
n

m
s
0
/
Lower-Tailed Test About a Population Mean:
s Known
 Critical Value Approach


0 z = 1.645
Reject H0
Do Not Reject H0
z
Sampling
distribution
of z
x
n

m
s
0
/
Upper-Tailed Test About a Population Mean:
s Known

Steps of Hypothesis Testing
Step 1. Develop the null and alternative hypotheses.
Step 2. Specify the level of significance .
Step 3. Collect the sample data and compute the
value of the test statistic.
p-Value Approach
Step 4. Use the value of the test statistic to compute the
p-value.
Step 5. Reject H0 if p-value < .

Critical Value Approach
Step 4. Use the level of significance to determine the
critical value and the rejection rule.
Step 5. Use the value of the test statistic and the rejection
rule to determine whether to reject H0.
Steps of Hypothesis Testing

 Example: Metro EMS
The EMS director wants to perform a hypothesis
test, with a .05 level of significance, to determine
whether the service goal of 12 minutes or less is
being achieved.
The response times for a random sample of 40
medical emergencies were tabulated. The sample
mean is 13.25 minutes. The population standard
deviation is believed to be 3.2 minutes.
One-Tailed Tests About a Population Mean:
s Known

1. Develop the hypotheses.
2. Specify the level of significance.  = .05
H0: m 
Ha: m
 p -Value and Critical Value Approaches
s Known
3. Compute the value of the test statistic.
m
s
 
  
13.25 12
2.47
/ 3.2/ 40
x
z
n

5. Determine whether to reject H0.
 p –Value Approach
s Known
4. Compute the p –value.
For z = 2.47, cumulative probability = .9932.
p–value = 1  .9932 = .0068
Because p–value = .0068 <  = .05, we reject H0.
There is sufficient statistical evidence
to infer that Metro EMS is not meeting
the response goal of 12 minutes.

p-value

0 z =
1.645
 = .05
z
z =
2.47
s Known
Sampling
distribution
of z
x
n

m
s
0
/

There is sufficient statistical evidence
to infer that Metro EMS is not meeting
the response goal of 12 minutes.
Because 2.47 > 1.645, we reject H0.
s Known
For  = .05, z.05 = 1.645
4. Determine the critical value and rejection rule.
Reject H0 if z > 1.645

p-Value Approach to
Two-Tailed Hypothesis Testing
 The rejection rule:
Reject H0 if the p-value < .
 Compute the p-value using the following three steps:
3. Double the tail area obtained in step 2 to obtain
the p –value.
2. If z is in the upper tail (z > 0), compute the
probability that z is greater than or equal to the
value of the test statistic. If z is in the lower tail
(z < 0), compute the probability that z is less than or
equal to the value of the test statistic.
1. Compute the value of the test statistic z.

Critical Value Approach to
Two-Tailed Hypothesis Testing
 The critical values will occur in both the lower and
upper tails of the standard normal curve.
 The rejection rule is:
Reject H0 if z < -z/2 or z > z/2.
 Use the standard normal probability distribution
table to find z/2 (the z-value with an area of /2 in
the upper tail of the distribution).

 Example: Glow Toothpaste
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.
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.
Two-Tailed Tests About a Population Mean:
s Known

Perform a hypothesis test, at the .03 level of
significance, to help determine whether the filling
process should continue operating or be stopped and
corrected.
Assume that a sample of 30 toothpaste tubes
provides a sample mean of 6.1 oz. The population
standard deviation is believed to be 0.2 oz.
s Known
 Example: Glow Toothpaste

1. Determine the hypotheses.
2. Specify the level of significance.
 = .03
 p –Value and Critical Value Approaches
H0: m 
Ha: 6m 
s Known
m
s
 
  0 6.1 6
2.74
/ .2/ 30
x
z
n

s Known
For z = 2.74, cumulative probability = .9969
p–value = 2(1  .9969) = .0062
Because p–value = .0062 <  = .03, we reject H0.
There is sufficient statistical evidence to
infer that the alternative hypothesis is true
(i.e. the mean filling weight is not 6 ounces).

s Known
/2 =
.015
0
z/2 = 2.17
z
/2 =
.015
-z/2 = -2.17
z = 2.74z = -2.74
1/2
p -value
= .0031
1/2
p -value
= .0031

s Known
There is sufficient statistical evidence to
infer that the alternative hypothesis is true
(i.e. the mean filling weight is not 6 ounces).
For /2 = .03/2 = .015, z.015 = 2.17
Reject H0 if z < -2.17 or z > 2.17

/2 = .015
0 2.17
Reject H0Do Not Reject H0
z
Reject H0
-2.17
Sampling
distribution
of z
x
n

m
s
0
/
s Known
/2 = .015

Confidence Interval Approach to
Two-Tailed Tests About a Population
Mean
 Select a simple random sample from the population
and use the value of the sample mean to develop
the confidence interval for the population mean m.
(Confidence intervals are covered in previous chapter)
x
 If the confidence interval contains the hypothesized
value m0, do not reject H0. Otherwise, reject H0.
(Actually, H0 should be rejected if m0 happens to be
equal to one of the end points of the confidence
interval.)

Confidence Interval Approach to
Two-Tailed Tests About a Population
Mean
The 97% confidence interval for m is
/2 6.1 2.17(.2 30) 6.1 .07924x z
n

s
    
Because the hypothesized value for the
population mean, m0 = 6, is not in this interval,
the hypothesis-testing conclusion is that the
null hypothesis, H0: m = 6, can be rejected.
or 6.02076 to 6.17924

Tests About a Population Mean:
s Unknown
 Test Statistic
t
x
s n

 m0
/
This test statistic has a t distribution
with n - 1 degrees of freedom.

 Rejection Rule: p -Value Approach
H0: m  m Reject H0 if t > t
Reject H0 if t < -t
Reject H0 if t < - t or t > t
H0: m  m
H0: mm
Tests About a Population Mean:
s Unknown
 Rejection Rule: Critical Value Approach
Reject H0 if p –value < 

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.

A State Highway Patrol periodically samples
vehicle speeds at various locations on a particular
roadway. The sample of vehicle speeds is used to
test the hypothesis H0: m < 65.
Example: Highway Patrol
 One-Tailed Test About a Population Mean: s Unknown
The locations where H0 is rejected are deemed the
best locations for radar traps. At Location F, a
sample of 64 vehicles shows a mean speed of 66.2
mph with a standard deviation of 4.2 mph. Use 
= .05 to test the hypothesis.

One-Tailed Test About a Population Mean:
s Unknown
 = .05
H0: m < 65
Ha: m > 65
m 
  0 66.2 65
2.286
/ 4.2/ 64
x
t
s n

s Unknown
For t = 2.286, the p–value must be less than .025
(for t = 1.998) and greater than .01 (for t = 2.387).
.01 < p–value < .025
Because p–value <  = .05, we reject H0.
We are at least 95% confident that the mean speed
of vehicles at Location F is greater than 65 mph.

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.
s Unknown
For  = .05 and d.f. = 64 – 1 = 63, t.05 = 1.669
Reject H0 if t > 1.669


0 t =
1.669
Reject H0
Do Not Reject H0
t
s Unknown

 The equality part of the hypotheses always appears
in the null hypothesis.
 In general, a hypothesis test about the value of a
population proportion p must take one of the
following three forms (where p0 is the hypothesized
value of the population proportion).
A Summary of Forms for Null and Alternative
Hypotheses About a Population Proportion
One-tailed
(lower tail)
One-tailed
(upper tail)
Two-tailed
H0: p > p0
Ha: p < p0
H0: p < p0
Ha: p > p0
H0: p = p0
Ha: p ≠ p0

 Test Statistic
Tests About a Population Proportion
where:
assuming np > 5 and n(1 – p) > 5
z
p p
p

 0
s
sp
p p
n

0 01( )

 Rejection Rule: p –Value Approach
H0: p  p Reject H0 if z > z
Reject H0 if z < -z
Reject H0 if z < -z or z > z
H0: p  p
H0: pp
Tests About a Population Proportion
Reject H0 if p –value < 
 Rejection Rule: Critical Value Approach

 Example: National Safety Council (NSC)
For a Christmas and New Year’s week, the
National Safety Council estimated that 500 people
would be killed and 25,000 injured on the nation’s
roads. The NSC claimed that 50% of the accidents
would be caused by drunk driving.
Two-Tailed Test About a
Population Proportion
A sample of 120 accidents showed that 67 were
caused by drunk driving. Use these data to test the
NSC’s claim with  = .05.

 = .05
0: .5H p 
: .5aH p 
0 0(1 ) .5(1 .5)
.045644
120
p
p p
n
s
 
  
s
 
  0 (67/120) .5
1.28
.045644p
p p
z
a common
error is using
in this
formula
p

 pValue Approach
4. Compute the p -value.
Because p–value = .2006 >  = .05, we cannot reject H0.
For z = 1.28, cumulative probability = .8997
p–value = 2(1  .8997) = .2006

For /2 = .05/2 = .025, z.025 = 1.96
4. Determine the criticals value and rejection rule.
Reject H0 if z < -1.96 or z > 1.96
Because 1.278 > -1.96 and < 1.96, we cannot reject H0.

Hypothesis Testing and Decision Making
 In many decision-making situations the decision
maker may want, and in some cases may be forced,
to take action with both the conclusion do not reject
H0 and the conclusion reject H0.
 In such situations, it is recommended that the
hypothesis-testing procedure be extended to include
consideration of making a Type II error.

Calculating the Probability of a Type II Error
in Hypothesis Tests About a Population Mean
1. Formulate the null and alternative hypotheses.
3. Using the rejection rule, solve for the value of the
sample mean corresponding to the critical value of
the test statistic.
2. Using the critical value approach, use the level of
significance  to determine the critical value and
the rejection rule for the test.

Calculating the Probability of a Type II Error
in Hypothesis Tests About a Population Mean
4. Use the results from step 3 to state the values of the
sample mean that lead to the acceptance of H0; this
defines the acceptance region.
5. Using the sampling distribution of for a value of m
satisfying the alternative hypothesis, and the acceptance
region from step 4, compute the probability that the
sample mean will be in the acceptance region. (This is
the probability of making a Type II error at the chosen
level of m.)
xx

 Example: Metro EMS (revisited)
The EMS director 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.
Recall that the response times for a random sample
of 40 medical emergencies were tabulated. The sample
mean is 13.25 minutes. The population standard
deviation is believed to be 3.2 minutes.
Calculating the Probability
of a Type II Error


 
12
1.645
3.2/ 40
x
z
 
   
 
3.2
12 1.645 12.8323
40
x
4. We will accept H0 when x < 12.8323
3. Value of the sample mean that identifies
the rejection region:
2. Rejection rule is: Reject H0 if z > 1.645
1. Hypotheses are: H0: m  and Ha: m
of a Type II Error

12.0001 1.645 .9500 .0500
12.4 0.85 .8023 .1977
12.8 0.06 .5239 .4761
12.8323 0.00 .5000 .5000
13.2 -0.73 .2327 .7673
13.6 -1.52 .0643 .9357
14.0 -2.31 .0104 .9896
12.8323
3.2/ 40
z
m

Values of mb1-b
5. Probabilities that the sample mean will be
in the acceptance region:
of a Type II Error

of a Type II Error
 Calculating the Probability of a Type II Error
 When the true population mean m is far above
the null hypothesis value of 12, there is a low
probability that we will make a Type II error.
 When the true population mean m is close to
the null hypothesis value of 12, there is a high
probability that we will make a Type II error.
Observations about the preceding table:
Example: m = 12.0001, b = .9500
Example: m = 14.0, b = .0104

Power of the Test
 The probability of correctly rejecting H0 when it is
false is called the power of the test.
 For any particular value of m, the power is 1 – b.
 We can show graphically the power associated
with each value of m; such a graph is called a
power curve. (See next slide.)

Power Curve
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
11.5 12.0 12.5 13.0 13.5 14.0 14.5
ProbabilityofCorrectly
RejectingNullHypothesis
m
H0 False

 The specified level of significance determines the
probability of making a Type I error.
 By controlling the sample size, the probability of
making a Type II error is controlled.
Determining the Sample Size for a Hypothesis Test
About a Population Mean

m0

ma
x
x
Sampling
distribution
of when
H0 is true
and m = m0
x
Sampling
distribution
of when
H0 is false
and ma > m0
x
Reject H0
b
c
c
H0: m  m
Ha: mm
Note:

where
z = z value providing an area of  in the tail
zb = z value providing an area of b in the tail
s = population standard deviation
m0 = value of the population mean in H0
ma = value of the population mean used for the
Type II error
n
z z
a



( )
( )
 b s
m m
2 2
0
2
n
z z
a



( )
( )
 b s
m m
2 2
0
2
Note: In a two-tailed hypothesis test, use z /2 not z

 Let’s assume that the director of medical
services makes the following statements about the
allowable probabilities for the Type I and Type II
errors:
•If the mean response time is m = 12 minutes, I am
willing to risk an  = .05 probability of rejecting H0.
•If the mean response time is 0.75 minutes over the
specification (m = 12.75), I am willing to risk a b = .10
probability of not rejecting H0.

 = .05, b = .10
z = 1.645, zb = 1.28
m0 = 12, ma = 12.75
s = 3.2
2 2 2 2
2 2
0
( ) (1.645 1.28) (3.2)
155.75 156
( ) (12 12.75)a
z z
n  b s
m m
 
   
 

Relationship Among , b, and n
 Once two of the three values are known, the other
can be computed.
 For a given level of significance , increasing the
sample size n will reduce b.
 For a given sample size n, decreasing  will increase
b, whereas increasing  will decrease b.

7 hypothesis testing

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