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Chapter 9
Hypothesis Testing
McGraw-Hill/Irwin
Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.
Hypothesis Testing
9.1
Null and Alternative Hypotheses and
Errors in Testing
9.2
z Tests about a Population Mean
s Known
9.3
t Tests about a Population Mean
s Unknown
9-2
Null and Alternative Hypotheses and
Errors in Hypothesis Testing
• The null hypothesis, denoted H0, is a
statement of the basic proposition being
tested. The statement generally represents
the status quo and is not rejected unless
there is convincing sample evidence that it is
false.
• The alternative or research hypothesis,
denoted Ha, is an alternative (to the null
hypothesis) statement that will be accepted
only if there is convincing sample evidence
that it is true.
9-3
Example 9.1: Trash Bag Case
• Tests show the trash bag has a mean
breaking strength µ close to but not
exceeding 50 lbs
– The null hypothesis H0 is that the new bag
has a mean breaking strength that is 50 lbs
or less
• The new bag’s mean breaking strength
is not known and is in question, but it is
hoped it is stronger than the current one
9-4
Example 9.1: Trash Bag Case
Continued
• The alternative hypothesis Ha is that the
new bag has a mean breaking strength
that exceeds 50 lbs
• One-sided, “greater than” alternative
– H0: µ ≤ 50
Ha: µ > 50
versus
9-5
Example 9.2: Payment Time Case
• With new billing system, the mean bill paying
time µ is hoped to be less than 19.5 days
– Alternative hypothesis Ha is the new billing system
has a mean payment time less than 19.5 days
• With old billing system, the mean bill paying
time µ was close to but not less than 39 days
– The null hypothesis H0 is that the new billing
system has a mean payment time close to but not
less than 39 days
• One-sided, “less than” alternative
– H0:   19.5 versus Ha:  < 19.5
9-6
Example 9.3: The Valentine’s Day
Chocolate Case
• They have designed a new box
• They hope sales increase by 10 percent
• Last year’s sales were 330 boxes
– The null hypothesis will be that sales this
year will not be 330 boxes
• Two-sided, “not equal to” alternative
– H0:  = 330 versus Ha:  ≠ 330
9-7
Types of Hypotheses
• One-Sided, “Greater Than” Alternative
H0:   0 vs.
Ha:  > 0
• One-Sided, “Less Than” Alternative
H0 :   0 vs.
Ha :  < 0
• Two-Sided, “Not Equal To” Alternative
H0 :  = 0 vs.
Ha :   0
where 0 is a given constant value (with the
appropriate units) that is a comparative value
9-8
Types of Decisions
• As a result of testing H0 vs. Ha, will have
to decide either of the following
decisions for the null hypothesis H0:
• Do not reject H0
– A weaker statement than “accepting H0”
– But you are rejecting the alternative Ha
• Or reject H0
– A weaker statement than “accepting Ha”
9-9
Test Statistic
• To “test” H0 vs. Ha, use the “test statistic”
x  0
x  0
z

sx
s n
where 0 is the given value (often the
“claimed to be true”) and x is the sample
mean
• z measures the distance between 0 and x
on the sampling distribution of the sample
mean
• If the population is normal or n is large*, the
test statistic z follows a normal distribution
*
n ≥ 30, by the Central Limit Theorem
9-10
Test Statistic and Trash Bag Case
• With µ0 = 50, use z  x  50  x  50
the test statistic:
sx
s n
• If z ≤ 0, then x ≤ µ0 and there is no evidence
to support rejecting H0 in favor of Ha
– The point estimate x indicates µ is probably less
than 50
• If z > 0, then x > µ0 and there is evidence to
support rejecting H0 in favor of Ha
– The point estimate x indicates that µ is probably
greater than 50
– The larger z (the farther x is above µ), the stronger
the evidence to support rejecting H0 in favor of Ha
9-11
Type I and Type II Errors
Type I Error: Rejecting H0 when it is true
Type II Error: Failing to reject H0 when it is false
State of Nature
Conclusion
Reject H0
Do not Reject H0
H0 True
Type I
Error
Correct
Decision
H0 False
Correct
Decision
Type II
Error
9-12
Error Probabilities
• Type I Error: Rejecting H0 when it is true
–  is the probability of making a Type I error
– 1 –  is the probability of not making a Type I error
• Type II Error: Failing to reject H0 when it is
false
– β is the probability of making a Type II error
– 1 – β is the probability of not making a Type II
error
9-13
Error Probabilities
State of Nature
Conclusion
Reject H0
Do not Reject H0
H0 True
H0 False

1- 
1-β
β
9-14
Typical Values
• Usually set  to a low value
– So there is a small chance of rejecting a
true H0
– Typically,  = 0.05
• Strong evidence is required to reject H0
• Usually choose  between 0.01 and 0.05
–  = 0.01 requires very strong evidence is to reject H0
– Tradeoff between  and β
• For fixed sample size, the lower , the higher β
– And the higher , the lower β
9-15
z Tests about a Population Mean:
σ Known
• Test hypotheses about a population mean
using the normal distribution
– Called z tests
– Require that the true value of the population
standard deviation σ is known
• In most real-world situations, σ is not known
– But often is estimated from s of a single sample
– When σ is unknown, test hypotheses about a population
mean using the t distribution
• Here, assume that we know σ
• Also use a “rejection rule”
9-16
Steps in Testing a “Greater Than”
Alternative
1.
2.
3.
4.
State the null and alternative hypotheses
Specify the significance level 
Select the test statistic
Determine the rejection rule for deciding
whether or not to reject H0
5. Collect the sample data and calculate the
value of the test statistic
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
7. Interpret the statistical results in managerial
terms and assess their practical importance
9-17
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #1
• State the null and alternative
hypotheses
H0:   50
Ha:  > 50
where μ is the mean breaking strength of
the new bag
• Specify the significance level 
–  = 0.05
9-18
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #2
3. Select the test statistic
– Use the test statistic
z
x  50
x  50

sx
s n
– A positive value of this this test statistic
results from a sample mean that is
greater than 50 lbs
•
Which provides evidence against H0 in favor
of Ha
9-19
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #3
4. Determine the rejection rule for deciding
whether or not to reject H0
– To decide how large the test statistic must be to
reject H0 by setting the probability of a Type I
error to , do the following:
– The probability  is the area in the right-hand tail
of the standard normal curve
– Use the normal table to find the point z (called
the rejection or critical point)
•
•
z is the point under the standard normal curve that
gives a right-hand tail area equal to 
Since  = 0.05 in the trash bag case, the rejection point
is z = z0.05 = 1.645
9-20
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #4
4. Continued
– Reject H0 in favor of Ha if the test statistic
z is greater than the rejection point z
•
This is the rejection rule
– In the trash bag case, the rejection rule is
to reject H0 if the calculated test statistic z
is > 1.645
9-21
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #5
5. Collect the sample data and calculate
the value of the test statistic
– In the trash bag case, assume that σ is
known and σ = 1.65 lbs
– For a sample of n = 40, x = 50.575 lbs.
Then
z
x  50
s
n

50 .575  50
1.65
 2.20
40
9-22
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #6
9-23
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #7
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
– Compare the value of the test statistic to the
rejection point according to the rejection rule
– In the trash bag case, z = 2.20 is greater than
z0.05 = 1.645
– Therefore reject H0: μ ≤ 50 in favor of Ha: μ > 50
at the 0.05 significance level
•
Have rejected H0 by using a test that allows only a 5%
chance of wrongly rejecting H0
•
This result is “statistically significant” at the 0.05 level
9-24
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #8
7. Interpret the statistical results in
managerial terms and assess their
practical importance
– Can conclude that the mean breaking
strength of the new bag exceeds 50 lbs
9-25
Steps in Testing a “Greater Than”
Alternative in Trash Bag Case #9
Testing H0:   50 versus Ha:  > 50
9-26
Effect of 
• At  = 0.01, the rejection point is z0.01 = 2.33
• In the trash example, the test statistic
z = 2.20 is < z0.01 = 2.33
• Therefore, cannot reject H0 in favor of Ha at
the  = 0.01 significance level
– This is the opposite conclusion reached with
=0.05
– So, the smaller we set , the larger is the rejection
point, and the stronger is the statistical evidence
that is required to reject the null hypothesis H0
9-27
The p-Value
• The p-value or the observed level of
significance is the probability of the obtaining
the sample results if the null hypothesis H0 is
true
– The p-value is used to measure the weight of the
evidence against the null hypothesis
• Sample results that are not likely if H0 is true
have a low p-value and are evidence that H0
is not true
– The p-value is the smallest value of  for which
we can reject H0
• The p-value is an alternative to testing with a
z test statistic
9-28
The p-Value for “Greater Than”
Alternative
• For a greater than alternative, the
p-value is the probability of observing a
value of the test statistic greater than or
equal to z when H0 is true
• Use p-value as an alternative to testing
a greater than alternative with a z test
statistic
9-29
The p-value for “Greater Than”
Example: Trash Bag Case
• Testing H0: µ ≤ 50
versus Ha: µ > 50
using rejection
points in (a) and pvalue in (b)
• Have to modify
steps 4, 5, and 6 of
the previous
procedure to use pvalues
9-30
Steps Using a p-value to Test a
“Greater Than” Alternative
4. Collect the sample data and compute the
value of the test statistic
– In the trash bag case, the value of the test
statistic was calculated to be z = 2.20
5. Calculate the p-value by corresponding to
the test statistic value
– In the trash bag case, the area under the
standard normal curve in the right-hand tail to the
right of the test statistic value z = 2.20
– The area is 0.5 – 0.4861 = 0.0139
– The p-value is 0.0139
9-31
Steps Using a p-value to Test a
“Greater Than” Alternative
Continued
5. Continued
– If H0 is true, the probability is 0.0139 of obtaining
a sample whose mean is 50.575 lbs or higher
– This is so low as to be evidence that H0 is false
and should be rejected
6. Reject H0 if the p-value is less than 
– In the trash bag case, a was set to 0.05
– The calculated p-value of 0.0139 is <  = 0.05
•
This implies that the test statistic z = 2.20 is greater
than the rejection point z0.05 = 1.645
– Therefore reject H0 at the  = 0.05 significance
level
9-32
Steps in Testing a “Less Than”
Alternative
1. State the null and alternative hypotheses
2. Specify the significance level 
3. Select the test statistic
4. Determine the rejection rule for deciding whether
or not to reject H0
5. Collect the sample data and calculate the value of
the test statistic
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
7. Interpret the statistical results in managerial terms
and assess their practical importance
9-33
Steps in Testing a “Less Than”
Alternative in Payment Time Case #1
1. State the null and alternative
hypotheses
– In the payment time case, H0:  ≥ 19.5
vs. Ha:  < 19.5, where  is the mean bill
payment time (in days)
2. Specify the significance level 
– In the payment time case, set  = 0.01
9-34
Steps in Testing a “Less Than”
Alternative in Payment Time Case #2
3. Select the test statistic
– In the payment time case, use the test
statistic
x  19 .5 x  19 .5
z

sx
s n
– A negative value of this this test statistic
results from a sample mean that is less
than 19.5 days
•
Which provides evidence against H0 in favor
of Ha
9-35
Steps in Testing a “Less Than”
Alternative in Payment Time Case #3
4. Determine the rejection rule for deciding
whether or not to reject H0
– To decide how much less than 0 the test statistic
must be to reject H0 by setting the probability of a
Type I error to , do the following:
– The probability  is the area in the left-hand tail
of the standard normal curve
– Use normal table to find the rejection point –z
•
–z is the negative of z
•
–z is the point on the horizontal axis under the
standard normal curve that gives a left-hand tail area
equal to 
9-36
Steps in Testing a “Less Than”
Alternative in Payment Time Case #3
4. Continued
– Since  = 0.01 in the payment time case,
the rejection point is –z = –z0.01 = –2.33
– Reject H0 in favor of Ha if the test statistic
z is calculated to be less than the
rejection point –z
•
This is the rejection rule
– In the payment time case, the rejection
rule is to reject H0 if the calculated test
statistic –z is less than –2.33
9-37
Steps in Testing a “Less Than”
Alternative in Payment Time Case #4
5. Collect the sample data and calculate
the value of the test statistic
– In the payment time case, assume that σ
is known and σ = 4.2 days
– For a sample of n=65, x = 18.1077 days:
z
x  19 .5
s
n

18 .1077  19 .5
4.2
 2.67
65
9-38
Steps in Testing a “Less Than”
Alternative in Payment Time Case #5
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
– Compare the value of the test statistic to the
rejection point according to the rejection rule
– In the payment time case, z = –2.67 is less than
z0.01 = –2.33
– Therefore reject H0: μ ≥ 19.5 in favor of
Ha: μ < 19.5 at the 0.01 significance level
•
Have rejected H0 by using a test that allows only a 1%
chance of wrongly rejecting H0
•
This result is “statistically significant” at the 0.01 level
9-39
Steps in Testing a “Less Than”
Alternative in Payment Time Case #6
7. Interpret the statistical results in
managerial terms and assess their
practical importance
– Can conclude that the mean bill payment
time of the new billing system is less than
19.5 days
9-40
Steps in Testing a “Less Than”
Alternative in Payment Time Case
Testing H0: µ ≥ 19.5 versus Ha: µ < 19.5 for  = 0.01
9-41
The p-value for “Less Than”
Have to modify steps 4, 5, and 6 of the
previous procedure to use p-values
9-42
Steps Using a p-value to Test a
“Less Than” Alternative
(Steps 1–3 are the same)
4. Collect the sample data and compute the
value of the test statistic
– In the payment time case, the value of the test
statistic was calculated to be z = –2.67
5. Calculate the p-value by corresponding to
the test statistic value
– In the payment time case, the area under the
standard normal curve in the left-hand tail to the
left of the test statistic z = –2.67
– The area is 0.5 – 0.4962 = 0.0038
– The p-value is 0.0038
9-43
Steps Using a p-value to Test a
“Less Than” Alternative
Continued
5. Continued
– If H0 is true, then the probability is 0.0038 of
obtaining a sample whose mean is as low as
18.1077 days or lower
– This is so low as to be evidence that H0 is false
and should be rejected
6. Reject H0 if the p-value is less than 
– In the payment time case,  was 0.01
– The calculated p-value of 0.0038 is <  = 0.01
•
This implies that the test statistic z = –2.67 is less than
the rejection point –z0.01 = –2.33
– Therefore, reject H0 at the  = 0.01 significance
level
9-44
Summary of Testing a One-Sided
Alternative Using a Test Statistic
If the population is normal and s is known*, we can
reject H0:  = 0 at the  level of significance
(probability of Type I error equal to ) if and only if
the appropriate rejection point rule holds
Test Statistic
z
x  0
s
Alternative
Ha :  > 0
Reject H0 if:
z > z
Ha :  < 0
z < z
n
* If s unknown and n is very large (n > 120), estimate s by s
9-45
Summary of Testing a One-Sided
Alternative Using a p-value
If the population is normal and s is known*, we can
reject H0:  = 0 at the  level of significance
(probability of Type I error equal to ) if and only if
the corresponding p-value is less than the specified 
Test Statistic
z
x  0
s
n
Alternative
Ha :  > 0
Reject H0 if:
p > z
Ha :  < 0
p < z
(1)  is the area under the standard normal curve to the right of z
(2)  is the area under the standard normal curve to the left of -z
* If s unknown and n is very large (n > 120), estimate s by s
9-46
Steps in Testing a “Not Equal To”
Alternative
1. State the null and alternative hypotheses
2. Specify the significance level 
3. Select the test statistic
4. Determine the rejection rule for deciding whether
or not to reject H0
5. Collect the sample data and calculate the value
of the test statistic
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
7. Interpret the statistical results in managerial
terms and assess their practical importance
9-47
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #1
1. State null and alternative hypotheses
– In the Valentine Day case, H0:  = 330 vs.
Ha:  ≠ 330
2. Specify the significance level 
– In the Valentine Day case, set  = 0.05
3. Select the test statistic
– In this case, use the test statistic
z
x  330
sx

x  330
s n
9-48
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #2
3. Continued
– A positive value of this test statistic results from a
sample mean that is greater than 330
•
Which provides evidence against H0 and for Ha
– A negative value of this test statistic results from
a sample mean that is less than 330
•
Which provides evidence against H0 and for Ha
– A very small value close to 0 (either slightly
positive or slightly negative) of this test statistic
results from a sample mean that is nearly 330
•
Which provides evidence in favor of H0 and against Ha
9-49
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #3
4. Determine the rejection rule for deciding
whether or not to reject H0
– To decide how different the test statistic must be
from zero (positive or negative) to reject H0 in
favor of Ha by setting the probability of a Type I
error to , do the following:
•
Divide the  in half to find /2; /2 is the tail area in
both tails of the standard normal curve
•
Under the standard normal curve, the probability /2 is
the area in the right-hand tail and probability /2 is the
area in the left-hand tail
9-50
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #4
4. Continued
– Use the normal table to find the rejection
points z/2 and –z/2
•
z/2 is the point on the horizontal axis under
the standard normal curve that gives a righthand tail area equal to /2
•
–z/2 is the point on the horizontal axis under
the standard normal curve that gives a lefthand tail area equal to /2
9-51
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #5
4. Continued
–
Because  = 0.05, /2=0.025
•
The area under the standard normal to the right of the
mean is 0.5 – 0.025 = 0.475
•
From Table A.3, the area is 0.475 for z = 1.96
–
Rejection points are z=1.96,–z=– 1.96
–
Reject H0 in favor of Ha if the test statistic z
satisfies either:
•
z greater than the rejection point z/2, or
•
–z less than the rejection point –z/2
•
This is the rejection rule
9-52
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #6
5. Collect the sample data and calculate
the value of the test statistic
– In the Valentine Day case, assume that σ
is known and σ = 40
– For a sample of n = 100, x = 326
– Then
x  330 326  330
z

 1.00
s n
40 100
9-53
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #7
6. Decide whether to reject H0 by using the test
statistic and the rejection rule
– Compare the value of the test statistic to the
rejection point according to the rejection rule
– In this case, – z = –1.00 is < – z0.025 = –1.96
– Therefore cannot reject H0: µ = 330 in favor of
Ha: µ ≠ 330 at the 0.05 significance level
•
Have not rejected H0 by using a test that allows only a
5% chance of wrongly rejecting H0
9-54
Steps in Testing a “Not Equal To”
Alternative in Valentine Day Case #8
7. Interpret the statistical results in
managerial terms and assess their
practical importance
– Cannot conclude that the mean order
quantity this year of the Valentine Day
box at large retail stores will differ from
330 boxes
9-55
Steps Using a p-value to Test a
“Not Equal To” Alternative
(Steps 1–3 are the same)
4. Collect the sample data and compute the
value of the test statistic
– In the Valentine Day case, the value of the test
statistic was calculated to be z = –1.00
5. Calculate the p-value by corresponding to
the test statistic value
– In the Valentine Day case, the area under the
standard normal curve in the left-hand tail to the
left of the test statistic value z = –1.00
– The area is 0.1587
– The p-value is 0.1587 · 2 = 0.3174
9-56
Steps Using a p-value to Test a
“Not Equal To” Alternative
Continued
5. Continued
– That is, if H0 is true, then the probability is 0.3174
of obtaining a sample whose mean is at least as
extreme as 326
– This probability is not so low as to be evidence
that H0 is false and should be rejected
6. Reject H0 if the p-value is less than a
– In the Valentine Day case,  was 0.05
– Calculated p-value of 0.3174 is greater than 
•
This implies that the test statistic z = –1.00 is greater
than the rejection point –z0.025 = –1.96
– Therefore do not reject H0 at the  = 0.05
significance level
9-57
Testing H0: µ = 330 Versus Ha: µ ≠ 330 by
Using Critical Values and the p-Value
9-58
Summary of Testing a Two-Sided Not
Equal Alternative Using a p-value
If the population is normal and s is known*, we can
reject H0:  = 0 at the  level of significance
(probability of Type I error equal to ) if and only if:
Test Statistic
z
x  0
s
n
Alternative
Ha :  > 0
Reject H0 if:
z > z/2 or p < 
(1)
z > –z/2 or p < 
(2)
(1)  is the area under the standard normal curve to the right of z
(2)  is the area under the standard normal curve to the left of -z
* If s unknown and n is very large (n > 120), estimate s by s
9-59
Weight of Evidence Against the Null
• Calculate the test statistic and the
corresponding p-value
• Rate the strength of the conclusion about the
null hypothesis H0 according to these rules:
– If p < 0.10, there is some evidence to reject H0
– If p < 0.05, there is strong evidence to reject H0
– If p < 0.01, there is very strong evidence to reject
H0
– If p < 0.001, there is extremely strong evidence
to reject H0
9-60
t Tests about a Population Mean:
σ Unknown
• Suppose the population being sampled
is normally distributed
• The population standard deviation σ is
unknown, as is the usual situation
– If the population standard deviation σ is
unknown, then it will have to estimated
from a sample standard deviation s
• Under these two conditions, have to use
the t distribution to test hypotheses
9-61
Defining the t Random Variable:
σ Unknown
• Define a new random variable t:
t
x 
s
n
– with the definition of symbols as before
• The sampling distribution of this random
variable is a t distribution with n – 1
degrees of freedom
9-62
Defining the t Statistic: σ Unknown
• Let x be the mean of a sample of size n with
standard deviation s
• Also, µ0 is the claimed value of the population
mean
• Define a new test statistic
x  0
t
s n
• If the population being sampled is normal,
and s is used to estimate σ, then …
• The sampling distribution of the t statistic is a
t distribution with n – 1 degrees of freedom
9-63
t Tests about a Population Mean:
σ Unknown
• Reject H0: µ = µ0 in favor of a particular
alternative hypothesis Ha at the a level
of significance if and only if the
appropriate rejection point rule or,
equivalently, the corresponding p-value
is less than 
• We have the following rules …
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t Tests about a Population Mean:
σ Unknown
Continued
Alternative Reject H0 if: p-value
Ha: µ > µ0
t > t
Area under t
distribution to right of t
Ha: µ < µ0
t < –t
Area under t
distribution to left of –t
Ha: µ  µ0
|t| > t /2 *
Twice area under t
distribution to right of |t|
t, t/2, and p-values are based on n – 1 degrees of
freedom (for a sample of size n)
* either t > t/2 or t < –t/2
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Example 9.4 #1
• The current mean credit card interest rate is
as it was in 1991, at the rate of 18.8%
– This is the null hypothesis H0: µ = 18.8
• The alternative to be tested is that the current
mean interest rate is not as it was back in
1991, but in fact has decreased
– This is the alternative hypothesis Ha: µ  18.8
• Test at the  = 0.05 level of significance
– If H0 can be rejected at the 5% level, then
conclude that the mean rate now is less than
18.8% rate charged in 1991
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Example 9.4 #2
• Randomly select n = 15 credit cards
– n = 15
– x = 16.828
– s = 1.538
–  = 0.05
• Then df = n – 1 = 14
• Assume the population of the rates of
all cards is normal
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Example 9.4 #3
• The rejection rule is reject
H0:  = 18.8 against
Ha:  < 18.8 if t < t
– For  = 0.05 and with df = 14,
t = t0.05 =1.761
– Calculate the value of the t statistic:
t
x  18.8
s n

16.827  18.8
1.538 15
 4.97
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Example 9.4 #4
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Example 9.4 #5
• Because t = 4.97 < t0.05 = 1.761,
reject H0
• Therefore, conclude at 5% significance
level that the mean interest rate is lower
than it was in 1991
– In fact it is µ0 – x = 18.8 – 16.827 = 1.973%
lower
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p-value for Example 9.4
• The p-value is the left-hand tail area under
the t curve with df = 14 to the left of t = 4.97
– This t value is off the t table, so use statistics
software on a computer to calculate the p-value
• p = 0.000103
– This says that if the claimed mean of 18.8 % is
true, then there is only a 0.01% chance of
randomly selecting a sample of 15 credit whose
mean rate would be as low as 16.827% or lower
– This p-value is less than 0.05, 0.01, and 0.001
• Reject H0 at the 0.05, 0.01, 0.001 significance
levels
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Computer Software Output for
Example 9.4
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