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CHAPTER 10
MORE ON HYPOTHESIS TESTING
Outline
• The p-value of a test of hypothesis
• The probability of a Type II error
1
THE p-VALUE
• For some given sample data the null hypothesis may be
rejected or not rejected depending on the significance
level.
• If the significance level is large, the rejection region is
large. So, the null hypothesis is more likely to be rejected.
• If the significance level is small, the rejection region is
small. So, the null hypothesis is less likely to be rejected.
• So, a natural question is for how large a level of
significance, the null hypothesis will be rejected and for
how small a level of significance the null hypothesis will not
be rejected.
• The p-value answers the above question.
2
THE p-VALUE
• If we know the p-value, then for any level of significance
larger than the p-value, the null hypothesis will be rejected
and for any level of significance smaller than the p-value,
the null hypothesis will not be rejected.
• The p-value of a test of hypothesis is the smallest value of
 that would lead to rejection of the null hypothesis.
• The p-value depends on the alternative hypothesis. For the
same sample data, the p-values are different for different
alternative hypothesis.
3
• The p-value for a two-tail ztest
Step 1: Using the sample
mean, standard deviation
and size of the sample,
compute the test statistic:
x 
z
/ n
Step 2: The p-value for a twotail test is the sum of two
areas - one on the right of
|z| and one on the left of -|z|
f(x)
THE p-VALUE
TWO-TAIL
p-value of a two-tail
z-test is the

x 
sum of two
n
areas
z

z
4
Step 2: The p-value for a
right-tail test is the area on
the right of z.
p-value of
a right-tail
z-test
x 

f(x)
• The p-value for a right-tail
z-test
Step 1: Using the sample
mean, standard deviation
and size of the sample,
compute the test statistic:
x 
z
/ n
f(x)
THE p-VALUE
RIGHT-TAIL
p-value of
a right-tail
z-test

n
z
x 
z


n
5
Step 2: The p-value for a lefttail test is the area on the
left of z.
p-value of
a left-tail
z-test
z
f(x)
• The p-value for a left-tail ztest
Step 1: Using the sample
mean, standard deviation
and size of the sample,
compute the test statistic:
x 
z
/ n
f(x)
THE p-VALUE
LEFT-TAIL
p-value of
a left-tail
z-test
x 

n

x 
 z

n
6
THE p-VALUE
TWO-TAIL
Example 7: A machine that produces ball bearings is set so
that the average diameter is 0.60 inch. In a sample of 49
ball bearings, the mean diameter was found to be 0.61 inch.
Assume that the standard deviation is 0.035. Can we
conclude that the mean diameter is not 0.60 inch? State the
null hypothesis and alternative hypothesis. Compute and
interpret the p-value of the test of hypothesis.
7
THE p-VALUE
TWO-TAIL
1. State H0 and HA
HO :
HA :
x 
2. Find test statistic z 
/ n
3. Compute the p-value
f(x)
x 

n
4. Interpret the p-value

8
THE p-VALUE
TWO-TAIL
Example 8: A machine that produces ball bearings is set so
that the average diameter is 0.60 inch. In a sample of 49
ball bearings, the mean diameter was found to be 0.61 inch.
Assume that the standard deviation is 0.035. Can we
conclude that the mean diameter is not 0.60 inch? Consider
following values of level of significance:
1.  = 1%

 
2.  = 3%
n
3.  = 5%
4.  = 10%
Use the p-value from Example 7.

f(x)
x
9
THE p-VALUE
RIGHT-TAIL
Example 9: A random sample of 100 observations from a
normal population whose standard deviation is 50 produced
a mean of 145. Does this statistic provide sufficient
evidence at the 5% significance level to infer that population
mean is more than 140? State the null hypothesis and
alternative hypothesis. Compute and interpret the p-value of
the test of hypothesis.
10
THE p-VALUE
RIGHT-TAIL
1. State H0 and HA
HO :
HA :
x 
2. Find test statistic z 
/ n
3. Compute the p-value
f(x)
x 

n
4. Interpret the p-value

11
THE p-VALUE
LEFT-TAIL
Example 10: A random sample of 100 observations from a
normal population whose standard deviation is 50 produced
a mean of 145. Does this statistic provide sufficient
evidence at the 5% significance level to infer that population
mean is less than 150? State the null hypothesis and
alternative hypothesis. Compute and interpret the p-value of
the test of hypothesis.
12
THE p-VALUE
LEFT-TAIL
1. State H0 and HA
HO :
HA :
x 
2. Find test statistic z 
/ n
3. Compute the p-value
f(x)
x 

n
4. Interpret the p-value

13
TYPE II ERROR
• There are two types of errors in a test of hypothesis
• The level of significance,  sets a limit on the Type I error.
• A Type I error is committed when the null hypothesis is
actually true, but the sample data leads to the rejection of
the null hypothesis.
• The Type II error is the opposite of Type I error.
• A Type II error is committed when the null hypothesis is
actually false, but the sample data does not lead to the
rejection of the null hypothesis.
14
TYPE II ERROR
• The probability of committing a Type II error is denoted by .
• Example: consider the manufacturer of the packaged cereal
again. Each cereal box is expected to have a net weight of
100 gm. But, due to some problems in the production
system, the average weight is shifted to 98 gm. A Type II
error is committed if a sample is collected with average
weight nearly 100 gm. Notice that in such a case, the
problem with the production system will not be detected by
the sample!
15
TYPE II ERROR
• Given
• The Type II error,  is obtained
as follows
– Null and alternative
hypothesis, H0 and
– Find the rejection region and
HA
state it in terms of x-values
explained in the next 3 slides
– Level of significance,

– Type II error,  is the
probability that the sample
– Actual population
mean will not lie in the
mean, 
rejection region when the
– Standard deviation, 
actual population mean is ,
– Sample size n
standard deviation is  and a
random sample of size n is
16
drawn
TYPE II ERROR
TWO-TAIL
Rejection Region

x   H O  z / 2
and
n

x   H O  z / 2
n
Type II Error





z

x
 HO

/2
n


P

 
 x   H O  z / 2

n

17
TYPE II ERROR
RIGHT-TAIL
Rejection Region

x   H O  z
n
Type II Error
 

  P x   H O  z 

n

18
TYPE II ERROR
LEFT-TAIL
Rejection Region

x   H O  z
n
Type II Error
 

  P x   H O  z 

n

19
TYPE II ERROR PROBABILITY
TWO-TAIL
Example 11: A machine that produces ball bearings is set so
that the average diameter is 0.60 inch. In a sample of 49
ball bearings, the mean diameter was found to be 0.61 inch.
Assume that the standard deviation is 0.035. Can we
conclude that the mean diameter is not 0.60 inch? If the
probability of the of a Type I error is chosen to be 0.05 (in
other words, the level of significance,  of the test of
hypothesis is 0.05), calculate the probability of a Type II
error, assuming that the true population mean is 0.605 inch.
(Notice that the sample mean of 0.61 inch is irrelevant)
20
TYPE II ERROR PROBABILITY
TWO-TAIL
1. State H O :
HA :
2. Find rejection region:
3. Convert rejection region in terms of x-values


x   H O  z / 2
, x   H O  z / 2
n
n
f(x)
x 

n
4. Find 

21
TYPE II ERROR PROBABILITY
RIGHT-TAIL
Example 12: A random sample of 100 observations from a
normal population whose standard deviation is 50 produced a
mean of 145. A hypothesis test is conducted to see if this
statistic provides sufficient evidence at the 5% significance
level to infer that population mean is more than 140.
Calculate the probability of a Type II error of the test of
hypothesis, assuming that the true population mean is 142.
(Notice that the sample mean of 145 is irrelevant)
22
TYPE II ERROR PROBABILITY
RIGHT-TAIL
1. State H O :
HA :
2. Find rejection region:
3. Convert rejection region in terms of x-values

x   H O  z
n
f(x)
x 

n
4. Find 

23
TYPE II ERROR PROBABILITY
LEFT-TAIL
Example 13: A random sample of 100 observations from a
normal population whose standard deviation is 50 produced
a mean of 145. A hypothesis test is conducted to see if this
statistic provide sufficient evidence at the 5% significance
level to infer that population mean is less than 150.
Calculate the probability of a Type II error of the test of
hypothesis, assuming that the true population mean is 147.
(Notice that the sample mean of 145 is irrelevant)
24
TYPE II ERROR PROBABILITY
LEFT-TAIL
1. State H O :
HA :
2. Find rejection region:
3. Convert rejection region in terms of x-values

x   H O  z
n
f(x)
x 

n
4. Find 

25
READING AND EXERCISES
• Sections 10.4-10.5:
– Reading: pp. 346-359
– Exercises: 10.14,10.16,10.20, 10.22, 10.24, 10.32
26
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