```Chapter
Tests of Hypotheses:
Large Samples
Rejection
region
Acceptance
region
1
GOALS
2
TO DEFINE HYPOTHESES AND
HYPOTHESIS TESTING.
TO DESCRIBE THE HYPOTHESIS TESTING
PROCEDURE.
TO DISTINGUISH BETWEEN ONE-TAILED
AND TWO-TAILED TEST OF HYPOTHESIS.
TO CONDUCT A TEST FOR THE
POPULATION MEAN OR PROPORTION.
GOALS
3
TO CONDUCT A TEST FOR THE
DIFFERENCE BETWEEN TWO
POPULATION MEANS OR PROPORTIONS.
TO DESCRIBE STATISTICAL ERRORS
ASSOCIATED WITH HYPOTHESIS TESTING.
WHAT IS A HYPOTHESIS?
4
 Hypothesis: A statement about the value of
a population parameter developed for the
purpose of testing.
 Examples of hypotheses, or statements,
» The mean monthly income from all sources for
systems analysts is \$3,625.
» Twenty percent of all juvenile offenders ultimately
are caught and sentenced to prison.
WHAT IS HYPOTHESIS TESTING?
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 Hypothesis testing: A procedure, 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.
 Following is a five-step procedure for
testing a hypothesis.
STEP 1
State null H0 and alternative
hypotheses H1
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 H0
Reject H0 and
accept H1
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 Null Hypothesis H0: A statement about
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the 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.
 Level of Significance: The probability of
rejecting the null hypothesis when it is
actually true.
 Type I Error: Rejecting the null hypothesis,
H0, when it is actually true.
 Type II Error: Accepting the null hypothesis,
H0, when it is actually false.
Researcher
Null hypothesis
Accepts H0
If H0 is true and
Correct
Decision
If H0 is false and
Type II
error
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Rejects H0
Type I
error
Correct
Decision
 Test statistic: A value, determined from
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sample information, used to determine
whether or not to reject the null hypothesis.
 Critical value: The dividing point between
the region where the null hypothesis is
rejected and the region where it is not
rejected.
ONE-TAILED TESTS OF SIGNIFICANCE
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 A test is one-tailed when the alternate
hypothesis, H1, states a direction such as:
 H0: The mean income of the females is less
than or equal to the mean income of the
males.
 H1: The mean income of the females is
greater than the mean income of the males.
 The region of rejection in this case is to the
right (upper) tail of the curve. An example
is shown next:
Sampling Distribution for the Statistic Z for a
One-Tailed Test, 0.05 Level of Significance
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Critical
value
1.645 = Z
Region of
rejection
0.95 Probability
0.05
z
TWO-TAILED TESTS OF SIGNIFICANCE
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 A test is two-tailed when no direction is
specified in the alternate hypothesis H1,
such as:
 H0: The mean income of the females is
equal to the mean income of the males.
 H1: The mean income of the females is not
equal to the mean income of the males.
 The region of rejection in this case is
divided equally into the two tails of the
curve. An example is shown next:
Sampling Distribution for the Statistic Z for a
Two-Tailed Test, 0.05 Level of Significance
Critical
value
-1.96 = Z
Do not
reject H0
Critical
value
1.96 = Z
Region of
rejection
Region of
rejection
0.025
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0.95
0.025
z
TESTING FOR THE POPULATION
MEAN: LARGE SAMPLE, POPULATION
STANDARD DEVIATION KNOWN
 The test statistic is given by:
X


z
/ n
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EXAMPLE
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 The processors of Mets Catsup indicate on the
label that the bottle contains 16 ounces of catsup.
Mets’ Quality Control Department is responsible
for monitoring the amount included in the bottle.
A sample of 36 bottles is selected hourly and the
contents weighed. Last hour a sample of 36
bottles had a mean weight of 16.12 ounces with a
standard deviation of 0.5 ounces. At the 0.05
significance level can we conclude that the
process is out of control? (i.e. not meeting weight
goals)
EXAMPLE (continued)
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 Step 1: State the null and the alternative
hypotheses.
 H0:  = 16
H1:  16
 Step 2: State the decision rule.
 H0 is rejected if z < -1.96 or z > 1.96.(2-side)
 Step 3: Compute the value of the test
statistic.
 z= [16.12 - 16]/[0.05/36] = 1.44.
 Step 4: What is the decision on H0?
 H0 is not rejected, because 1.44 is less than
the critical value of 1.96.
Sampling Distribution for the Statistic Z for a
Two-Tailed Test, 0.05 Level of Significance
Critical
value
-1.96 = Z
Do not
reject H0
0.025
Critical
value
1.96 = Z
Test Stat.
1.44
Region of
rejection
0.95
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Region of
rejection
0.025
z
TESTING FOR THE POPULATION
MEAN: LARGE SAMPLE, POPULATION
STANDARD DEVIATION UNKNOWN
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is unknown, so we estimate it with
the sample standard deviation s.
 Here
As long as the sample size n 30, z can be
approximated with
X


z
s/ n
EXAMPLE
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 The Thompson’s Discount Store chain issues
its own credit card. The credit manager wants
to find out if the mean monthly unpaid balance
is more than \$400. The level of significance is
set at 0.05. A random check of 172 unpaid
balances revealed the sample mean to be \$407
and the sample standard deviation to be \$38.
Should the credit manager 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 (continued)
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 Step 1: State the null and the alternative
hypotheses.
 H0:   400
H1: > 400
 Step 2: State the decision rule.
 H0 is rejected if z > 1.645. (one-sided)
 Step 3: Compute the value of the test statistic.
 z = [407 - 400]/[38/172] = 2.42.
 Step 4: What is the decision on H0?
 H0 is rejected. The manager can conclude that
mean unpaid balance is greater than \$400.
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Computed
z = 2.42
Rejection
region
0
1.645
z
HYPOTHESIS TESTING: TWO
POPULATION MEANS
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Assume the parameters for the two populations
are 1,2,1,and 2.
Case I: When 1,2 are known, the test statistic
is:
z
X1  X 2
2
1
n1

2
2
n2
HYPOTHESIS TESTING: TWO
POPULATION MEANS
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 Case II: When 1,2 are unknown but the
sample sizes n1 and n2 are greater than or
equal to 30, the test statistic is :
z
X1  X 2
2
s1
n1

2
s2
n2
EXAMPLE
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 A study was conducted to compare the mean
years of service for those retiring in 1975 with
those retiring last year Kentucky
Manufacturing Co. The following sample data
was obtained. At the 0.01 significance level
can we conclude that the workers retiring last
EXAMPLE (continued)
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 State the null and the alternative
hypotheses:
 Let population 2 refer to those that retired
last year.
H0:2  1
H1:2 > 1
 State the decision rule.
 Reject H0 if z > 2.33. (one-sided)
 Compute the value of the test statistic.
.  256
.  680
z  304
.
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. 2  2.92
45
40
EXAMPLE (continued)
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 What is the decision on the null
hypothesis? Interpret the results.
 Since z = 6.80 > 2.33, H0 is rejected. Those
retiring last year had more years of service.
TESTS CONCERNING PROPORTION
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 Proportion: A fraction or percentage that
indicates the part of the population or
sample having a particular trait of interest.
Sample proportion is denoted by p where
Number
of
successes
in
the
sample
p
Number sampled
TEST STATISTIC FOR TESTING A
SINGLE POPULATION PROPORTION
p

p
z
p(1 p)
n
p  population proportion
p  sample proportion
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EXAMPLE
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 In the past 15% of the mail order
solicitations for a certain charity resulted in
a financial contribution. A new solicitation
letter has been drafted. Will this new letter
increase the solicitation rate? The new
letter is sent to a sample of 200 people and
45 responded with a contribution. At the
0.05 significance level can it be concluded
that the new letter is more effective?
EXAMPLE (continued)
 State the null and the alternative
hypotheses:
 H0:p  0.15
H1:p > 0.15.
 State the decision rule.
 H0 is rejected if z > 1.645. (one-sided)
 Compute the value of the test statistic.
45  015
.
z  200
 2.97
(015
. )(085
. )
200
30
EXAMPLE (continued)
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 What is the decision on the null
hypothesis? Interpret the results.
 Since z = 2.97 > 1.645, H0 is rejected. The
new letter is more effective.
A TEST INVOLVING THE DIFFERENCE
BETWEEN TWO POPULATION
PROPORTIONS
 The test statistic in this case is
z
p1  p2
pc (1  pc )
n1

pc (1  pc )
n2
n1 is the sample size from population 1.
n2 is the sample size from population 2.
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TWO PROPORTIONS (continued)
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pc is the weighted mean of the two sample
proportions, computed by:
Total number of successes X 1  X 2
pc 
=
Total number in samples
n1  n2
X1 is the number of successes in n1.
X2 is the number of successes in n2.
EXAMPLE
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 Are unmarried workers more likely to be
absent from work than married workers? A
sample of 250 married workers showed 22
missed more than 5 days last year for any
reason. A sample of 300 unmarried workers
showed 35 missed more than 5 days. Use
the 0.05 significance level.
 State the null and alternative hypotheses.
 H0:p2  p1
H1:p2 > p1 where subscript 2
refers to the population of unmarried
workers.
EXAMPLE (continued)
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 State the decision rule.
 H0 is rejected if z > 1.645. (one-sided)
 Compute the value of the test statistic.
p  22  35  01036
.
250 300
01167
.
 00880
.
Z
110
.
01036
.
(1 01036
.
)  01036
.
(1 01036
.
)
300
250
c
EXAMPLE (continued)
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 What is the decision on the null
hypothesis?
 H0 is not rejected. There is no difference in
the proportion of married and unmarried
workers missing more than 5 days of work.
For Final Exam
Chapters 12 and 13:
Regression, Inference, and
Model Building/ ANOVA
James S. Hawkes
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