Example
Example 9.2. In the past, average productivity of worker
is 120. To improve productivity, new management
procedure is applied. Recently survey of 25 workers
shows sample mean is 126.2, sample variance is 225.
Assumed that productivity is Normality,
(a) Test the hypothesis that average productivity has
increased, at significant level of 5% and 1%
Sample: 9 = 25; 7̅ = 126.2; 8< = 225
Then 8 = 15
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Lecture 9. HYPOTHESIS TESTING
Statistical Hypothesis
Error Types
T-test for Mean
Z-test for Population
Chi-sq test for Variance
[1] Chapter 9 + 11
pp. 382 – 432; 488 - 491
PROBABILITY & STATISTICS – Bui Duong Hai – NEU – www.mfe.edu.vn/buiduonghai 1
9.1. Statistical Hypothesis
Hypothesis: statement about statistical issue
Testing: test for the “True” or “False” of hypothesis
Example:
Government reports that average income is $2400
Expected value of price is $10
Mean of consumption was 2 mil.VND
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Hypotheses Pair
For parameter
∶Null hypothesis
:Alternative hypothesis
There are 3 types
: ≤
: >
one-sided, upper tail
: ≥
: <
one-sided, lower tail
: =
: ≠
two-sided, two tails
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Example
Ex. Testing the statement that “average income of
habitant is $2000”
0: = 2000 : the statement is True
1: ≠ 2000 : the statement is False
Example 9.1. What is hypotheses pair for the following
statements:
(a) The average income of habitant is greater than 2000
(b) The proportion of male in customers is less than 50%
(c) The variability of price is more than 20 USD2
(d) The dispersion of price is less than 20 USD
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Decision & Errors
Decision: Reject or Accept H0
Two types of Errors
Error type 1: Reject the True hypothesis
Error type 2: Accept the False hypothesis
Allow a given Type 1 error probability, minimize Type 2
error probability
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Decision H0 is True H0 is False
Accept H0 Correct decision Type 2 error
Reject H0 Type 1 error Correct decision
Testing Procedure
Probability for Type 1 Error: Significant Level,
Calculate Critical value correspond to
Indicate Reject Area for H0
From sample: calculate Statistical value
Rule:
Statistical value is in the Reject Area: Reject H0
Stat. value is not in the Reject Area: Not Reject H0
(Not enough evident to reject H0)
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Testing Procedure
Step 1: Setting up Hypothesis pair, significant level
Step 2: Data gathering
Step 3: Calculating statistical value
Step 4: Comparing statistical value with critical value
Step 5: Concluding about the hypothesis
There is Probability value (P-value) of the test
− < :reject
− > : not reject
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9.2. T-test for Mean
Testing population mean with a value
Significant level:
Hypotheses pairs
: ≤
: >
: ≥
: <
: =
: ≠
Only one Statistical value: =
̅
/
Critical value and Reject Area are different
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T-test for Mean
Statistical
value
Hypotheses
pair
Critical
value
Reject H0
=
−̅
/
: ≤
: >
>
: ≥
: <
− < −
: =
: ≠
/ | |> /
PROBABILITY & STATISTICS – Bui Duong Hai – NEU – www.mfe.edu.vn/buiduonghai 9
Example
Example 9.2. In the past, average productivity of worker
is 120. To improve productivity, new management
procedure is applied. Recently survey of 25 workers
shows sample mean is 126.2, sample variance is 225.
Assumed that productivity is Normality,
(a) Test the hypothesis that average productivity has
increased, at significant level of 5% and 1%
Sample: = 25; =̅ 126.2; = 225
Then = 15
PROBABILITY & STATISTICS – Bui Duong Hai – NEU – www.mfe.edu.vn/buiduonghai 10
Example
Hyp. Pair:
: ≤ 120
: > 120
Stat. value: =
.
/
=
At 5%:
Critical value: . =
At 1%:
Critical value: . =
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P-value of the Test
There is one Probability value ∗ that Statistical value
equal Critical value
Probability value of the test: −
The rule
− < Reject
− > Not Reject
Accurate − is calculated by software
Using table, find the interval of −
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Example
Example 9.2 (cont.)
(b) At 5%, test the hypothesis that recent average
productivity equals 130
(c) Which interval contains the − of the test
in question (b): (0 – 1%); (1% – 5%); (5% – 10%); > 10%
(d) At 10%, test the hypothesis that recent average
productivity is less than 135, and estimate the
− of the test
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9.3. Z-test for Proportion
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Statistical value
Hypotheses
pair
Critical
value
Reject H0
=
−̅
1 − /
: ≤
: >
>
: ≥
: <
− < −
: =
: ≠
/ | |> /
Example
Example 9.3. Last year, the proportion of visitors buying
goods is 20%. Recent year, in observed 200 visitors, 52 of
them buy at least one item.
(a) With significant level at 5%, testing the hypothesis
that the buying proportion has increased
(b) Estimate the − of the test
(c) At significant level of 1%, test the claim that
Proportion is 30%, and estimate the −
PROBABILITY & STATISTICS – Bui Duong Hai – NEU – www.mfe.edu.vn/buiduonghai 15
9.3. Chisq-test for Variance
PROBABILITY & STATISTICS – Bui Duong Hai – NEU – www.mfe.edu.vn/buiduonghai 16
Statistical value
Hypotheses
pair
Reject H0
=
( − 1)
:
≤
:
>
>
:
≥
:
<
<
:
=
:
≠
> /
Or < /
Example
Example 9.4. The required limit for the variability of
fruits’ weight is 4g2. For sample of 20 fruits, sample
variance is 5g2.
(a) With the significant level of 5%, test the claim that the
variability of fruits’ weight exceeds the required limit,
assumed that the it is normal distributed.
(b) Test the hypothesis that Population Standard
deviation is less than 4g
PROBABILITY & STATISTICS – Bui Duong Hai – NEU – www.mfe.edu.vn/buiduonghai 17
Key Concepts
Hypotheses pair
Types of Error
Significant level
Critical value & Statistical value
Reject and Not reject H0
− of the test
T-test, Z-test, Chisq-test
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Exercise
[1] Chapter 9, 11
(387) 1, 2, (389) 5, 7
(402) 10, 15, 17
(408) 23, 25, 27, 29, 34
(414) 36, 38, 40, 43
(491) 4, 11
Case Problem 1
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