2. Introduction:
Hypothesis testing is the process of deciding statistically whether the findings of an
investigating reflect chance effects or real effects at a given level of probability.
If the results represent real effects, then we say that the results are statistically significant.
That is, when we say that our results are statistically significant, we mean that the patterns
or differences seen in the sample data are generalizable to the population.
2
3. The logic of hypothesis testing:
Scientifically is called โTesting of Hypothesesโ. This procedure is used commonly for all cases
where one is not sure about the true situation. There are two types of hypotheses namely:
Alternative hypothesis ๐ป๐ด : Statement which we want to retain or (prove) or (accept)
Null hypothesis ๐ป0 : Statement which we want to reject or (disprove) or (nullify)
๐ป0 is always opposite of ๐ป๐ด
3
4. For example,
1. If ๐ป๐ด: ๐ < 60, then ๐ป0: ๐ โฅ 60 (one-tail hypothesis, ๐ป๐ด to the left) (directional)
2. If ๐ป๐ด: ๐ > 60, then ๐ป0: ๐ โค 60 (one-tail hypothesis, ๐ป๐ด to the right) (directional)
3. If ๐ป๐ด: ๐ โค 60, then ๐ป0: ๐ > 60 (one-tail hypothesis, ๐ป๐ด to the left) (directional)
4. If ๐ป๐ด: ๐ โฅ 60, then ๐ป0: ๐ < 60 (one-tail hypothesis, ๐ป๐ด to the right) (directional)
5. If ๐ป๐ด: ๐ โ 60, then ๐ป0: ๐ = 60 (two-tail hypothesis, ๐ป๐ด to both sides) (non-directional)
6. If ๐ป๐ด: ๐ = 60, then ๐ป0: ๐ โ 60 (two-tail hypothesis, ๐ป๐ด to both sides) (non-directional)
4
5. Remarks:
1) Both, ๐ป๐ด and ๐ป0 are statements about population parameters.
2) From 1 to 4 above are called directional hypotheses (๐ค๐ ๐ข๐ ๐ ๐ผ)
3) 5 and 6 above are called are non-directional hypotheses. ๐ค๐ ๐ข๐ ๐
๐ผ
2
4) ๐ป๐ด and ๐ป0 are mutual exclusive, means that if we reject ๐ป0 then we accept ๐ป๐ด
and vice versa, if we accept ๐ป0 then we reject ๐ป๐ด (We canโt reject both or accept both)
5
6. Errors in Hypothesis Testing:
Possible hypothesis testing outcomes are shown in the table below:
The probability of Type (I) error is ๐ผ (always 0 โค ๐ผ โค 1)
The probability of Type (II) error is ๐ฝ = 1 โ ๐ผ (always 0 โค ๐ฝ โค 1)
Since ๐ฝ = 1 โ ๐ผ, one can observe that,
As ๐ถ increases, then ๐ท decreases
As ๐ถ decreases, then ๐ท increases
6
Decision:
Reject ๐ฏ๐
Decision:
Accept ๐ฏ๐
๐ฏ๐ ๐๐ ๐๐๐๐ Type (I) error Correct Decision
๐ฏ๐ ๐๐ ๐๐๐๐๐ Correct Decision Type (II) error
7. Remarks:
1. By convention, ๐ผ is usually set as either ๐ผ = 0.01 or ๐ผ = 0.05
2. The smaller ๐ผ, the less the chance of making a Type (I) error and consequently the more chance of
making a correct decision
3. The smaller ๐ฝ, the less the chance of making a Type (II) error and consequently the more chance of
making a correct decision
4. ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) โค ๐ผ; ๐๐๐๐๐๐ก ๐ป0 (i.e. statistically significant)
5. ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) > ๐ผ; ๐๐๐ก๐๐๐ ๐ป0 (i.e. statistically not significant)
7
8. 6. If ๐ผ increases (same as ๐ฝ decreases ), means that is made less significant
7. If ๐ผ decreases (same as ๐ฝ increases ), means that is made more significant
8. If ๐ผ = 0.05 (๐๐ 5%), then ๐ฝ = 0.95 (๐๐ 95%) means that:
(a) ๐ผ = 0.05 โ The probability of rejecting a true ๐ฏ๐ is 0.05 (making Type (I) error) & Consequently, the
probability of making correct decision is 0.95
(b) ๐ฝ = 0.95 โ The probability of accepting a false ๐ฏ๐ is 0.95 (making Type (II) error) & Consequently, the
probability of making correct decision is 0.05
9. ๐ผ + ๐ฝ = 1 (๐๐๐ค๐๐ฆ๐ )
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9. Comparisons between ๐ฏ๐จ and ๐ฏ๐
๐ฏ๐จ ๐ฏ๐
1 Prediction intended for evaluation Opposite of HA
2 HA claims that the results are โrealโ or โsignificantโ effect,
which means that the independent variable influenced
the dependent variable.
H๐ claims that the results are โnot realโ or โnot significantโ
effect, which means that the independent variable did not
influence the dependent variable.
3 โrealโ or โsignificantโ effect, means that the results in the
sample data can be generalized to the population.
โNot realโ or โnot significantโ effect, means that the results in
the sample data cannot be generalized to the population.
4 There are differences in the data in each direction There are no differences in the data in each direction
9
10. Steps in hypothesis Testing:
1. Formulate alternative hypothesis ๐ป๐ด and appropriate null hypothesis ๐ป0
2. Specify significance level ๐ผ (always is given)
3. Find the critical ๐ง ๐ง๐๐๐๐ก from the
4. Calculate the statistic ๐ง ๐ง๐๐๐ก (๐ obtained) by using ๐ง๐๐๐ก =
าง
๐ฅโ๐
เต
๐
๐
, whenever ๐ โฅ 30,
๐๐ข๐ก, we calculate the statistic ๐ก (๐ก๐๐๐ก) (๐ obtained) by using ๐ก๐๐๐ก =
าง
๐ฅโ๐
เต
๐
๐
whenever ๐ < 30
5. Decide if to reject or retain ๐ป0 (by using the decision rules as follows)
(a) If ๐ง๐๐๐ก โฅ ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0 & If ๐ง๐๐๐ก < ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0 ๐ โฅ 30
(b) If ๐ก๐๐๐ก โฅ ๐ก๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0 & If ๐ก๐๐๐ก < ๐ก๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0 ๐ < 30
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11. Example 1
A rehabilitation therapist devises an exercise program which is expected to reduce the time taken for
people to leave hospital following orthopedic surgery. Previous records show that the recovery time for
patients has been ๐ = 30, with ๐ = 8.
A sample of 64 patients are treated with exercise program, and their mean recovery time is found to be
เดค
๐ = 24.
Do these results show that patients who had treatment recovered significantly faster than previous
patients?
11
12. Solution
We can apply the steps for hypothesis testing to make our decision.
1. ๐ป๐ด: ๐ < 30 , therefore ๐ป0: ๐ โฅ 30 (one-tail hypothesis, ๐ป๐ด to the left) (directional)
(๐ป๐ด ๐ ๐ข๐๐๐๐๐ก๐ ๐กโ๐ ๐๐๐ ๐๐๐๐โ๐๐ / ๐๐๐ฃ๐๐ ๐ก๐๐๐๐ก๐๐ ๐๐๐๐๐ ๐กโ๐๐ก ๐กโ๐ ๐๐ฅ๐๐๐๐๐ ๐ ๐๐๐๐๐๐๐ ๐๐๐๐ข๐๐๐
๐กโ๐ ๐ก๐๐๐ ๐ก๐ ๐๐๐๐ฃ๐ โ๐๐ ๐๐๐ก๐๐ ๐๐๐๐๐๐ 30 ๐๐๐ฆ๐ )
2. Let ๐ผ = 0.01 (๐๐๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ ๐๐๐๐โ๐๐ ๐๐๐๐๐๐๐๐๐๐, ๐๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐ก๐ ๐ข๐ )
3. ๐ง๐๐๐๐ก = 2.33, because ๐ = 64 โฅ 30 we use the ๐ง-table
(By looking at the second column where ๐ผ = 0.01, then the corresponding ๐ง = 2.33)
4. ๐ง๐๐๐ก =
๐ฅางโ๐
๐
๐
เต
โ ๐ง๐๐๐ก =
24โ30
8
64
เต
= โ 6
5. ๐ง๐๐๐ก โฅ ๐ง๐๐๐๐ก โ โ6 โฅ 2.33 โ 6 โฅ 2.33 โ ๐๐๐๐๐๐ก ๐ป0 โ ๐๐๐ก๐๐๐ (๐๐๐๐๐๐ก) ๐ป๐ด
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13. Conclusion:
The researcher/investigator can reject ๐ป0 and accept ๐ป๐ด at a 0.01 level of significance.
That is, patients who treated with the exercise program, recover earlier than the population
of untreated patients.
Note that, the independent variable is the exercise program and the dependent variable
is the time to recover, because the recovery time to leave hospital earlier depends
on the treatment by the exercise program.
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HW 1: Do example 1 above with ๐ผ = 0.05 ๐ป๐๐๐ก: ๐ง๐๐๐๐ก = 1.65
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14. Example 2
A researcher hypothesizes that males now weigh more than in the previous years. To investigate this
hypothesis, he randomly selects 100 adult males and recorded their weights. The measurements for the
sample have a mean of เดค
๐ = 70 ๐พ๐. In a census taken several years ago, the mean weight of weight of
males was ๐ = 68 ๐๐, with ๐ = 8 ๐๐.
Do these results show that adult males now have more weight than previous?
14
15. Solution
We can apply the steps for hypothesis testing to make our decision.
1. ๐ป๐ด: ๐ > 68 , therefore ๐ป0: ๐ โค 68 (one-tail hypothesis, ๐ป๐ด to the right) (directional)
(๐ป๐ด ๐ ๐ข๐๐๐๐๐ก๐ ๐กโ๐ ๐๐๐ ๐๐๐๐โ๐๐ ๐๐๐๐๐ ๐กโ๐๐ก ๐๐๐ข๐๐ก ๐๐๐๐๐ ๐ค๐๐๐โ๐ก ๐๐๐ค ๐๐ ๐๐๐๐ ๐กโ๐๐ ๐๐๐๐ฃ๐๐๐ข๐ = 68)
2. Let ๐ผ = 0.01 (๐๐๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ ๐๐๐๐โ๐๐ ๐๐๐๐๐๐๐๐๐๐, ๐๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐ก๐ ๐ข๐ )
3. ๐ง๐๐๐๐ก = 2.33, because ๐ = 100 โฅ 30 we use the ๐ง-table
(By looking at the second column where ๐ผ = 0.01, then the corresponding ๐ง = 2.33)
4. ๐ง๐๐๐ก =
๐ฅางโ๐
๐
๐
เต
โ ๐ง๐๐๐ก =
70โ68
8
100
เต
= 2.5
5. ๐ง๐๐๐ก โฅ ๐ง๐๐๐๐ก โ 2.5 โฅ 2.33 โ 2.5 โฅ 2.33 โ ๐๐๐๐๐๐ก ๐ป0 โ ๐๐๐ก๐๐๐ (๐๐๐๐๐๐ก) ๐ป๐ด
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16. Conclusion:
The researcher/investigator can reject ๐ป0 and accept ๐ป๐ด at a 0.01 level of significance. That is,
adult malesโ weight now has increased more than previous weight 68 kg.
Note that, the independent variable is time, and the dependent variable is weight,
because the weight depends on the running time (The more time the more weight)
----------------------------------------------------------------------------------------------------------------
HW 2: do the example 2 above with ๐ผ = 0.05 ๐ป๐๐๐ก: ๐ง๐๐๐๐ก = 1.65
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17. Example 3
A researcher hypothesizes that males now have different weights than in the previous years. To investigate
this hypothesis, he randomly selects 100 adult males and recorded their weights. The measurements for
the sample have a mean of เดค
๐ = 70 ๐พ๐. In a census taken several years ago, the mean weight of weight of
males was ๐ = 68 ๐๐, with ๐ = 8 ๐๐.
Do these results show that adult males now have more weight than previous?
17
18. Solution
We can apply the steps for hypothesis testing to make our decision.
1. ๐ป๐ด: ๐ โ 68 , therefore ๐ป0: ๐ = 68 (one-tail hypothesis, ๐ป๐ด to the right) (non-directional)
(HA supports the researcher claim that adult males weight now is different than previous = 68)
2. Let ๐ผ = 0.01 (depends on the researcher preference, and is given to us)
Now, since we have non-directional hypothesis, we divide ๐ผ by 2 โ
0.01
2
= 0.005
3. ๐ง๐๐๐๐ก = 2.58, because ๐ = 100 โฅ 30 we use the ๐ง-table
(By looking at the second column where ๐ผ = 0.005, then the corresponding ๐ง = 2.58)
4. ๐ง๐๐๐ก =
๐ฅางโ๐
๐
๐
เต
โ ๐ง๐๐๐ก =
70โ68
8
100
เต
= 2.5
5. ๐ง๐๐๐ก < ๐ง๐๐๐๐ก โ 2.5 < 2.58 โ 2.5 < 2.58 โ ๐๐๐ก๐๐๐ (๐๐๐๐๐๐ก)๐ป0 โ ๐๐๐๐๐๐ก ๐ป๐ด
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19. Conclusion:
The researcher/investigator can accept ๐ป0 and reject ๐ป๐ด at a 0.01 level of significance. That is, the
mean adult malesโ weight now is same as previous mean weight = 68 kg.
Note that, the independent variable is time, and the dependent variable is weight,
because the weight depends on the running time.
----------------------------------------------------------------------------------------------------------------
HW 3 (a): Show that if we worked example 3 above with ๐ผ = 0.05, เดฅ
๐ฟ = ๐๐
Answer: reject ๐ป0 and accept (retain) ๐ป๐ด ๐ป๐๐๐ก: ๐ง๐๐๐๐ก = 1.96
HW 3 (b): Show that if we worked example 3 above with ๐ผ = 0.05, เดฅ
๐ฟ = ๐๐
Answer: reject ๐ป0 and accept (retain) ๐ป๐ด ๐ป๐๐๐ก: ๐ง๐๐๐๐ก = 1.96
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20. Example 4
Work example 2 but with small sample size ๐ = 16 ๐๐๐ก๐๐๐ ๐กโ๐๐ก ๐ = 16 < 30
20
21. Solution
We can apply the steps for hypothesis testing to make our decision.
1. ๐ป๐ด : ๐ > 68 , therefore ๐ป0: ๐ โค 68 (one-tail hypothesis, ๐ป๐ด to the right) (directional)
(๐ป๐ด ๐ ๐ข๐๐๐๐๐ก๐ ๐กโ๐ ๐๐๐ ๐๐๐๐โ๐๐ ๐๐๐๐๐ ๐กโ๐๐ก ๐๐๐ข๐๐ก ๐๐๐๐๐ ๐ค๐๐๐โ๐ก ๐๐๐ค ๐๐ ๐๐๐๐ ๐กโ๐๐ ๐๐๐๐ฃ๐๐๐ข๐ = 68)
2. Let ๐ผ = 0.01 (๐๐๐๐๐๐๐ ๐๐ ๐กโ๐ ๐๐๐ ๐๐๐๐โ๐๐ ๐๐๐๐๐๐๐๐๐๐, ๐๐๐ ๐๐ ๐๐๐ฃ๐๐ ๐ก๐ ๐ข๐ )
3. ๐ก๐๐๐๐ก = 2.602, because ๐ = 16 < 30 we use the ๐ก-table (we have small sample)
(By looking at the directional row where ๐ = ๐ผ = 0.01,
and using ๐. ๐. = ๐ โ 1 โ 16 โ 1 = 15)
(๐. ๐. ๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐ค๐๐ฆ๐ = ๐ โ 1)
4. ๐ก๐๐๐ก =
๐ฅางโ๐
๐
๐
เต
โ ๐ก๐๐๐ก =
70โ68
8
16
เต
= 1
5. ๐ง๐๐๐ก < ๐ก๐๐๐๐ก โ 1 < 2.602 โ 1 < 2.602 โ ๐๐๐ก๐๐๐ ๐๐ (๐๐๐๐๐๐ก)๐ป0 โ ๐๐๐๐๐๐ก ๐ป๐ด
Clearly, when ๐ = 16 (small sample), the investigation did not show a significant weight
increase for the males.
21
22. Conclusion:
The researcher/investigator can accept ๐ป0 and reject ๐ป๐ด at a 0.01 level of significance. That is, adult malesโ weight now
has not increased more than previous weight 68 kg.
Note that, the independent variable is time, and the dependent variable is weight,
because the weight depends on the running time.
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HW 4: Do the example 4 above with เดค
๐ = 72 and use the following information:
(1) Sample size ๐ = 25, ๐ผ = 0.01, เดค
๐ = 72 ๐ป๐๐๐ก: ๐ก๐๐๐๐ก = 2.492 & ๐ก๐๐๐ก = 2 Answer: Accept ๐ป0
(2) Sample size ๐ = 25, ๐ผ = 0.05, เดค
๐ = 72 ๐ป๐๐๐ก: ๐ก๐๐๐๐ก = 1.711 & ๐ก๐๐๐ก = 2 Answer: Reject ๐ป0
(3) Sample size ๐ = 16, ๐ผ = 0.05, เดค
๐ = 72 ๐ป๐๐๐ก: ๐ก๐๐๐๐ก = 1.753 & ๐ก๐๐๐ก = 2 Answer: Reject ๐ป0
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23. 23
Example 5
Work example 3 but with small sample size ๐ = 16 ๐๐๐ก๐๐๐ ๐กโ๐๐ก ๐ = 16 < 30
Solution
We can apply the steps for hypothesis testing to make our decision.
1. ๐ป๐ด: ๐ โ 68 , therefore ๐ป0: ๐ = 68
(Two-tail hypothesis)โ (non-directional)
(HA supports the researcher claim that adult maleโฒs weight now is different than previous = 68)
2. Let ๐ผ = 0.01 (depends on the researcher preference, and is given to us)
3. ๐ก๐๐๐๐ก = 2.947, because ๐ = 16 < 30 we use the ๐ก-table (we have small sample)
Now, since we have non-directional hypothesis, we divide ๐ผ ๐๐ฆ 2 โ
๐ผ
2
=
0.01
2
= 0.005
and using ๐. ๐. = ๐ โ 1 โ 16 โ 1 = 15 ๐. ๐. ๐๐๐๐๐ ๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐๐๐๐ ๐๐๐ ๐๐๐ค๐๐ฆ๐ = ๐ โ 1
4. ๐ก๐๐๐ก =
าง
๐ฅโ๐
เต
๐
๐
โ ๐ก๐๐๐ก =
70โ68
เต
8
16
= 1
5. ๐ง๐๐๐ก < ๐ก๐๐๐๐ก โ 1 < 2.947 โ 1 < 2.947 โ ๐๐๐ก๐๐๐ ๐๐ ๐๐๐๐๐๐ก ๐ป0 โ ๐๐๐๐๐๐ก ๐ป๐ด
Clearly, when ๐ = 16 (small sample), the investigation did not show a significant weight change for the males
Conclusion:
The researcher/investigator can accept ๐ป0 and reject ๐ป๐ด at a 0.01 level of significance. That is, adult malesโ weight now has not
increased more than previous weight 68 kg.
24. HW 5: Do the example 5 above with เดค
๐ = 73 and use the following information:
(1) Sample size ๐ = 25, ๐ผ = 0.01 ๐ป๐๐๐ก: ๐ก๐๐๐๐ก = 2.797 & ๐ก๐๐๐ก = 2.5 Answer: Accept ๐ป0
(2) Sample size ๐ = 25, ๐ผ = 0.05 ๐ป๐๐๐ก: ๐ก๐๐๐๐ก = 2.064 & ๐ก๐๐๐ก = 2.5 Answer: Reject ๐ป0
(3) Sample size ๐ = 16, ๐ผ = 0.05 ๐ป๐๐๐ก: ๐ก๐๐๐๐ก = 2.131 & ๐ก๐๐๐ก = 2.5 Answer: Reject ๐ป0
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25. Important facts
1 Hypothesis A proposition advanced by the researcher which is evaluated by using the data collected from a sample.
2 Alternative Hypothesis (๐ป๐ด) This is the hypothesis for which the researcher is trying to gain support in a statistical analysis by rejecting the null
hypothesis (๐ป0)
3 Null Hypothesis (๐ป0) This is the hypothesis for which the researcher is trying to reject in a statistical analysis by accepting the alternative
hypothesis (๐ป๐ด)
4 Directional Hypothesis Is the hypothesis that asserts (assures) that differences between groups in the data will occur in one direction,
either to left or right direction.
5 Non-directional Hypothesis Is the hypothesis that asserts (assures) that differences between groups in the data will occur in two directions, to
left and right directions at the same time.
6 ๐ป๐ด and ๐ป0 are mutual exclusive
If we reject, ๐ป0 then we must accept ๐ป๐ด, conversely,
if we accept ๐ป0 then we must reject ๐ป๐ด.
(We canโt reject both or accept both)
25
26. 26
7 Type (I) Error Rejecting ๐ป0 while it is true, and its probability is ฮฑ
8 Type (II) Error Accepting ๐ป0 while it is false, and its probability is ๐ฝ = 1 โ ๐ผ
9
Degrees of Freedom ๐. ๐. ๐. ๐. = ๐ โ 1 (๐ ๐๐ ๐กโ๐ ๐ ๐๐๐๐๐ ๐ ๐๐ง๐)
10 Statistical Significance
1. Demonstrates that the result obtained is probably not due to chance but is โrealโ
2. The independent variable must have had a very large effect on the dependent variable.
3. If the obtained probability p is โค ๐ผ โ can reject ๐ป0 (means accept ๐ป๐ด)
27. Important graphs to illustrate directional and non-directional of ๐ป๐ด
1. ๐ผ = 0.05, (one-tail hypothesis, ๐ป๐ด to the right) (directional hypothesis ๐ป๐ด)
(a) The gray region is the rejection region of the true ๐ป0 at level ๐ผ = 0.05
(b) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) โค 0.05; ๐๐๐๐๐๐ก ๐ป0 (Type (I) error = 5%) ๐๐ ๐ง๐๐๐ก โฅ ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0
(c) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) > 0.05; ๐๐๐ก๐๐๐ ๐ป0 (Correct decision = 95%) ๐๐ ๐ง๐๐๐ก < ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0 27
28. 1. ๐ผ = 0.05, (two-tail hypothesis, ๐ป๐ด to both sides) (non-directional hypothesis ๐ป๐ด)
In the two-tail hypothesis we divide ๐ผ ๐๐ฆ 2 โ 0.05/2 = 0.025
(a) The two gray regions are the rejection regions of the true ๐ป0 at level ๐ผ = 0.05
(b) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) โค 0.025; ๐๐๐๐๐๐ก ๐ป0 (Type (I) error = 5%) ๐๐ ๐ง๐๐๐ก โฅ ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0
(c) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) > 0.025; ๐๐๐ก๐๐๐ ๐ป0 (Correct decision = 95%) ๐๐ ๐ง๐๐๐ก < ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0
28
29. 1. ๐ผ = 0.01, (one-tail hypothesis, ๐ป๐ด to the right) (directional hypothesis ๐ป๐ด)
(a) The gray region is the rejection region of the true ๐ป0 at level ๐ผ = 0.01
(b) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) โค 0.01; ๐๐๐๐๐๐ก ๐ป0 (Type (I) error = 1%) ๐๐ ๐ง๐๐๐ก โฅ ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0
(c) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) > 0.01; ๐๐๐ก๐๐๐ ๐ป0 (Correct decision = 99%) ๐๐ ๐ง๐๐๐ก < ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0
29
30. 1. ๐ผ = 0.01, (two-tail hypothesis, ๐ป๐ด to both sides) (non-directional hypothesis ๐ป๐ด)
In the two-tail hypothesis we divide ๐ผ ๐๐ฆ 2 โ 0.01/2 = 0.005
(a) The two gray regions are the rejection regions of the true ๐ป0 at level ๐ผ = 0.01
(b) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) โค 0.005; ๐๐๐๐๐๐ก ๐ป0 (Type (I) error = 1%) ๐๐ ๐ง๐๐๐ก โฅ ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0
(c) If ๐(๐ป0 ๐๐ ๐ก๐๐ข๐) > 0.005; ๐๐๐ก๐๐๐ ๐ป0 (Correct decision = 99%) ๐๐ ๐ง๐๐๐ก < ๐ง๐๐๐๐ก ; ๐๐๐๐๐๐ ๐ป0
30