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 “Testing of Hypotheses”. 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 𝐻𝐴 : Statement which we want to retain or (prove) or (accept)
Null hypothesis 𝐻0 : Statement which we want to reject or (disprove) or (nullify)
𝐻0 is always opposite of 𝐻𝐴
3
4. For example,
1. If 𝐻𝐴: 𝜇 < 60, then 𝐻0: 𝜇 ≥ 60 (one-tail hypothesis, 𝐻𝐴 to the left) (directional)
2. If 𝐻𝐴: 𝜇 > 60, then 𝐻0: 𝜇 ≤ 60 (one-tail hypothesis, 𝐻𝐴 to the right) (directional)
3. If 𝐻𝐴: 𝜇 ≤ 60, then 𝐻0: 𝜇 > 60 (one-tail hypothesis, 𝐻𝐴 to the left) (directional)
4. If 𝐻𝐴: 𝜇 ≥ 60, then 𝐻0: 𝜇 < 60 (one-tail hypothesis, 𝐻𝐴 to the right) (directional)
5. If 𝐻𝐴: 𝜇 ≠ 60, then 𝐻0: 𝜇 = 60 (two-tail hypothesis, 𝐻𝐴 to both sides) (non-directional)
6. If 𝐻𝐴: 𝜇 = 60, then 𝐻0: 𝜇 ≠ 60 (two-tail hypothesis, 𝐻𝐴 to both sides) (non-directional)
4
5. Remarks:
1) Both, 𝐻𝐴 and 𝐻0 are statements about population parameters.
2) From 1 to 4 above are called directional hypotheses (𝑤𝑒 𝑢𝑠𝑒 𝛼)
3) 5 and 6 above are called are non-directional hypotheses. 𝑤𝑒 𝑢𝑠𝑒
𝛼
2
4) 𝐻𝐴 and 𝐻0 are mutual exclusive, means that if we reject 𝐻0 then we accept 𝐻𝐴
and vice versa, if we accept 𝐻0 then we reject 𝐻𝐴 (We can’t 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 𝛼 (always 0 ≤ 𝛼 ≤ 1)
The probability of Type (II) error is 𝛽 = 1 − 𝛼 (always 0 ≤ 𝛽 ≤ 1)
Since 𝛽 = 1 − 𝛼, one can observe that,
As 𝜶 increases, then 𝜷 decreases
As 𝜶 decreases, then 𝜷 increases
6
Decision:
Reject 𝑯𝟎
Decision:
Accept 𝑯𝟎
𝑯𝟎 𝒊𝒔 𝒕𝒓𝒖𝒆 Type (I) error Correct Decision
𝑯𝟎 𝒊𝒔 𝒇𝒂𝒍𝒔𝒆 Correct Decision Type (II) error
7. Remarks:
1. By convention, 𝛼 is usually set as either 𝛼 = 0.01 or 𝛼 = 0.05
2. The smaller 𝛼, the less the chance of making a Type (I) error and consequently the more chance of
making a correct decision
3. The smaller 𝛽, the less the chance of making a Type (II) error and consequently the more chance of
making a correct decision
4. 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) ≤ 𝛼; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 (i.e. statistically significant)
5. 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) > 𝛼; 𝑟𝑒𝑡𝑎𝑖𝑛 𝐻0 (i.e. statistically not significant)
7
8. 6. If 𝛼 increases (same as 𝛽 decreases ), means that is made less significant
7. If 𝛼 decreases (same as 𝛽 increases ), means that is made more significant
8. If 𝛼 = 0.05 (𝑜𝑟 5%), then 𝛽 = 0.95 (𝑜𝑟 95%) means that:
(a) 𝛼 = 0.05 → The probability of rejecting a true 𝑯𝟎 is 0.05 (making Type (I) error) & Consequently, the
probability of making correct decision is 0.95
(b) 𝛽 = 0.95 → The probability of accepting a false 𝑯𝟎 is 0.95 (making Type (II) error) & Consequently, the
probability of making correct decision is 0.05
9. 𝛼 + 𝛽 = 1 (𝑎𝑙𝑤𝑎𝑦𝑠)
8
9. Comparisons between 𝑯𝑨 and 𝑯𝟎
𝑯𝑨 𝑯𝟎
1 Prediction intended for evaluation Opposite of HA
2 HA claims that the results are ‘real’ or ‘significant’ effect,
which means that the independent variable influenced
the dependent variable.
H𝟎 claims that the results are ‘not real’ or ‘not significant’
effect, which means that the independent variable did not
influence the dependent variable.
3 ‘real’ or ‘significant’ effect, means that the results in the
sample data can be generalized to the population.
‘Not real’ or ‘not significant’ 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 𝐻𝐴 and appropriate null hypothesis 𝐻0
2. Specify significance level 𝛼 (always is given)
3. Find the critical 𝑧 𝑧𝑐𝑟𝑖𝑡 from the
4. Calculate the statistic 𝑧 𝑧𝑜𝑏𝑡 (𝒛 obtained) by using 𝑧𝑜𝑏𝑡 =
ҧ
𝑥−𝜇
ൗ
𝜎
𝑛
, whenever 𝑛 ≥ 30,
𝑏𝑢𝑡, we calculate the statistic 𝑡 (𝑡𝑜𝑏𝑡) (𝒕 obtained) by using 𝑡𝑜𝑏𝑡 =
ҧ
𝑥−𝜇
ൗ
𝑠
𝑛
whenever 𝑛 < 30
5. Decide if to reject or retain 𝐻0 (by using the decision rules as follows)
(a) If 𝑧𝑜𝑏𝑡 ≥ 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒋𝒆𝒄𝒕 𝐻0 & If 𝑧𝑜𝑏𝑡 < 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒕𝒂𝒊𝒏 𝐻0 𝑛 ≥ 30
(b) If 𝑡𝑜𝑏𝑡 ≥ 𝑡𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒋𝒆𝒄𝒕 𝐻0 & If 𝑡𝑜𝑏𝑡 < 𝑡𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒕𝒂𝒊𝒏 𝐻0 𝑛 < 30
10
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 𝜇 = 30, with 𝜎 = 8.
A sample of 64 patients are treated with exercise program, and their mean recovery time is found to be
ത
𝑋 = 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. 𝐻𝐴: 𝜇 < 30 , therefore 𝐻0: 𝜇 ≥ 30 (one-tail hypothesis, 𝐻𝐴 to the left) (directional)
(𝐻𝐴 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠 𝑡ℎ𝑒 𝑟𝑒𝑠𝑒𝑎𝑟𝑐ℎ𝑒𝑟 / 𝑖𝑛𝑣𝑒𝑠𝑡𝑖𝑔𝑎𝑡𝑜𝑟 𝑐𝑙𝑎𝑖𝑚 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑒𝑥𝑒𝑟𝑐𝑖𝑠𝑒 𝑝𝑟𝑜𝑔𝑟𝑎𝑚 𝑟𝑒𝑑𝑢𝑐𝑒𝑠
𝑡ℎ𝑒 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑙𝑒𝑎𝑣𝑒 ℎ𝑜𝑠𝑝𝑖𝑡𝑎𝑙 𝑏𝑒𝑓𝑜𝑟𝑒 30 𝑑𝑎𝑦𝑠)
2. Let 𝛼 = 0.01 (𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑒𝑠𝑒𝑎𝑟𝑐ℎ𝑒𝑟 𝑝𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒, 𝑎𝑛𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡𝑜 𝑢𝑠)
3. 𝑧𝑐𝑟𝑖𝑡 = 2.33, because 𝑛 = 64 ≥ 30 we use the 𝑧-table
(By looking at the second column where 𝛼 = 0.01, then the corresponding 𝑧 = 2.33)
4. 𝑧𝑜𝑏𝑡 =
𝑥ҧ−𝜇
𝜎
𝑛
ൗ
→ 𝑧𝑜𝑏𝑡 =
24−30
8
64
ൗ
= − 6
5. 𝑧𝑜𝑏𝑡 ≥ 𝑧𝑐𝑟𝑖𝑡 → −6 ≥ 2.33 → 6 ≥ 2.33 → 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 → 𝑟𝑒𝑡𝑎𝑖𝑛 (𝑎𝑐𝑐𝑒𝑝𝑡) 𝐻𝐴
12
13. Conclusion:
The researcher/investigator can reject 𝐻0 and accept 𝐻𝐴 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.
----------------------------------------------------------------------------------------------------------------
HW 1: Do example 1 above with 𝛼 = 0.05 𝐻𝑖𝑛𝑡: 𝑧𝑐𝑟𝑖𝑡 = 1.65
13
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 ത
𝑋 = 70 𝐾𝑔. In a census taken several years ago, the mean weight of weight of
males was 𝜇 = 68 𝑘𝑔, with 𝜎 = 8 𝑘𝑔.
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. 𝐻𝐴: 𝜇 > 68 , therefore 𝐻0: 𝜇 ≤ 68 (one-tail hypothesis, 𝐻𝐴 to the right) (directional)
(𝐻𝐴 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠 𝑡ℎ𝑒 𝑟𝑒𝑠𝑒𝑎𝑟𝑐ℎ𝑒𝑟 𝑐𝑙𝑎𝑖𝑚 𝑡ℎ𝑎𝑡 𝑎𝑑𝑢𝑙𝑡 𝑚𝑎𝑙𝑒𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 𝑛𝑜𝑤 𝑖𝑠 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 = 68)
2. Let 𝛼 = 0.01 (𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑒𝑠𝑒𝑎𝑟𝑐ℎ𝑒𝑟 𝑝𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒, 𝑎𝑛𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡𝑜 𝑢𝑠)
3. 𝑧𝑐𝑟𝑖𝑡 = 2.33, because 𝑛 = 100 ≥ 30 we use the 𝑧-table
(By looking at the second column where 𝛼 = 0.01, then the corresponding 𝑧 = 2.33)
4. 𝑧𝑜𝑏𝑡 =
𝑥ҧ−𝜇
𝜎
𝑛
ൗ
→ 𝑧𝑜𝑏𝑡 =
70−68
8
100
ൗ
= 2.5
5. 𝑧𝑜𝑏𝑡 ≥ 𝑧𝑐𝑟𝑖𝑡 → 2.5 ≥ 2.33 → 2.5 ≥ 2.33 → 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 → 𝑟𝑒𝑡𝑎𝑖𝑛 (𝑎𝑐𝑐𝑒𝑝𝑡) 𝐻𝐴
15
16. Conclusion:
The researcher/investigator can reject 𝐻0 and accept 𝐻𝐴 at a 0.01 level of significance. That is,
adult males’ 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 𝛼 = 0.05 𝐻𝑖𝑛𝑡: 𝑧𝑐𝑟𝑖𝑡 = 1.65
16
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 ത
𝑋 = 70 𝐾𝑔. In a census taken several years ago, the mean weight of weight of
males was 𝜇 = 68 𝑘𝑔, with 𝜎 = 8 𝑘𝑔.
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. 𝐻𝐴: 𝜇 ≠ 68 , therefore 𝐻0: 𝜇 = 68 (one-tail hypothesis, 𝐻𝐴 to the right) (non-directional)
(HA supports the researcher claim that adult males weight now is different than previous = 68)
2. Let 𝛼 = 0.01 (depends on the researcher preference, and is given to us)
Now, since we have non-directional hypothesis, we divide 𝛼 by 2 →
0.01
2
= 0.005
3. 𝑧𝑐𝑟𝑖𝑡 = 2.58, because 𝑛 = 100 ≥ 30 we use the 𝑧-table
(By looking at the second column where 𝛼 = 0.005, then the corresponding 𝑧 = 2.58)
4. 𝑧𝑜𝑏𝑡 =
𝑥ҧ−𝜇
𝜎
𝑛
ൗ
→ 𝑧𝑜𝑏𝑡 =
70−68
8
100
ൗ
= 2.5
5. 𝑧𝑜𝑏𝑡 < 𝑧𝑐𝑟𝑖𝑡 → 2.5 < 2.58 → 2.5 < 2.58 → 𝑟𝑒𝑡𝑎𝑖𝑛 (𝑎𝑐𝑐𝑒𝑝𝑡)𝐻0 → 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻𝐴
18
19. Conclusion:
The researcher/investigator can accept 𝐻0 and reject 𝐻𝐴 at a 0.01 level of significance. That is, the
mean adult males’ 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 𝛼 = 0.05, ഥ
𝑿 = 𝟕𝟎
Answer: reject 𝐻0 and accept (retain) 𝐻𝐴 𝐻𝑖𝑛𝑡: 𝑧𝑐𝑟𝑖𝑡 = 1.96
HW 3 (b): Show that if we worked example 3 above with 𝛼 = 0.05, ഥ
𝑿 = 𝟔𝟔
Answer: reject 𝐻0 and accept (retain) 𝐻𝐴 𝐻𝑖𝑛𝑡: 𝑧𝑐𝑟𝑖𝑡 = 1.96
19
21. Solution
We can apply the steps for hypothesis testing to make our decision.
1. 𝐻𝐴 : 𝜇 > 68 , therefore 𝐻0: 𝜇 ≤ 68 (one-tail hypothesis, 𝐻𝐴 to the right) (directional)
(𝐻𝐴 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑠 𝑡ℎ𝑒 𝑟𝑒𝑠𝑒𝑎𝑟𝑐ℎ𝑒𝑟 𝑐𝑙𝑎𝑖𝑚 𝑡ℎ𝑎𝑡 𝑎𝑑𝑢𝑙𝑡 𝑚𝑎𝑙𝑒𝑠 𝑤𝑒𝑖𝑔ℎ𝑡 𝑛𝑜𝑤 𝑖𝑠 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 = 68)
2. Let 𝛼 = 0.01 (𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑟𝑒𝑠𝑒𝑎𝑟𝑐ℎ𝑒𝑟 𝑝𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒, 𝑎𝑛𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑡𝑜 𝑢𝑠)
3. 𝑡𝑐𝑟𝑖𝑡 = 2.602, because 𝑛 = 16 < 30 we use the 𝑡-table (we have small sample)
(By looking at the directional row where 𝑝 = 𝛼 = 0.01,
and using 𝑑. 𝑓. = 𝑛 − 1 → 16 − 1 = 15)
(𝑑. 𝑓. 𝑚𝑒𝑎𝑛𝑠 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑎𝑛𝑑 𝑎𝑙𝑤𝑎𝑦𝑠 = 𝑛 – 1)
4. 𝑡𝑜𝑏𝑡 =
𝑥ҧ−𝜇
𝑠
𝑛
ൗ
→ 𝑡𝑜𝑏𝑡 =
70−68
8
16
ൗ
= 1
5. 𝑧𝑜𝑏𝑡 < 𝑡𝑐𝑟𝑖𝑡 → 1 < 2.602 → 1 < 2.602 → 𝑟𝑒𝑡𝑎𝑖𝑛 𝑜𝑟 (𝑎𝑐𝑐𝑒𝑝𝑡)𝐻0 → 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻𝐴
Clearly, when 𝑛 = 16 (small sample), the investigation did not show a significant weight
increase for the males.
21
22. Conclusion:
The researcher/investigator can accept 𝐻0 and reject 𝐻𝐴 at a 0.01 level of significance. That is, adult males’ 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.
------------------------------------------------------------------------------------------------------------------
HW 4: Do the example 4 above with ത
𝑋 = 72 and use the following information:
(1) Sample size 𝑛 = 25, 𝛼 = 0.01, ത
𝑋 = 72 𝐻𝑖𝑛𝑡: 𝑡𝑐𝑟𝑖𝑡 = 2.492 & 𝑡𝑜𝑏𝑡 = 2 Answer: Accept 𝐻0
(2) Sample size 𝑛 = 25, 𝛼 = 0.05, ത
𝑋 = 72 𝐻𝑖𝑛𝑡: 𝑡𝑐𝑟𝑖𝑡 = 1.711 & 𝑡𝑜𝑏𝑡 = 2 Answer: Reject 𝐻0
(3) Sample size 𝑛 = 16, 𝛼 = 0.05, ത
𝑋 = 72 𝐻𝑖𝑛𝑡: 𝑡𝑐𝑟𝑖𝑡 = 1.753 & 𝑡𝑜𝑏𝑡 = 2 Answer: Reject 𝐻0
22
23. 23
Example 5
Work example 3 but with small sample size 𝑛 = 16 𝑛𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑎𝑡 𝑛 = 16 < 30
Solution
We can apply the steps for hypothesis testing to make our decision.
1. 𝐻𝐴: 𝜇 ≠ 68 , therefore 𝐻0: 𝜇 = 68
(Two-tail hypothesis)→ (non-directional)
(HA supports the researcher claim that adult male′s weight now is different than previous = 68)
2. Let 𝛼 = 0.01 (depends on the researcher preference, and is given to us)
3. 𝑡𝑐𝑟𝑖𝑡 = 2.947, because 𝑛 = 16 < 30 we use the 𝑡-table (we have small sample)
Now, since we have non-directional hypothesis, we divide 𝛼 𝑏𝑦 2 →
𝛼
2
=
0.01
2
= 0.005
and using 𝑑. 𝑓. = 𝑛 − 1 → 16 − 1 = 15 𝑑. 𝑓. 𝑚𝑒𝑎𝑛𝑠 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚 𝑎𝑛𝑑 𝑎𝑙𝑤𝑎𝑦𝑠 = 𝑛 – 1
4. 𝑡𝑜𝑏𝑡 =
ҧ
𝑥−𝜇
ൗ
𝑠
𝑛
→ 𝑡𝑜𝑏𝑡 =
70−68
ൗ
8
16
= 1
5. 𝑧𝑜𝑏𝑡 < 𝑡𝑐𝑟𝑖𝑡 → 1 < 2.947 → 1 < 2.947 → 𝑟𝑒𝑡𝑎𝑖𝑛 𝑜𝑟 𝑎𝑐𝑐𝑒𝑝𝑡 𝐻0 → 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻𝐴
Clearly, when 𝑛 = 16 (small sample), the investigation did not show a significant weight change for the males
Conclusion:
The researcher/investigator can accept 𝐻0 and reject 𝐻𝐴 at a 0.01 level of significance. That is, adult males’ weight now has not
increased more than previous weight 68 kg.
24. HW 5: Do the example 5 above with ത
𝑋 = 73 and use the following information:
(1) Sample size 𝑛 = 25, 𝛼 = 0.01 𝐻𝑖𝑛𝑡: 𝑡𝑐𝑟𝑖𝑡 = 2.797 & 𝑡𝑜𝑏𝑡 = 2.5 Answer: Accept 𝐻0
(2) Sample size 𝑛 = 25, 𝛼 = 0.05 𝐻𝑖𝑛𝑡: 𝑡𝑐𝑟𝑖𝑡 = 2.064 & 𝑡𝑜𝑏𝑡 = 2.5 Answer: Reject 𝐻0
(3) Sample size 𝑛 = 16, 𝛼 = 0.05 𝐻𝑖𝑛𝑡: 𝑡𝑐𝑟𝑖𝑡 = 2.131 & 𝑡𝑜𝑏𝑡 = 2.5 Answer: Reject 𝐻0
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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 (𝐻𝐴) This is the hypothesis for which the researcher is trying to gain support in a statistical analysis by rejecting the null
hypothesis (𝐻0)
3 Null Hypothesis (𝐻0) This is the hypothesis for which the researcher is trying to reject in a statistical analysis by accepting the alternative
hypothesis (𝐻𝐴)
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 𝐻𝐴 and 𝐻0 are mutual exclusive
If we reject, 𝐻0 then we must accept 𝐻𝐴, conversely,
if we accept 𝐻0 then we must reject 𝐻𝐴.
(We can’t reject both or accept both)
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26. 26
7 Type (I) Error Rejecting 𝐻0 while it is true, and its probability is α
8 Type (II) Error Accepting 𝐻0 while it is false, and its probability is 𝛽 = 1 − 𝛼
9
Degrees of Freedom 𝐝. 𝐟. 𝑑. 𝑓. = 𝑛 − 1 (𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒)
10 Statistical Significance
1. Demonstrates that the result obtained is probably not due to chance but is ‘real’
2. The independent variable must have had a very large effect on the dependent variable.
3. If the obtained probability p is ≤ 𝛼 → can reject 𝐻0 (means accept 𝐻𝐴)
27. Important graphs to illustrate directional and non-directional of 𝐻𝐴
1. 𝛼 = 0.05, (one-tail hypothesis, 𝐻𝐴 to the right) (directional hypothesis 𝐻𝐴)
(a) The gray region is the rejection region of the true 𝐻0 at level 𝛼 = 0.05
(b) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) ≤ 0.05; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 (Type (I) error = 5%) 𝑂𝑅 𝑧𝑜𝑏𝑡 ≥ 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒋𝒆𝒄𝒕 𝐻0
(c) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) > 0.05; 𝑟𝑒𝑡𝑎𝑖𝑛 𝐻0 (Correct decision = 95%) 𝑂𝑅 𝑧𝑜𝑏𝑡 < 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒕𝒂𝒊𝒏 𝐻0 27
28. 1. 𝛼 = 0.05, (two-tail hypothesis, 𝐻𝐴 to both sides) (non-directional hypothesis 𝐻𝐴)
In the two-tail hypothesis we divide 𝛼 𝑏𝑦 2 → 0.05/2 = 0.025
(a) The two gray regions are the rejection regions of the true 𝐻0 at level 𝛼 = 0.05
(b) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) ≤ 0.025; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 (Type (I) error = 5%) 𝑂𝑅 𝑧𝑜𝑏𝑡 ≥ 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒋𝒆𝒄𝒕 𝐻0
(c) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) > 0.025; 𝑟𝑒𝑡𝑎𝑖𝑛 𝐻0 (Correct decision = 95%) 𝑂𝑅 𝑧𝑜𝑏𝑡 < 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒕𝒂𝒊𝒏 𝐻0
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29. 1. 𝛼 = 0.01, (one-tail hypothesis, 𝐻𝐴 to the right) (directional hypothesis 𝐻𝐴)
(a) The gray region is the rejection region of the true 𝐻0 at level 𝛼 = 0.01
(b) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) ≤ 0.01; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 (Type (I) error = 1%) 𝑂𝑅 𝑧𝑜𝑏𝑡 ≥ 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒋𝒆𝒄𝒕 𝐻0
(c) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) > 0.01; 𝑟𝑒𝑡𝑎𝑖𝑛 𝐻0 (Correct decision = 99%) 𝑂𝑅 𝑧𝑜𝑏𝑡 < 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒕𝒂𝒊𝒏 𝐻0
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30. 1. 𝛼 = 0.01, (two-tail hypothesis, 𝐻𝐴 to both sides) (non-directional hypothesis 𝐻𝐴)
In the two-tail hypothesis we divide 𝛼 𝑏𝑦 2 → 0.01/2 = 0.005
(a) The two gray regions are the rejection regions of the true 𝐻0 at level 𝛼 = 0.01
(b) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) ≤ 0.005; 𝑟𝑒𝑗𝑒𝑐𝑡 𝐻0 (Type (I) error = 1%) 𝑂𝑅 𝑧𝑜𝑏𝑡 ≥ 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒋𝒆𝒄𝒕 𝐻0
(c) If 𝑃(𝐻0 𝑖𝑠 𝑡𝑟𝑢𝑒) > 0.005; 𝑟𝑒𝑡𝑎𝑖𝑛 𝐻0 (Correct decision = 99%) 𝑂𝑅 𝑧𝑜𝑏𝑡 < 𝑧𝑐𝑟𝑖𝑡 ; 𝒓𝒆𝒕𝒂𝒊𝒏 𝐻0
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