For two independent samples, the following table presents the structure of the null hypothesis, the test statistics to be used, the different forms of alternative hypotheses and their corresponding critical region.
For two independent samples, the following table presents the structure of the null hypothesis, the test statistics to be used, the different forms of alternative hypotheses and their corresponding critical region.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Problem 1P
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From this two photo it shows 3 different forms of formula for my biostatistic subject. I want to ask in this 3 different formula how can I differentitate on which formula to use on what type of question.
![Tests on Means for Two sample cases
For two independent samples, the following table presents the structure of the null hypothesis,
the test statistics to be used, the different forms of alternative hypotheses and their
corresponding critical region.
Case 1: 0₁² and ₂² are known
Ho
H1 - H2 = do
Test statistic
Z=
(x₁-x₂)-do
01²
n1
02²
n2
+
Ha
H1-H2 < do
H1 - H2> do
H1 - H2 # do
Critical region
(the average weights of girls and boys do not differ)
Z<-Za
Z> Za
IZI> Za/2
Example: A weight loss management program was given to obese children; 50 girls and 75 boys.
After the program, the girls made an average weight of 80 kilos with a standard deviation of 4 and
the boys had an average of 86 kilos with a standard deviation of 6. Is there sufficient evidence at
0.05 level of significance that the average weight of girls and boys differ?
Solution: Let μ₁ = average weight of girls and μ₂= average weight of boys
Step 1. Set the null and alternative hypotheses
Ho H₁ H₂
OR (can be written this way)
H₁ H₁ - H₂ = 0 (there is no significant difference between the average weights of girls and boys)
Ha μ₁ μ2 (the average weights of girls and boys differ)
OR (can be written this way)
Ha : μ1 - U2 #0 (there is a significant difference between the average weights of girls and boys)
Step 2. The level of significance is set at a = 0.05
This means that we are willing to commit a 5% chance of committing a Type I error; that is,
rejecting the null hypothesis when it is true.
Since, the alternative hypothesis is non-directional, used the two-tailed test, thus divide alpha value
by 2, %/2 = 0.05/2 = 0.025.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe569dae1-82ce-45a6-b2ad-14cffb482849%2Ff90d104a-f54c-45d1-8e87-e4c195ca2793%2F0wao5z_processed.png&w=3840&q=75)
Transcribed Image Text:Tests on Means for Two sample cases
For two independent samples, the following table presents the structure of the null hypothesis,
the test statistics to be used, the different forms of alternative hypotheses and their
corresponding critical region.
Case 1: 0₁² and ₂² are known
Ho
H1 - H2 = do
Test statistic
Z=
(x₁-x₂)-do
01²
n1
02²
n2
+
Ha
H1-H2 < do
H1 - H2> do
H1 - H2 # do
Critical region
(the average weights of girls and boys do not differ)
Z<-Za
Z> Za
IZI> Za/2
Example: A weight loss management program was given to obese children; 50 girls and 75 boys.
After the program, the girls made an average weight of 80 kilos with a standard deviation of 4 and
the boys had an average of 86 kilos with a standard deviation of 6. Is there sufficient evidence at
0.05 level of significance that the average weight of girls and boys differ?
Solution: Let μ₁ = average weight of girls and μ₂= average weight of boys
Step 1. Set the null and alternative hypotheses
Ho H₁ H₂
OR (can be written this way)
H₁ H₁ - H₂ = 0 (there is no significant difference between the average weights of girls and boys)
Ha μ₁ μ2 (the average weights of girls and boys differ)
OR (can be written this way)
Ha : μ1 - U2 #0 (there is a significant difference between the average weights of girls and boys)
Step 2. The level of significance is set at a = 0.05
This means that we are willing to commit a 5% chance of committing a Type I error; that is,
rejecting the null hypothesis when it is true.
Since, the alternative hypothesis is non-directional, used the two-tailed test, thus divide alpha value
by 2, %/2 = 0.05/2 = 0.025.
![and boys.
● Case 2: 0₁² = 0₂² but unknown
H₂
Name:
Section:
μ₁-μ₂= do
sp=
H₁-H₂= do
(x₁-x₂)-do
1 1
sp n₁ 1₂
df=v=n₁ + n₂-2
Schedule:
df
Test statistic
t =
Case 3: 0₁²0₂² and unknown
H₂
(n₁−1)s₁²+(n₂−1)S₂²
n₁+n₂-2
t =
Test statistic
(x₁-x₂)-do
721 722
n1
n₁-1
+
+
$₂²
n₂
n₂-1
H₂
H1-H₂<do
H1-H₂> do
μ1-μ₂ #do
H₂
H₁-H₂ < do
H1-H₂> do
H₁-H₂ # do
Critical region
t<-ta
t> ta
|t|> ta/2
Class number:
Date:
Critical region
t<-ta
t> to
|t|>ta/2
Example: A group of researchers selected 50 countries and classified them according to three GNP levels:
Low (less than $50B), Moderate ($50B - $174B) and High (above $175B). For each classification, the average
life expectancy was determined:](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe569dae1-82ce-45a6-b2ad-14cffb482849%2Ff90d104a-f54c-45d1-8e87-e4c195ca2793%2Fmhlwan_processed.png&w=3840&q=75)
Transcribed Image Text:and boys.
● Case 2: 0₁² = 0₂² but unknown
H₂
Name:
Section:
μ₁-μ₂= do
sp=
H₁-H₂= do
(x₁-x₂)-do
1 1
sp n₁ 1₂
df=v=n₁ + n₂-2
Schedule:
df
Test statistic
t =
Case 3: 0₁²0₂² and unknown
H₂
(n₁−1)s₁²+(n₂−1)S₂²
n₁+n₂-2
t =
Test statistic
(x₁-x₂)-do
721 722
n1
n₁-1
+
+
$₂²
n₂
n₂-1
H₂
H1-H₂<do
H1-H₂> do
μ1-μ₂ #do
H₂
H₁-H₂ < do
H1-H₂> do
H₁-H₂ # do
Critical region
t<-ta
t> ta
|t|> ta/2
Class number:
Date:
Critical region
t<-ta
t> to
|t|>ta/2
Example: A group of researchers selected 50 countries and classified them according to three GNP levels:
Low (less than $50B), Moderate ($50B - $174B) and High (above $175B). For each classification, the average
life expectancy was determined:
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