Hi
What is the appropriate two sample t procedure for the summary below milking periods of cow breeds pooled/unpooled (Welch) using R?
Frisian cow Xbar=342.5 S1=10.2 n1=126
Jersey cow Xbar =348.6 S2=12.2 n2=39
Construct 95% confidence interval for µ1-µ2
Give proper interpretation of the interval
Thanks
@Alma1 , I do not know what data you have, an example would be helpful.
In the below script (when you execute) you see a dataset and both t-test performed, giving different results.
Generally, statistical Tests have preconditions, and t-test assumes normal Distribution of the dataset,
pooled t-test assumes equal variance,
t-test works also with different variance.
This is because they work with mean and variance, and if that two numbers do not represent the data correctly, the method doesn't work.
If you perform a t-test, and the assumptions are not valid, you cannot rely on the result.
In the below example you can see different results for both t-Tests, but be Aware, that normal Distribution is also not valid in this case.
You can find These results in the documentation also, and I found STIPS a quite good training to understand this stuff.
Names Default To Here( 1 );
cdt = Open( "$SAMPLE_DATA\companies.jmp" );
cdt << Oneway(
Y( :Name( "Profits ($M)" ) ),
X( :Type ),
Means( 1 ),
t Test( 1 ),
Mean Diamonds( 1 ),
SendToReport(
Dispatch( {"Oneway Anova"}, "Summary of Fit", OutlineBox, {Close( 1 )} ),
Dispatch( {"Oneway Anova"}, "Analysis of Variance", OutlineBox, {Close( 1 )} ),
Dispatch( {"Oneway Anova"}, "Means for Oneway Anova", OutlineBox, {Close( 1 )} )
)
);
Nop! it is not a homework. we are conducting a research and I didnt know whether to use pooled /unpooled test whats wrong asking for ideas? you are free not to answer if you wish
@Alma1 , I do not know what data you have, an example would be helpful.
In the below script (when you execute) you see a dataset and both t-test performed, giving different results.
Generally, statistical Tests have preconditions, and t-test assumes normal Distribution of the dataset,
pooled t-test assumes equal variance,
t-test works also with different variance.
This is because they work with mean and variance, and if that two numbers do not represent the data correctly, the method doesn't work.
If you perform a t-test, and the assumptions are not valid, you cannot rely on the result.
In the below example you can see different results for both t-Tests, but be Aware, that normal Distribution is also not valid in this case.
You can find These results in the documentation also, and I found STIPS a quite good training to understand this stuff.
Names Default To Here( 1 );
cdt = Open( "$SAMPLE_DATA\companies.jmp" );
cdt << Oneway(
Y( :Name( "Profits ($M)" ) ),
X( :Type ),
Means( 1 ),
t Test( 1 ),
Mean Diamonds( 1 ),
SendToReport(
Dispatch( {"Oneway Anova"}, "Summary of Fit", OutlineBox, {Close( 1 )} ),
Dispatch( {"Oneway Anova"}, "Analysis of Variance", OutlineBox, {Close( 1 )} ),
Dispatch( {"Oneway Anova"}, "Means for Oneway Anova", OutlineBox, {Close( 1 )} )
)
);
by the way jmp background is R.
@Alma1 wrote:
by the way jmp background is R.
I'm not sure exactly what you're thinking of here but JMP has no background with R.
JMP is a product from SAS and is developed solely by us.
There is integration with R through JSL, providing the ability to call R and submit R code but that's the only association between JMP and R.