Test at the α = 0.05 level of significance

[Ramd]

The following table shows the observed distribution of A, B, AB, and O blood types in three samples of African Americans living in different locations.
Test at the α = 0.05 level of significance, whether the distribution of blood type for African Americans is different across the three regions.

1 (Florida) II( Iowa) III (Missouri)
A 122 1781 353
B 117 1351 269
AB 19 289 60
O 244 3301 713

Could you guys please let me know how to start with this, I am reachable at [email protected]

 

[ShaktiRathore]

I would advise to perform F test or
t-test taking two samples at a time and compare them
compare florida and iowa than; H0:f=i and Ha:f!=i
than compare florida and Missouri than; H0:f=m and Ha:f!=m and than compare the results from the above tests

thanks

 

[aadityafrm]

Its a typical test of goodness of fit where we are required to establish whether or not an observed frequency distribution differs from a theoretical distribution. so chi-square test is suggested.
The null hypothesis here would be H0: pA(Florida) = p(Iowa) = p(Missouri) we have similar null hypothesis H0 for rows pB, pAB and pO. test statistics is arrived at by computing X^2 = Sum of (observed value- expected value)^2/ expected value for each cell

Observed value is the value in each cell of the matrix, and the expected value = (sum of columns for each row)* (sum of rows for each column)/total observations. for e.g. first cell's expected value would be Row(A)*Col(Florida)/Total observations.

After calculating X^2 and degrees of freedom (The null distribution with j rows and k columns is approximated by the chi-squared distribution with (k − 1)(j − 1) degrees of freedom), we are interested in the probability P( X^2> calculated stat) = (value from chi stat table for (k − 1)(j − 1) degrees of freedom)

based on this we decide whether to reject or accept null.

Hope this helps

 

[ShaktiRathore]

I would advise to perform F test or
t-test taking two samples at a time and compare them
compare florida and iowa than; H0:f=i and Ha:f!=i
than compare florida and Missouri than; H0:f=m and Ha:f!=m and than compare the results from the above tests

thanks

Oh i misread the term distributions.If the above are distributions to be compared (i mistook for comparing means)than we have to use chi-square test as explained by aadityafrm
It s the chi square test that finds the goodness of fit for each of the data of each region and compares if distributions are same for the three region. if the data has been identified with certain distributions than that is the best distribution for data. If all the three regions have same distributions as identified by chi-square test as explained by aadityafrm than they have same distribution otherwise not.

thanking you

 

[Ramd]

Hi Shakti,

I have tested the data with both F and T test. I have few doubts mentioned in "Comparision between F and TTest" and results in "F-Test (A10) and T-Test (A10)".
You can find Original question in tab "Question 10".

Could you kindly advice me.

Best Regards,
Ashutosh Singh

 

[ShaktiRathore]

check this file. I have made some changes.

 

[Ramd]

Thanks to both (Aaditya and Shakti) of you.

I have doubt in using chi-square or anova....

The chi-square test of association is used to test the null hypothesis that there is no association between two nominal scale variables. Chi-square gives me 46.3% over 5%, so that I can strongly fail the null hypothesis

Whereas,

ANOVA is a statistical test of whether the means of several groups are all equal. In anova, I am using 'Two-Factor Without Replication' in Excel which gives me below table:

Here F cric value has its own story...
Please advice.... which way to go...
Many thanks in advance

 

[ShaktiRathore]

We are Comparing columns here: Here F crit<F and this suggests that we reject the null hypothesis that blood type for African Americans is same across the three regions.p value<.05 so that at 95% CL we conclude that the blood type for African Americans is different across the three regions.
F=Variance B/w regions of blood samples/variance of blood samples within the region=> greater F value there is greater variance of blood samples between the regions compare to within the region so higher F value suggests greater variation between the groups or regions(columns).
hope it makes sense,
thanks

 

[Ramd]

yea, thanks a lot again for your kind help!