I have completed parts A & B but I am not sure if it is correct, and I am not su

Question

I have completed parts A & B but I am not sure if it is correct, and I am not sure how to answer part C. Please help to answer part C. Thank you.

table shows the data for these patients and for an independently chosen group of 10,000 healthy controls from same city UP O Blood types were determined for 1,655 ulcer patients. The Blood type Ulcer patients Controls 911 579 124 4,578 4,219 890 (a) The value of the chi-square statistic for this contingency table is x2 = 49.0. Carry out the chi-square test at = 0.01 AB Total 1.655 10,000 (b) Construct a table showing the percentage distributions of blood type for patients and for controls. (c) Verify the value of given in part (a) Ao : Blce type distrb tion i's the sane for ulcer atiet and control atients A Bloodl type dstrution is not the sane for olcer pahierte an ind test stats and the P valu control patients ale O.o0 is lessthan the given level of significance e peject the Ho the blood distrbution in eve tanc of ool s W no+ the ame for vlcer patients and comtrol patiets Blood Type O AB 55.05 3a8 Contigency Halale cer Patien Controls 145794a. 19

Explanation / Answer

(a) Here in part (a), you can also calculated the critical value of chi-square for dF = 3 ;

(c) Here we have to verify the value of chi-square X2 we have to calculate the value of chi-square

Observed Table :

Expected Table

Here we have to find the expected values. i.e.

for Blood Group O and Ulcer Patients = (1655 * 5489)/ 11655 = 779.43

similarly, all expected values will be calculated. THe table is given below.

Now for each cell, we will calculated the chi-square valeu

X2  = (observed - expected)2/expected

SO we will get value of chi- square X2 = 49

and critical value of X2critical  = 11.345

so X2 > X2critical

Blood Type Ulcer Patients Controls Total O 911 4578 5489 A 579 4219 4798 B 124 890 1014 AB 41 313 354 Total 1655 10000 11655

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