Respuesta :
Answer:
A)The Standard error for difference in proportion is 0.0354
B)The Confidence interval with 95 % for difference in proportion is (0.356 ,0.495)
C)The conclusion is that proportion of herbicides and no herbicides and having them diagnosed with malignant lymphoma is between 35.6 % and 49.5 % And higher than the proportion to pets in homes where no pesticides were used.
Step-by-step explanation:
Given:
Total dogs which are herbicide =827=n1
diagnosing malignant lymphoma Dogs=473
Total no herbicides Dogs=130=n2
19 were found to have lymphoma.
To Find:
a)What’s the standard error of the difference in the two proportions?
b) Construct a 95% confidence interval for this difference.
c) State an appropriate conclusion
Solution:
The standard error of difference is given by ,
Consider S1 for case 1 (herbicides) and S2 for case 2 (no herbicides)
So
[tex]S1-S2=Sqrt[p1(1-p1)/n1 +p2(1-P2)/n2)][/tex]
Here[tex]p1=473/827[/tex] =0.578 and [tex]p2=19/130[/tex]=0.146
[tex]S1-S2=Sqrt[0.572(0.428)/827 +0.146(0.854)/130][/tex]
[tex]S1-S2=Sqrt[2.96e-4 +9.59e-4][/tex]
[tex]S1-S2=Sqrt[0.001255][/tex]
[tex]S1-S2=0.0354[/tex]
This gives the difference between two proportions 0.0354
Now,
For 95 % C.I it is given by for above difference at Z=1.96
[tex]C.I.=[p1-p2-Z*Sqrt[p1(1-p1)/n1 +p2(1-P2)/n2) ,p1-p2+Z*Sqrt[p1(1-p1)/n1 +p2(1-P2)/n2)][/tex]
[tex]C.I=0.426-1.96*Sqrt[0.57(0.42)/827 +0.146(0.85)/130],0.426+1.96*Sqrt[0.57(0.42)/827 +0.146(0.85)/130][/tex]
[tex]C.I.=(0.356,0.495)[/tex]
The conclusion is that proportion of herbicides and no herbicides and having them diagnosed with malignant lymphoma is between 35.6 % and 49.5 % And higher than the proportion to pets in homes where no pesticides were used.