Find Z test proportion 2. In Hospital A, tracheotomy had been done in 60 patients out of 150 while in the other Hospital B it was done in 50 out of 150. Find if the difference observed in the two Hospitals is by chance or due to some influence. Find the overall sample proportioni 60 P1 = nl = 150 150 50 p, = 150 , n2 = 150 Find the overall sample proportion X1+ X2 = 60 + 50 %3D n +n2 150 + 150 ,1-p Null Hypothesis H: There is no significant difference in the tracheotomy in both hospitals H: There a significant difference the tracheotomy both hospitals • We Use a 5% alpha level. • Z test statistics is given as: Page 3 of 4 • Z = "P-P = P-Pa %3D %3D %3D x( Z= • The critical value associated with a 5% alpha level is1.96. • Analysis: Computed value is than the critical value 1.9 • Decision: Ho Conclusion: There is - -- in the tracheotomy rate at 95% significance level between Hospital A and Hospital B

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150)
Find Z test proportion
2. In Hospital A, tracheotomy had been done in 60 patients out of 150 while
in the other Hospital B it was done in 50 out of 150. Find if the difference
observed in the two Hospitals is by chance or due to some influence.
Find the overall sample proportion _:
60
P, =
150
nl = 150
50
p, =
150
- , n2 = 150
Find the overall sample proportion
X1 + X2
60 + 50
=
n1 + n2
%3D
150 + 150
,1-p
Null Hypothesis
H: There is no significant difference in the tracheotomy in both hospitals
H: There is a significant difference
e tracheotomy in both hospitals
We Use a 5% alpha level.
• Ztest statistics is given as:
Page 3 of 4
P1-P2
%3D
%3D
x(
150
Z=
The critical value associated with a 5% alpha level is1.96.
• Analysis: Computed value is
than the critical value 1.9
• Decision: Ho
Conclusion: There is - ------ in the tracheotomy rate at 95%
significance level between Hospital A and Hospital B
Transcribed Image Text:150) Find Z test proportion 2. In Hospital A, tracheotomy had been done in 60 patients out of 150 while in the other Hospital B it was done in 50 out of 150. Find if the difference observed in the two Hospitals is by chance or due to some influence. Find the overall sample proportion _: 60 P, = 150 nl = 150 50 p, = 150 - , n2 = 150 Find the overall sample proportion X1 + X2 60 + 50 = n1 + n2 %3D 150 + 150 ,1-p Null Hypothesis H: There is no significant difference in the tracheotomy in both hospitals H: There is a significant difference e tracheotomy in both hospitals We Use a 5% alpha level. • Ztest statistics is given as: Page 3 of 4 P1-P2 %3D %3D x( 150 Z= The critical value associated with a 5% alpha level is1.96. • Analysis: Computed value is than the critical value 1.9 • Decision: Ho Conclusion: There is - ------ in the tracheotomy rate at 95% significance level between Hospital A and Hospital B
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