Answer:
s e([tex]p^{-} _{1} - p^{-} _{2}[/tex] ) = 0.07576
Step-by-step explanation:
Step(i):-
The proportion of the first sample
[tex]p^{-} _{1} = \frac{x_{1} }{n_{1} } = \frac{23}{53} = 0.43396[/tex]
The proportion of the second sample
[tex]p^{-} _{2} = \frac{x_{2} }{n_{2} } = \frac{48}{178} = 0.2696[/tex]
Step(ii):-
Standard Error of ([tex]p^{-} _{1} - p^{-} _{2}[/tex]) is determined by
[tex]se(p^{-} _{1} - p^{-} _{2}) = \sqrt{\frac{p^{-} _{1} (1-p^{-} _{1} )}{n_{1} }+\frac{p^{-} _{2}(1-p^{-} _{2} ) }{n_{2} } }[/tex]
[tex]se(p^{-} _{1} - p^{-} _{2}) = \sqrt{\frac{0.43396) (1-0.43396 )}{53 }+\frac{0.2696(1-0.2696 ) }{178} }[/tex]
on simplification, we get
s e([tex]p^{-} _{1} - p^{-} _{2}[/tex] ) = 0.07576
