a)
population
proportion ,p= 0.7
n= 100
std error
, SE = √( p(1-p)/n ) = 0.0458
we need to
compute probability for
0.62 < p̂ < 0.79
Z1 =( p̂1
- p )/SE= ( 0.62 -
0.7 ) / 0.0458 =
-1.746
Z2 =( p̂2
- p )/SE= ( 0.79 -
0.7 ) / 0.0458 =
1.964
P( 0.62 < p̂ <
0.79 ) = P[( p̂1-p )/SE< Z
<(p̂2-p)/SE ] =P( -1.746
< Z < 1.964 )
= P ( Z < 1.964 ) - P (
-1.746 ) = 0.9752
- 0.040 =
0.9348
b)
population proportion ,p= 0.7
n= 100
std error , SE = √( p(1-p)/n ) = 0.0458
we need to compute probability for
P(p̂1<p̂<p̂2)= 0.9
proportion left 0.10 is equally
distributed both left and right side of normal
curve
P(p̂ < p̂ 1) = 0.05 and P(p̂ >
p̂2)= 0.95
Z value at 0.05 = -1.645
(excel formula =NORMSINV( 0.10 / 2 )
)
Z value at 0.95 = 1.645 (excel
formula =NORMSINV( 0.10 / 2 )
)
Z =( p̂ - p )/SE
so, p̂ = Z*SE+p
p̂ 1 = Z*SE+p = -1.645 *
0.0458 + 0.7 =
0.6246
p̂ 2 = Z*SE+p = 1.645 *
0.0458 + 0.7 =
0.7754
so, answer : above 62.46% and below 77.54%
c)
population proportion ,p= 0.7
n= 100
std error , SE = √( p(1-p)/n ) = 0.0458
we need to compute probability for
P(p̂1<p̂<p̂2)= 0.91
proportion left 0.09 is equally
distributed both left and right side of normal
curve
P(p̂ < p̂ 1) = 0.045 and P(p̂ >
p̂2)= 0.955
Z value at 0.045 =
-1.695 (excel formula =NORMSINV(
0.09 / 2 ) )
Z value at 0.955 =
1.695 (excel formula =NORMSINV(
0.09 / 2 ) )
Z =( p̂ - p )/SE=
so, p̂ = Z*SE+p
p̂ 1 = Z*SE+p = -1.695 *
0.0458 + 0.7 =
0.6223
p̂ 2 = Z*SE+p = 1.695 *
0.0458 + 0.7 =
0.7777
above 62.23% and below 77.77%
d)
option d)
b. The probability is 90% that the sample percentage will be oontained within what symmetrical limits of the population percentage? The probability is 90% that the sample percentage will be conta...
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