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Q1
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"
Ans
#$%&#%!'()*
#$&#!('(+)*
,!-.(/)01
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;)(0"
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@:
#:?;?A211"(B
C?;?)(0"D?;?)(B
EFG
:?;?H?;?
There Is No Signicance between them
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#
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,&:'##-/&'-,+)*4410/)110
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)*"&*03-*3=-(B//
)*
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@:
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C??)*D??)(8"
EFG
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,&:'Q#&'-,R*/)114(8
!!*0!
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#!('R4
#
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@:
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Q1 A researcher at a major clinic wishes to estimate the proportion of the adult population of the United States that has sleep deprivation. How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 5%. Ans There Is No Significance between them - Under The Null Hypothesis Ho: p1 = p2 There Is Significance between them - Under The Alternate Hypothesis H1: p1 != p2 Probability Success( X1 )=40 Number of Observed (n1)=200 P1= X1/n1=0.2 Probability Success(X2)=45 No. of Observed (n2)=300 P2= X2/n2=0.15 Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2) P^=0.17 Q^ Value For Proportion= 1-P^=0.83 we use Test Statistic (Z) = (P1-P2)/?(P^Q^(1/n1+1/n2)) Z cal=(0.2-0.15)/Sqrt((0.17*0.83(1/200+1/300)) Z cal=1.458 | Z cal | =1.458 Critical Value The Value of |Z tab| at LOS 0.05% is 1.96 We got |Z cal| =1.458 & | Z tab | =1.96 Make Decision Hence Value of |Z cal | < | Z tab | and Here we Accept Ho There Is No Significance between them a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=297 Sample Size(n)=540 Sample proportion = x/n =0.55 Confidence Interval = [ 0.55 ?Z a/2 ( Sqrt ( 0.55*0.45) /540) ] = [ 0.55 - 2.58* Sqrt(0.0005) , 0.55 + 2.58* Sqrt(0.0005) ] = [ 0.495,0.605] Yes, it is affective Test Statistic Population Mean(U)=7 Given That X(Mean)=5.6 Standard Deviation(S.D)=2.1 Number (n)=20 we use Test Statistic (t) = x-U/(s.d/Sq ...
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