Problems on Normal Distribution,

Feb 3rd, 2012
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Head-shot on a) What is the null hypothesis? b) What is the alternative hypothesis? c) How may degrees of freedom are there? d) What is the sample standard deviation of the difference in the test scores? e) What is the value of the test statistic? f) What is the p-value? g) How to find a C.I

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Q1A 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 != p2Probability Success( X1 )=40Number of Observed (n1)=200P1= X1/n1=0.2Probability Success(X2)=45No. of Observed (n2)=300P2= X2/n2=0.15Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2) P^=0.17Q^ Value For Proportion= 1-P^=0.83we 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 | =1.458Critical ValueThe Value of |Z tab| at LOS 0.05% is 1.96We got |Z cal| =1.458 & | Z tab | =1.96Make DecisionHence Value of |Z cal | < | Z tab | and Here we Accept HoThere Is No Significance between them a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=297Sample Size(n)=540Sample proportion = x/n =0.55Confidence 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 affectiveTest StatisticPopulation Mean(U)=7Given That X(Mean)=5.6Standard Deviation(S.D)=2.1Number (n)=20we use Test Statistic (t) = x-U/(s.d/Sq

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