Hypothesis Z, t, C.I

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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Q1.AnsPooled variance t test is AnswerValue of Combined Standard deviation is S= (n1*s1^2 + n2*s2^2 )/(n1+n2-2)we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))Q2.AnsConfidence Interval CI = x Z a/2 * (sd/ Sqrt(n))Where, x = Mean sd = Standard Deviation a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=42144Standard deviation( sd )=3000Sample Size(n)=10Confidence Interval = [ 42144 Z a/2 ( 3000/ Sqrt ( 10) ) ] = [ 42144 - 1.96 * (948.6833) , 42144 + 1.96 * (948.6833) ] = [ 40284.5807,44003.4193 ]None of the above is AnswerQ3.Ans x = Mean n = Sample Size a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=54.4Sample Size(n)=80Sample proportion = x/n =0.68Confidence Interval = [ 0.68 Z a/2 ( Sqrt ( 0.68*0.32) /80) ] = [ 0.68 - 1.96* Sqrt(0.0027) , 0.68 + 1.96* Sqrt(0.0027) ] = [ 0.578,0.782] ~ [ 57.8% , 78.2% ]58; 78 is AnswerQ4.CI = x Z a/2 * (sd/ Sqrt(n))Where, x = Mean sd = Standard Deviation a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=15.23Standard deviation( sd )=1.65Sample Size(n)=24379Confidence Interval = [ 15.23 Z a/2 ( 1.65/ Sqrt ( 24379) ) ] = [ 15.23 - 1.96 * (0.0106) , 15.23 + 1.96 * (0.0106) ] = [ 15.2093,15.2507 ]15.21 ; 15.25 is AnswerQ5.Under The Null Hypothesis H0:P=0.77 Under The Alternate Hypothesis H

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