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Unformatted text preview: ECO2121: Methods of Economic Statistics TA12 – 8 April 2009 1. Hypothesis Testing (Two populations) (continued) Referring to TA10, test at α = 0.05 that H0: pX – pY = 0, or μX – μY = 0. Example 2: Given the following: nX = 51, ̅ = 67, sX = 15, and nY = 75, = 78, sY = 15. We still consider several cases. Case 0: if σX2 and σY2 are known, says σX2 = 102 and σY2 = 122, then − ~ N(μX – μY, σX2/nX+σY2/nY). Under H0, − ~ N(0, σX2/nX+σY2/nY). Test statistic = ( − ) H0 is rejected. + = (67 − 78) + = 5.58 < 1.96 = zα/2 Case 1: if σX2 and σY2 are unknown but equal, we use the “pooled sample variance”, then ( − )−( − ) = ( − )−( 1 + 1 − ) ~ + ( − 1) + ( − 2) = 4.04 < 1.98 = t0.05/2(124) − 1) Under H0, + (−) + = (−) 1 + 1 ~(
( We calculated Sp = 15, so the test statistic = ) H0 is rejected. Recall a 95% C.I. is [16.38, 5.61] does not cover 0. Case 2: if σX2 and σY2 are unknown and unequal, then
( )( ) ~ ( ) and = .
( ) Under H0, we can construct the test statistic = H0 is rejected. = 4.04 < 2 = t0.05/2(ν) Recall a 95% C.I. is [16.46, 5.54] does not cover 0. Example 2: Given the following: nX = 165, = 137,and nY = 159, = 96. Note that the test statistic should be derived under H0: pX – pY = 0! By CLT, as nX and nY are both “large”, ̂ − ̂ ~ N(pX – pY, pXqX/nX + pYqY/nY). ⁄ +̂ ⁄ When constructing C.I., we use ̂ to estimate pXqX/nX + pYqY/nY, but now we should derive an estimator under H0, i.e. pX = pY = p, we “pool” the data together and use ̂ ⁄ + ̂ ⁄ to estimate pXqX/nX + pYqY/nY, where In our example, Hence, the test statistic = ( ̂ − ̂ ) +∑ + = + + ̂ = (137 + 96) / (165 + 159) = 233 / 324 ̂= ∑ = 4.54 > 1.96 = z0.05/2 + = − + H0 is rejected. NO recall for C.I. even it does not cover 0. (why?) Taylor 10 March 2009 ...
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 Fall '08
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