Course Notes - Department of Mathematics and Statistics

CALCULATING THE CONFIDENCE INTERVAL (not assuming independence)Mean difference = 6610 = 6.6 Variance of difference = Σ(d i − d) 2304.4n−1=9= 33.82 Degrees of freedom, ν = n − 1 = 9 And t (0.025,9) = 2.262The 95% C.I. for the difference becomes:√33.826.6 ± 2.262 × √10= 6.6 ± 4.2That is, 2.4 < µ d < 10.8INTERPRETING THE CONFIDENCE INTERVAL (not assuming independence)There is evidence that the weight loss programme has reducedweights since the difference of 0 is not in this interval.The profile of each person is constant in this study because thesame person has produced the two values.Thus this is called the PAIRED T-TEST.6.4 Confidence Intervals for ProportionsSAMPLING DISTRIBUTION FOR A PROPORTION• If X is a binomial distribution with parameters n and π we knowthat µ X = nπ and σ X = √ nπ(1 − π).• Suppose one sample produces a proportion of successes p = X n inn trials.• If we take many samples we get different p’s.• Using the central limit theorem, the resulting distribution P ofthese proportions is normal.MEANIf P = X n : µ P = µ Xn= nπ n= π110