## Answers

solution D Here the null hypothesis is Hor=0.25 2) the alternative hypothers is his 80.25 3) level of significance d=5% 20.05 u Given n=51, Sample S.D(5)=0.37 5 test itatutic x = (n-D8 = 50(837) = 109.52 g2 d'f=n-1=50, P-value=0.0000 6 ny 7 Here P-value ca so, reject the no null hypother 1 there is a evidence to conclude that the true deviation in the percentage of titanium different from 0.25

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