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2 = 34.7, df = 21 0.75 <P<0.90 0.025 <P<0.05 0.90 < P<0.95 О 0.05<P<0.10 UESTIONS Given the test statistic and degrees of freQuestion 5 and 6

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Answer #1

Question 5

P value for \chi^2 = 2.76 with degree of freedom = 16 is 0.8383

0.75 < P vaue < 0.90

Question 6

Test Statistic :-

d̅ = Σdi/n = -60 / 5 = -12

S(d) = √(Σ (di - d̅)2 / n-1)
S(d) = √( 158 / 4) = 6.2849

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