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Sagot :
Answer:
99% of the confidence level of the mean WR score for all convicted drug dealers
(37.4544, 40.5456)
Step-by-step explanation:
Step(i):-
Given that the sample size 'n' = 100
Given that the mean of the sample(x⁻) = 39
Given a standard deviation of the sample (σ) = 6
Level of significance = 0.01 or 99%
Step(ii):-
99% of the confidence level of the mean WR score for all convicted drug dealers
[tex](x^{-} - Z_{0.01} \frac{S.D}{\sqrt{n} } ,x^{-} + Z_{0.01} \frac{S.D}{\sqrt{n} })[/tex]
[tex](39 -2.576 \frac{6}{\sqrt{100} } ,39 + 2.576 \frac{6}{\sqrt{100} })[/tex]
(39 - 1.5456 , 39+1.5456)
(37.4544, 40.5456)
Final answer:-
99% of the confidence level of the mean WR score for all convicted drug dealers
(37.4544, 40.5456)
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