To determine the confidence interval a left parenthesis 1 minus alpha right

To determine the confidence interval a left

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2. While either technology or a cumulative standardized normal distribution table can be used to find
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or sample mean =norm.s.inv standard normal distribution margin of error norm.s.inv at alpha over 2 [email protected]/2=-(Norm.s.inv(Zscore)) 1.644854 2.5758293035489 x α/ x α/ σ z σ z 2 2 n σ σ x x α/ x x α/ x σ z x LCL σ z x UCL 2 2 σ x x α/ σ z 2
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Margin Error=([email protected]/2)*(Std Error) 10.84321 11570.63 nfidence Level=sampl Mean+MoE= 258.6432 93927.63 nfidence Level=sampl Mean-MoE= 236.9568 70786.37 We have 90% confidence, Pop Mean of lexury hotel cost per person is between 237$ and $259 per night. x α/ σ z 2
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Over StartRoot n EndRoot EndFraction σ n
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n σ
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B. a. sample size=n 30 60 sample mean=xbar 507 507 Pop Std Dev 63 63 Confidence interval 90/100 0.9 0.9 Alph=1-confidence interval 0.1 0.1 Zscore=Alpha/2 0.05 0.05 we have: std error=pop std dev/SQRT(sample size) to calculate the =39/Sqrt(35) 11.50217 8.133265 STD Error Statement= A. n σ σ x
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