excel chapter 8.xlsx - common confidence levels 90 95 99 level of significance is called the alpha value outside of rang does not contain the true

excel chapter 8.xlsx - common confidence levels 90 95 99...

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quiz1 49 sample size 35 49 52 sample mean 247.8 52 Pop Std Dev 39 8 b. Confidence interval 90/100 0.9 0.95 Alph=1-confidence interval 0.1 0.05 Zscore=Alpha/2 0.05 0.025 we have: std error=pop std dev/SQRT(sample size) MoE= to calculate the =39/Sqrt(35) 6.592203 1.142857 My Upper Confidence L STD Error My Lower Confidence L Statement= We have 9 Construct a 95% x overbarxequals=52 nequals=49 With confidence interval to estimate the population mean using the data below. sigma σ equals=88 n σ σ x α x z ME
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sample mean =norm.s.inv standard normal distribution margin of erro norm.s.inv at alpha over 2 [email protected]/2=-(Norm.s.inv(Zscore)) 1.64485362695147 1.95996398454 Margin Error=([email protected]/2)*(Std Error) 10.84321 2.239958839 Level=sampl Mean+MoE= 258.6432 54.23996 Level=sampl Mean-MoE= 236.9568 49.76004 90% confidence, Pop Mean of lexury hotel cost per person is between 237$ and $259 per night. quiz1,2 a. 35 1 20 247.8 82357 81964 39 4492 4556 b. Confidence interval 90/100 0.9 0.99 Alph=1-confidence interval 0.1 0.01 critical zscore Zscore=Alpha/2 0.05 0.005 sample size= n sample mean= xbar Pop Std Dev= Q or S x x x LCL x UCL x α/ σ 2 x z ME
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My Upper ConSTD ErrorMy Lower ConStatement= W9595% confidence interval to estimate the population mean using the following data. What assumptions need to be made to construct this interval? What assumptions need to be made to construct this interval?
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