A student took a random sample of 181 properties to examine how much more waterf

Question

A student took a random sample of 181 properties to examine how much more waterfront property is worth. Her summaries and boxplots of the two groups of prices are shown. Construct and interpret a 90% confidence interval for the mean additional amount that waterfront property is worth.

non-waterfront prop:----------- Waterfront Prop:

n=106 ----------------------------------n=75

= 218,582.16-------------------- -- =320,525.36

s=102,708.88 ------------------------s=161,222.29

The 90% confidence interval is _ ± _

A student took a random sample of 181 properties to examine how much more waterfront property is worth. Her summaries and boxplots of the two groups of prices are shown. Construct and interpret a 90% confidence interval for the mean additional amount that waterfront property is worth.

non-waterfront prop:----------- Waterfront Prop:

Explanation / Answer

x1(bar) 320525.36 x2(bar) 218582.16 s1 161222.29 s2 102708.88 n1 75 n2 106 SE = sqrt[ (s12/n1) + (s22/n2) ] (s12/n1) 346568357.2379 (s22/n2) 99519943.6873 SE 21120.8026 DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] } [ (s12 / n1)2 / (n1 - 1) ] 1623103057278260.000 [ (s22 / n2)2 / (n2 - 1) ] 94325897062138.00 (s12/n1 + s22/n2)2 198994772222356000.00 DF = 116

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