Question

In: Math

Assume the branch manager is not satisfied with the widths of the obtained confidence intervals, and...

  1. Assume the branch manager is not satisfied with the widths of the obtained confidence intervals, and she requests estimates of the mean selling price of Gulf View condominiums with a margin of error of $40,000 and the mean selling price of No Gulf View condominiums with a margin of error of $15,000. Using 95% confidence, how should the sample sizes be?
    Gulf View Condominiums No Gulf View Condominiums
    List Price Sale Price List Price Sale Price
    495.0 475.0 227.0 227.0
    379.0 350.0 158.0 145.5
    529.0 519.0 196.5 189.0
    552.5 534.5 249.0 240.0
    334.9 334.9 289.0 277.5
    550.0 505.0 225.0 224.0
    169.9 165.0 289.0 269.0
    210.0 210.0 189.9 186.5
    975.0 945.0 159.9 154.9
    314.0 314.0 245.0 240.0
    315.0 305.0 209.8 202.0
    885.0 800.0 220.0 205.0
    975.0 975.0 236.0 222.0
    469.0 445.0 159.9 156.5
    329.0 305.0 170.0 170.0
    365.0 330.0 332.0 302.5
    332.0 312.0 197.5 189.0
    520.0 495.0 257.0 237.0
    425.0 405.0
    675.0 669.0
    409.0 400.0
    649.0 649.0
    319.0 305.0
    425.0 410.0
    359.0 340.0
    469.0 449.0
    895.0 875.0
    439.0 430.0
    435.0 400.0
    235.0 227.0
    638.0 618.0
    629.0 600.0
    329.0 309.0
    595.0 555.0
    339.0 315.0
    215.0 200.0
    395.0 375.0
    449.0 425.0
    499.0 465.0
    439.0 428.5

Solutions

Expert Solution

Gulf View Condominiums

I am assuming that sales price are in thousands

std dev for gulf view sale price=192.51775

Standard Deviation ,   =    192.5177534
sampling error ,    E =   40
Confidence Level ,   CL=   95%
      
alpha =   1-CL =   5%
Z value =    Zα/2 =    1.9600
      
Sample Size,n =    (Z*std dev/ E)² =   88.9852
      
      
So,Sample Size needed=       89

---------------------------------------------------------

No Gulf View Condominiums

Standard Deviation ,   =    43.89172358
sampling error ,    E =   15
Confidence Level ,   CL=   95%
      
alpha =   1-CL =   5%
Z value =    Zα/2 =    1.9600
      
Sample Size,n =    (Z*std dev / E)² =   32.8911
      
      
So,Sample Size needed=       33



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