Question

The May 1, 2009 issue of The Montclarian reported the following home sale
amounts for a sample of homes in Alameda, CA that were sold the previous
month (1000s of $): 590 815 575 608 350 1285 408 540 555 679
a) Calculate and interpret the samplemean and median.
b) Suppose the 6th observation had been 985 rather than
1285. How would themean and median change?
c) Calculate a 20% trimmed mean by first trimming the two
smallest and two largest observations.
d) Calculate a 15% trimmed mean

Answer 1

350, 408, 540, 555, 575, 590, 608, 679, 815, 1285

n = 10

a) $\bar x=\frac{1}{n}\sum x$

$\bar x=\frac{590+815+575+608+350+1285+408+540+555+679}{10}=640.5$

$median=\frac{575+590}{2}=582.5$

Both the mean of 640.5 and the median of 582.5 indicate the central tendency, thus the typical home sale amount is about 640.

b) $\bar x=\frac{590+815+575+608+350+985+408+540+555+679}{10}=610.5$

the median wouldn't change $median=\frac{575+590}{2}=582.5$

Outliers affect the median less than they affect the mean.

c) $\bar x_{.20}=\frac{540+555+575+590+608+679}{6}=591.2$

d) $\alpha=0.15, \ n=10,\ k=n\alpha=1.5$

$R=n-2k=10-2\cdot1.5=7$

Since integer part of k is 1, we throw out the smallest 350 and the highest 1285. k=1.5 has fractional part 0.5, so we throw out only 0.5 part of 408 and 0.5 part of the next largest 815.

$\bar x_{.15}=\frac{0.5\cdot408+540+555+575+590+608+679+0.5\cdot815}{7}=594.1$