Solutions for Tutorial Exercise 1NOT TO BE DISTRIBUTED

January 20, 2015

1. The sum of the observations is 1 + 7 + 4 + 9 + 5 + 3 + 6 = 35, and thereare 7 observations, so that x = 5. The sum of squares of the observationsis

12 + 72 + 42 + 92 + 52 + 32 + 62 = 217, (1)

so that (using the short cut formula

s2 =1

n

nX

i=1

x2i nx2!

,

s2 = (1/7)217 7 52 = 6

Ordering the observations into increasing order,

1, 3, 4, 5, 6, 7, 9 (2)

the middle observation is the sample median = 5.

2. A small percentage of very expensive homes makes the mean large, butdoes not aect the median much.

3. the upper quartile

4. The InterQuartile Range is the dierence between p75 and p25, so thatIQR = 5.887002 5.193727 = 0.693275.

5. All the observations in the sample are equal. The sample variance isdefined as the average of the squared mean deviations,

s2 =1

n

nXi=1

(xi x)2 . (3)

To make this zero, each mean deviation has to be zero, so each observationmust equal the sample mean - this is choice 4.

6. Solutions for 6

1

(a) From the description as the ages in years, it is count data.(b) 17.(c) 444.

(d) x =1

n

nPi=1

xi = 444/17 = 26.11765.

(e) The sum of the mean deviations isnP

i=1(xi x) =

nPi=1

xinx = 44717 26.11765 = 447 447 = 0. This is a result which is true ingeneral, not just in this example.

(f)nP

i=1x2i = 12234.

(g) s2 =1

17

12234 17 26.117652 = 1

17(12234 17 682.13164) =

1

17. (12234 11596.23791) = 1

17637.76209 = 37.51542. The alternate

version with divisor n 1 = 16 here is (637.76209) /16 = 39.86013(h) The sample standard deviation is either

ps2 =

p37.51542 = 6.12498

orp39.86013 = 6.31349.

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