You want to investigate the rental fees. You gather the data shown in the table below. Area A represents the area of the city where tenants are unhappy about their monthly rents. The data represent the monthly rents paid by a random sample of tenants in Area A and three other areas of similar size. Assume all the apartments represented are approximately the same size with the same amenities.
Area A |
Area B |
Area C |
Area D |
1275 |
1124 |
1085 |
928 |
1110 |
954 |
827 |
1096 |
975 |
815 |
793 |
862 |
862 |
1078 |
1170 |
735 |
1040 |
843 |
919 |
798 |
997 |
745 |
943 |
812 |
1119 |
796 |
756 |
1232 |
908 |
816 |
765 |
1036 |
890 |
938 |
809 |
998 |
1055 |
1082 |
1020 |
914 |
860 |
750 |
710 |
1005 |
975 |
703 |
775 |
930 |
Solution :
a)
For measure of central tendency we use mean as a best measure and standard deviation is the best measure for measures of dispersion for this data set.
b)
Area | A | B | C | D |
Mean | 1005.5 | 887 | 881 | 945.5 |
Standard Deviation | 123.0724 | 144.9125 | 146.2091 | 138.7014 |
c)
Yes. We can draw mean-sd chart for explaination:
d)
Area | A | B | C | D |
Mean | 1005.5 | 887 | 881 | 945.5 |
Standard Deviation | 123.0724 | 144.9125 | 146.2091 | 138.7014 |
CV(%) | 0.12 | 0.16 | 0.17 | 0.15 |
Since mean of Area A is largest and CV (coefficient of variation) of Area is smallest so the complaints in Area A are legitimate.
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