In: Statistics and Probability
The following data represent the daily hotel cost and rental car cost for 20 U.S cities during a week in October 2003
CITY HOTEL CARS
San Francisco 205 47
Los Angeles 179 41
Seattle 185 49
Phoenix 210 38
Denver 128 32
Dallas 145 48
Houston 177 49
Minneapolis 117 41
Chicago 221 56
St. Louis 159 41
New Orleans 205 50
Detroit 128 32
Cleveland 165 34
Atlanta 180 46
Orlando 198 41
Miami 158 40
Pittsburgh 132 39
Boston 283 67
New York 269 69
Washington DC 204 40
FOR EACH VARIABLE ( hotel cost and car cost)
a. Compute the mean, median, first quartile, and third quartile)
b. Compute the variance, standard deviation, range, interquartile range, coefficient of Variation
c. Are the data skewed? If so, how?
d. Base don’t he results a) through c), what conclusions can you reach concerning the daily costs of a hotel and rental car
Using excel we get an output as ,
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c. Are the data skewed? If so, how?
Note : If the mean is greater than the median, the distribution is positively skewed.
If the mean is less than the median, the distribution is negatively skewed.
For Hotel cost data : mean = 185.4 , median =179.5
Mean > Median .
Since Hotel cost data is Positive Skewed .
For Car cost : Mean = 43.5 , median =41.
Mean > median.
Since car cost data also positively skewed.
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Here you can also sue TI-83/84 calculator .
First plug both data in calculation in L1 and L2 , STAT >>> Edit , then you will get L1,L2 .. column .
Plug values in this column .
Thne use command as , STAT >>> CALC >>> 1 : 1-Var Stats.
First select L1 column then hit enter on calculate, you will get answer for Hotel cost data .
similarly run command again , then use cars cost data .