Question

find the range variance and standard deviation for each of the two samples, then compare the two set of test results.

when investigating times required for drive theu servuce, the following reaults were obtained.

restaurant a 120, 123, 153, 128, 124, 118, 154, 110

restaurant b : 115, 126 147, 156, 118, 110, 145, 137

Answer 1

restaurant a

X	(X - X̄)²
120	76.56
123	33.06
153	588.06
128	0.56
124	22.56
118	115.56
154	637.563
110	351.563

	X	(X - X̄)²
total sum	1030	1825.50
n	8	8

mean =    ΣX/n =    1030.000   /   8   =   128.7500

maximum =    154              
minimum=   110              
                  
range=max-min =    154   -   110   =   44

                      
sample variance =    Σ(X - X̄)²/(n-1)=   1825.5000   /   7   =   260.786
                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (1825.5/7)   =       16.1489

restaurant b:

X	(X - X̄)²
115	280.56
126	33.06
147	232.56
156	588.06
118	189.06
110	473.06
145	175.563
137	27.563

	X	(X - X̄)²
total sum	1054	1999.50
n	8	8

mean =    ΣX/n =    1054.000   /   8   =   131.7500
                      
sample variance =    Σ(X - X̄)²/(n-1)=   1999.5000   /   7   =   285.643
                      
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   √   (1999.5/7)   =       16.9010

maximum =    156              
minimum=   110              
                  
range=max-min =    156   -   110   =   46

Conclusiom: As we can see range, varinace and standard deviation for restaurant A is less than restaurant B. Hence Restaurant A data is less variable and may be need more data point for restaurant B to compare the other statistics.

Thanks in advance!

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