Question

3. A sample of 25 freshmen and senior students at a major university completed a survey that extent of satisfaction with the parking situation on the campus (the higher the score, the greater the satisfaction).

Freshmen

11              43                30                30                   45 41              12                40                42                   35       45              25                10                33                   50 42              32                38                11                   47     22              26                37                38                   10

Seniors

10             45               35                  28                  52 35             12               50                  40                  30 40             10               10                  38                  12 40             15               30                  20                  43 23             25               30                  40                  12

Compute the median, mean and standard deviation scores for these data. Based on the measures of central tendency and dispersion, describe the change in the satisfaction with the parking service between freshman and seniors. Make sure that you make a reference to the possible shape of the underlying distributions for the two groups and the change in them over the period of four years?                                                       5 Points

Answer 1

freshman

X	(X - X̄)²
11	432.640
30	3.240
41	84.640
42	104.040
45	174.240
33	1.440
42	104.040
11	432.640
22	96.040
38	38.440
43	125.440
45	174.240
12	392.040
35	10.240
25	46.240
50	331.240
32	0.040
47	231.040
26	33.640
10	475.240
30	3.240
40	67.240
10	475.240
38	38.440
37	27.040

	X	(X - X̄)²
total sum	795	3902
n	25	25

mean =    ΣX/n =    31.8
            
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   12.7508

Median=0.5(n+1)th value =    13 th value from ascending order=   35  

-------------------------------------------------

senior

X	(X - X̄)²
10	361.000
28	1.000
35	36.000
40	121.000
40	121.000
38	81.000
40	121.000
20	81.000
23	36.000
40	121.000
45	256.000
52	529.000
12	289.000
30	1.000
10	361.000
12	289.000
15	196.000
43	196.000
25	16.000
12	289.000
35	36.000
50	441.000
10	361.000
30	1.000
30	1.000

	X	(X - X̄)²
total sum	725	4342
n	25	25

mean =    ΣX/n =    29
            
sample std dev =   √ [ Σ(X - X̄)²/(n-1)] =   13.4505

Median=0.5(n+1)th value =    13   th value=   30  

__________________________________________________________________________________

freshman		senior
Mean	31.8	Mean	29
Standard Error	2.550163	Standard Error	2.690105
Median	35	Median	30
Mode	11	Mode	40
Standard Deviation	12.75082	Standard Deviation	13.45053
Sample Variance	162.5833	Sample Variance	180.9167
Kurtosis	-0.84554	Kurtosis	-1.23346
Skewness	-0.61812	Skewness	-0.08029
Range	40	Range	42
Minimum	10	Minimum	10
Maximum	50	Maximum	52
Sum	795	Sum	725
Count	25	Count	25

in frshman data, mean is smaller than median , so data is mild skew to left

here, median will represent center of measure of data

in junior data, mean is approximately equal to median , so mean will represent center of measure of data

dispersion of data is less in case of freshmen as compared to junior , because std dev is less in freshman data