Question

(1). Find the variance and standard deviation of the response time data. Treat it as a sample from a larger population. [0.12,0.3, 0.35,0.37,0.44,0.57,0.61,0.62,0.71,0.8,0.88,1.02,1.08,1.12,1.13,1.17,1.21,1.23,1.35,1.41,1.42,1.42,1.46,1.5,1.52,1.54,1.6,1.61,1.68,1.72,1.86,1.9,1.91,2.07,2.09,2.16,2.17,2.2,2.29,2.32,2.39,2.47,2.6,2.86,3.43,3.43,3.77,3.97,4.54,4.73]

(2). Find the interquartile range and the median absolute deviation for the response time data.

(3). In the response time data, replace the value X40=2.32 by 232.0. Recalculate the standard deviation, the interquartile range and the median absolute deviation and compare with the answers from problems 1 and

Answer 1

1. Mean for the given data is

Create the following table.

data	data-mean	(data - mean)2
0.12	-1.6224	2.63218176
0.3	-1.4424	2.08051776
0.35	-1.3924	1.93877776
0.37	-1.3724	1.88348176
0.44	-1.3024	1.69624576
0.57	-1.1724	1.37452176
0.61	-1.1324	1.28232976
0.62	-1.1224	1.25978176
0.71	-1.0324	1.06584976
0.8	-0.9424	0.88811776
0.88	-0.8624	0.74373376
1.02	-0.7224	0.52186176
1.08	-0.6624	0.43877376
1.12	-0.6224	0.38738176
1.13	-0.6124	0.37503376
1.17	-0.5724	0.32764176
1.21	-0.5324	0.28344976
1.23	-0.5124	0.26255376
1.35	-0.3924	0.15397776
1.41	-0.3324	0.11048976
1.42	-0.3224	0.10394176
1.42	-0.3224	0.10394176
1.46	-0.2824	0.07974976
1.5	-0.2424	0.05875776
1.52	-0.2224	0.04946176
1.54	-0.2024	0.04096576
1.6	-0.1424	0.02027776
1.61	-0.1324	0.01752976
1.68	-0.0624	0.00389376
1.72	-0.0224	0.00050176
1.86	0.1176	0.01382976
1.9	0.1576	0.02483776
1.91	0.1676	0.02808976
2.07	0.3276	0.10732176
2.09	0.3476	0.12082576
2.16	0.4176	0.17438976
2.17	0.4276	0.18284176
2.2	0.4576	0.20939776
2.29	0.5476	0.29986576
2.32	0.5776	0.33362176
2.39	0.6476	0.41938576
2.47	0.7276	0.52940176
2.6	0.8576	0.73547776
2.86	1.1176	1.24902976
3.43	1.6876	2.84799376
3.43	1.6876	2.84799376
3.77	2.0276	4.11116176
3.97	2.2276	4.96220176
4.54	2.7976	7.82656576
4.73	2.9876	8.92575376

Find the sum of numbers in the last column to get.

Calculate variance

b.  The first quartile (or lower quartile or 25th percentile) is the median of the bottom half of the numbers. So, to find the first quartile, we need to place the numbers in value order and find the bottom half.

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.32   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   

So, the bottom half is

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   

The median of these numbers is 1.08.

The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.32   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   

As you can see, we do not have just one middle number but we have a pair of middle numbers, so the median is the average of these two numbers:

Median=

The third quartile (or upper quartile or 75th percentile) is the median of the upper half of the numbers. So, to find the third quartile, we need to place the numbers in value order and find the upper half.

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.32   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   

So, the upper half is

1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.32   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   

The median of these numbers is 2.2.

The interquartile range is the difference between the third and first quartiles.

The third quartile is 2.2.

The first quartile is 1.08.

The interquartile range = 2.2 - 1.08 = 1.12.

Mean Absolute Deviation

c. Replacing the value X40=2.32 by 232.0

Create the following table.

data	data-mean	(data - mean)2
0.12	-6.216	38.638656
0.3	-6.036	36.433296
0.35	-5.986	35.832196
0.37	-5.966	35.593156
0.44	-5.896	34.762816
0.57	-5.766	33.246756
0.61	-5.726	32.787076
0.62	-5.716	32.672656
0.71	-5.626	31.651876
0.8	-5.536	30.647296
0.88	-5.456	29.767936
1.02	-5.316	28.259856
1.08	-5.256	27.625536
1.12	-5.216	27.206656
1.13	-5.206	27.102436
1.17	-5.166	26.687556
1.21	-5.126	26.275876
1.23	-5.106	26.071236
1.35	-4.986	24.860196
1.41	-4.926	24.265476
1.42	-4.916	24.167056
1.42	-4.916	24.167056
1.46	-4.876	23.775376
1.5	-4.836	23.386896
1.52	-4.816	23.193856
1.54	-4.796	23.001616
1.6	-4.736	22.429696
1.61	-4.726	22.335076
1.68	-4.656	21.678336
1.72	-4.616	21.307456
1.86	-4.476	20.034576
1.9	-4.436	19.678096
1.91	-4.426	19.589476
2.07	-4.266	18.198756
2.09	-4.246	18.028516
2.16	-4.176	17.438976
2.17	-4.166	17.355556
2.2	-4.136	17.106496
2.29	-4.046	16.370116
232	225.664	50924.240896
2.39	-3.946	15.570916
2.47	-3.866	14.945956
2.6	-3.736	13.957696
2.86	-3.476	12.082576
3.43	-2.906	8.444836
3.43	-2.906	8.444836
3.77	-2.566	6.584356
3.97	-2.366	5.597956
4.54	-1.796	3.225616
4.73	-1.606	2.579236

The first quartile (or lower quartile or 25th percentile) is the median of the bottom half of the numbers. So, to find the first quartile, we need to place the numbers in value order and find the bottom half.

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   232   

So, the bottom half is

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   

The median of these numbers is 1.08.

The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   232   

As you can see, we do not have just one middle number but we have a pair of middle numbers, so the median is the average of these two numbers:

Median=

The third quartile (or upper quartile or 75th percentile) is the median of the upper half of the numbers. So, to find the third quartile, we need to place the numbers in value order and find the upper half.

0.12   0.3   0.35   0.37   0.44   0.57   0.61   0.62   0.71   0.8   0.88   1.02   1.08   1.12   1.13   1.17   1.21   1.23   1.35   1.41   1.42   1.42   1.46   1.5   1.52   1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   232   

So, the upper half is

1.54   1.6   1.61   1.68   1.72   1.86   1.9   1.91   2.07   2.09   2.16   2.17   2.2   2.29   2.39   2.47   2.6   2.86   3.43   3.43   3.77   3.97   4.54   4.73   232   

The median of these numbers is 2.2.

The interquartile range is the difference between the third and first quartiles.

The third quartile is 2.2.

The first quartile is 1.08.

The interquartile range = 2.2 - 1.08 = 1.12.

Median absolute deviation is

We can see thqat adding outlier will increase the variation abmong the variables.