Question

Given the following data [92, 100, 106, 110, 114, 125, 115, 110, 108, 104, 94, 84], Find the average and standard deviation of the useful data after you apply the modified Thompson technique

Answer 1

Iteration I

Step 1:

Data: 84,92,94,100,104,106,108,110,110,114,115,125

Sample mean =

105.2727

Step 2:

min(x)- 84 105.2727| 21.2727

and,

таz(?) — а 125 105.2727| 19.7273

Since 21.2727 is greater, so we test 84 for being an outlier.

Step 3:

Using Tau table, for sample size 12,

T1.829

And, sample standard deviation =

>(zi-피)2=1.26613 n 1

Step 4:

Now,

T X S1.829 x 11.26613=20.60575

Now, Absolute difference 21.2727 > 20.60575, this point is an outlier and so we remove the observation 84.

Iteration II

Step 1:

Data: 92,94,100,104,106,108,110,110,114,115,125

Sample mean =

107.4 n

Step 2:

min(x)- 92 107.4 15.4

and, $|max(x)-\overline{x}|=|125-107.4|=17.6$

Since 17.6 is greater, so we test 125 for being an outlier.

Step 3:

Using Tau table, for sample size 11,

T=1.8153 7

And, sample standard deviation = $s=\sqrt{\frac{1}{n-1}\sum (x_{i}-\overline{x})^{2}}=9.4784$

Step 4:

Now,

T X S1.8153 x 9.4784 17.2061

Now, Absolute difference 17.6 > 17.2061, this point is an outlier and so we remove the observation 125.

Iteration III

Step 1:

Data: 92,94,100,104,106,108,110,110,114,115

Sample mean = $\frac{\sum x}{n}=105.4444$

Step 2:

$|min(x)-\overline{x}|=|92-105.4444|=13.4444$

and,

таz(?) — а 115 105.4444 9.5556

Since 13.4444 is greater, so we test 92 for being an outlier.

Step 3:

Using Tau table, for sample size 10, $\tau=1.7984$

And, sample standard deviation =

>(2i-피)2= 7.8473 n 1

Step 4:

Now, $\tau \times s=1.7984 \times 7.8473=14.1126$

Now, Absolute difference 13.4444 < 14.1126, this point is not an outlier and so we do not remove any observation.

Final data:

Data: 92,94,100,104,106,108,110,110,114,115

Sample mean = $\frac{\sum x}{n}=105.4444$

And, Sample standard deviation =

>(2i-피)2= 7.8473 n 1