13. les of size n 4 are given 316 .315 315 317 314 316 314 314 318 .312 .315 314

Question

13. les of size n 4 are given 316 .315 315 317 314 316 314 314 318 .312 .315 314 .315 .315 .315 a. Code the measurements using the base .315 b. Find the average measurement using the coded values. c. Check the answer by averaging the measurements d. Find the five range values, R, from the coded values and then from the measurements e. Using 2.059 ô from the coded val find f. Code the measurements using the base .310 and repeat (b), (d), and (e) ues and then from the measurements 14. Given the samples in Figure 7.27

Explanation / Answer

3. (a) To code the measurements using the base 0.315, let us consider a transformation u=(x-0.315)*1000; where u is the new score or code and x is the previous score as given in the problem. Then, new values of the measurements are

-1 -3 1 2 -2

3 0 0 -1 0

-3 -1 0 1 0

0 -1 1 -1 -3

(b) Average of measurements using the coded values are as follows:

Sample: 1 2 3 4 5

Mean(u) : -1/4 -5/4 1/2 1/4 -5/4

Mean(x): 0.31475 0.31375 0.3155 0.31525 0.31375

= Mean(u)/1000 + 0.315:

(c) Average of the sample observations directly are listed below:

Sampe: 1 2 3 4 5

Mean: 0.31475 0.31375 0.3155 0.31525 0.31375

(d) Ranges from the 5 samples are given in the following table for u, then x and x itself:

Sample: 1 2 3 4 5

Range(u): 6 3 1 3 3

Range(x): 0.006 0.003 0.001 0.003 0.003 [ Range(x) = Range(u)/1000 ]

R(x) from data: 0.006 0.003 0.001 0.003 0.003.

(e) R_bar (u) = (6+3+1+3+3)/5 = 3.2. Hence R_bar (x) = 3.2/1000 = 0.0032. Therefore, sigma = 0.0032/2.059 = 0.001554.

(f) New coded values (v) are listed below:

4 2 6 7 3

8 5 5 4 5

2 4 5 6 5

5 4 6 4 2

Average of measurements using the coded values are as follows:

Sample: 1 2 3 4 5

Mean(u) : 19/4 15/4 22/4 21/4 15/4

Mean(x): 0.31475 0.31375 0.3155 0.31525 0.31375

= Mean(u)/1000 + 0.310

  

Ranges from the 5 samples are given in the following table for u, then x and x itself:

Sample: 1 2 3 4 5

Range(u): 6 3 1 3 3

Range(x): 0.006 0.003 0.001 0.003 0.003 [ Range(x) = Range(u)/1000 ]

R(x) from data: 0.006 0.003 0.001 0.003 0.003.

R_bar (u) = (6+3+1+3+3)/5 = 3.2. Hence R_bar (x) = 3.2/1000 = 0.0032. Therefore, sigma = 0.0032/2.059 = 0.001554.

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