Question

using the following data, show all your work:

0.1329, 0.1347, 0.1328, 0.1325, 0.1329, 0.1331, 0.1323

1. use q test to find any value that must e rejected

2. find the mean of the values accepted by q test

3. find the standard deviation

4. find the 99% confidence limit

Answer 1

(1)

Arranging data in ascending order we get:

0.1323 , 0.1325, 0.1328, 0.1329, 0.1329, 0.1331, 0.1347

outlier in question = 0.1347

gap = absolute difference between the outlier in question and the closest number to it

    = 0.1347 - 0.1331 = 0.0016

Range = Maximum - Minimum = 0.1347 - 0.1323 = 0.0024

Q = 0.0016/0.0024 = 0.6667

From Table, for =0.05 and n = 7, we get:

Q_table = 0.568

Since Q > Q_table, we reject the questionable point = 0.1347.

So, Answer is:

0.1347 must be rejected.

(2)

Values accepted by Q test:

0.1323 , 0.1325, 0.1328, 0.1329, 0.1329, 0.1331

Mean () is given by:

(3)

Standard Deviation (s) is got as follows:

x	x -	(x- )2
0.1323	-0.00045	0.0000002025
0.1325	-0.00025	0.00000006250
0.1328	0.00005	0.00000000250
0.1329	0.00015	0.00000002250
0.1329	0.00015	0.00000002250
0.1331	0.00035	0.00000012250
Total =		0.00000043500

Standard Deviation (s) is given by:

(4)

n = 6

= 0.13275

s = 0.0002950

= 0.01

df = 6 - 1 = 5

From Table, critical values of t = 4.032

Confidence Interval:

= ( 0.1322644     ,0.1332356)

So,

Answer is:

( 0.1322644     ,0.1332356)