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Answer:
The Pooled Standard Deviation for these data sets is 1.09
Step-by-step explanation:
Given that;
Standard Deviation 1 = 1.11
Standard Deviation 2 = 0.99
Standard Deviation 3 = 1.15
pooled standard deviation is given as;
= (n1 - 1 )s1² + ..... + ( nk - 1 )sk² / n1 + ..... + nk - 1
where n1, n2....., nk are the sample sizes of k groups and s1, s2, ....., sk are the standard deviations of k groups respectively
so for the 3 groups;
² = (n1 - 1 )s1² + (n2 - 1 )s2² + ( n3 - 1)s3² / n1 + n2 + n3 - 3
we substitute
² = (5 - 1 )1.11² + (5 - 1 )0.99 ² + ( 5 - 1)1.15² / 5 + 5 + 5 - 3
² = (4.9284 + 3.9204 + 5.29) / 12
² = 14.1388 / 12
² = 1.1782
= √1.1782
= 1.08544 ≈ 1.09
Therefore, The Pooled Standard Deviation for these data sets is 1.09