A sample of 125 pieces of yarn had mean breaking strength 6.1 N and standard dev

Question

A sample of 125 pieces of yarn had mean breaking strength 6.1 N and standard deviation 0.9 N. A new batch of yarn was made, using new raw materials from a different vendor. In a sample of 75 pieces of yarn from the new batch, the mean breaking strength was 5.8 N and the standard deviation was 1.5 N. Additional pieces of yarn will be sampled in order to improve the precision of the confidence interval. Which would increase the precision the most: sampling 50 additional pieces of yarn from the old batch, 50 additional pieces from the new batch, or 25 additional pieces from each batch? Sampling 50 additional pieces from the new batch Sampling 50 additional pieces of yarn from the old batch Sampling 25 additional pieces from each batch

Explanation / Answer

the standard error is a measure of precision

sampling 50 pieces from new batch -

standard error = s * sqrt(n) = 1.5*sqrt(50) = 10.6066

standard error of 50 pieces from old batch - s*sqrt(n) = 0.9*sqrt(50) = 6.36396

standard error of 25 pieces from each batch = sqrt( s1^2*n1 + s2^2 *n2)

= sqrt(1.5^2 *25 +0.9^2 *25) = 8.7464

since 6.36396 is the minimum of standard error ,

hence

sampling 50 additional pieces of yarn from the old batch is correct answer

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