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Builder A and Builder B both take random samples of nails, and the width of the nails are measured in centimetres - Builder A estimates that a 95% confidence interval for the width of the nails is 1.2 +/- 1.96 * 0.05. Builder B estimates that a 95% confidence interval for the width of the nails is 1.2 +/- 1.96 * 0.005. Neither builder had knowledge of the population variance of the widths of the nauks, but the variance of the samples are equal. Assuming that no errors in calculations are made, find the size of Builder As sample and Builder Bs sample.

It's a MCQ - with options as follows:

A) Builder A took a sample of size n = 40, and Builder B took a sample of size n = 4000

B) Builder A took a sample of size n = 4000, and Builder B took a sample of size n = 40

C) Builder A took a sample of size n = 10, and Builder B took a sample of size n = 1000

D) Builder A took a sample of size n = 1000, and Builder B took a sample of size n = 10

E) Builder A took a sample of size n = 100, and Builder B took a sample of size n = 100.

Since the variances are equal, I thought that maybe s/rootx would be equal to s/rooty, for x equal to the size of Builder As sample and y equal to the size of Builder Bs sample, but this turned out to be wrong - since this method yields two possible answers.

Any help is appreciated.