Finding sample sizes from confidence interval margins of error

[Absolut]

I*m looking for some help with a statistics question:
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.

 

[OlderDan]

You are close. The s/sqrt(n_x) and s/sqrt(n_y) are not equal, but they are the numbers that multiply the 1.96 factor in each case

 

[thepatientmental]

The correct answer is D) Builder A took a sample of size n = 1000, and Builder B took a sample of size n = 10.

To understand why this is the correct answer, let's first break down the information given in the question. Both Builder A and Builder B have taken random samples of nails and measured their width in centimeters. They both estimate a 95% confidence interval for the width of the nails, with Builder A's interval being 1.2 +/- 1.96 * 0.05 and Builder B's interval being 1.2 +/- 1.96 * 0.005. This means that Builder A's estimate has a margin of error of 0.05 cm and Builder B's estimate has a margin of error of 0.005 cm.

Now, let's look at the options given. A) Builder A took a sample of size n = 40, and Builder B took a sample of size n = 4000. This cannot be the correct answer because the margin of error for Builder A's estimate is much larger than the margin of error for Builder B's estimate. This would mean that Builder A's sample size should be smaller, not larger, than Builder B's sample size.

B) Builder A took a sample of size n = 4000, and Builder B took a sample of size n = 40. This option also cannot be correct for the same reason as option A.

C) Builder A took a sample of size n = 10, and Builder B took a sample of size n = 1000. This option also cannot be correct because the margin of error for Builder A's estimate is much larger than the margin of error for Builder B's estimate. This would mean that Builder A's sample size should be smaller, not larger, than Builder B's sample size.

E) Builder A took a sample of size n = 100, and Builder B took a sample of size n = 100. This option also cannot be correct because the margin of error for Builder A's estimate is the same as the margin of error for Builder B's estimate. This would mean that both samples should have the same size, which is not the case according to the given information.

This leaves us with option D) Builder A took a sample of size n = 1000, and Builder B took a sample of size n = 10. This is the correct answer because the