An experimenter wishes to obtain a confidence interval for the difference betwee

Question

An experimenter wishes to obtain a confidence interval for the difference between true average breaking strength for cables manufactured by company A and company B. Suppose breaking strength is normally distributed for both types of cable with sigma_A 30 psi and sigma_B = 20 psi. If costs dictate that sample size for type A cable should be three times the sample size for the type B cable, how many observations are required if the length of 99% confidence interval is to be at most 20 psi? Suppose 400 observations are to be made. How many of the observations should be made on type A cable samples if the length of the resulting 99% confidence interval is to be minimum?

Explanation / Answer

1.

1.

z – std deviation

nA = (Z(a/2)*zA)^2/Error

Substitute values = 299.5 = 300.

nB = (Z(a/2)*zB)^2/Error

Substitute values = 133.

Given nA = 3*nB =

Take nA = 340

2.

Total observation = 400 .

Given nA = 3*nB

There fore , nA = 300 ( approximately )

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