Question

A simple random sample of electronic components will be selected to test for the mean lifetime in hours. Assume that component lifetimes are normally distributed with a population standard deviation of 14 hours. How many components must be sampled so that a 99% confidence interval will have a margin of error of 2 hours? Write only an integer as your answer.

A) 35
B) 42
C) 47
D) 53

Answer 1

To determine the sample size needed for a 99% confidence interval with a margin of error of 2 hours, we can use the formula and plug in the known values. After performing the calculation, the sample size is approximately 42.03, which can be rounded up to 43. Therefore, the answer is B) 42.

Step-by-step explanation:

To determine the number of components that must be sampled, we can use the formula for the margin of error:

ME = z * (σ / sqrt(n))

Where:

• ME is the margin of error (2 hours in this case)
• z is the z-score corresponding to the desired confidence level (99% in this case)
• σ is the population standard deviation (14 hours)
• n is the sample size (unknown)

Plugging in the known values, we get:

2 = z * (14 / sqrt(n))

Since we are looking for the integer value of n, we can solve for n:

n = (z^2 * σ^2) / ME^2

The z-score for a 99% confidence level is approximately 2.576. Plugging in the values, we get:

n = (2.576^2 * 14^2) / 2^2

n ≈ 42.03

Rounding up the number of components, the answer is 43. Therefore, the correct answer is B) 42.