Question

A tablet PC manufacturer wishes to estimate the proportion of people who want to purchase tablet PCs which cost more than $700. Find the required sample size to yield a 90% confidence interval whose length is below 0.04.

Answer 1

Answer 2

To find the required sample size for estimating the proportion of people who want to purchase tablet PCs costing more than $700, we can use the formula for the sample size of a proportion with a specified margin of error.

The formula for the sample size (n) needed to estimate a proportion with a desired margin of error (E) is given by:

n = (Z^2 * p * (1 - p)) / E^2

Where: Z is the Z-score corresponding to the desired confidence level (90% confidence level corresponds to a Z-score of approximately 1.645) p is the estimated proportion (we don't know this yet, so we use 0.5 as a conservative estimate since 0.5 gives the largest sample size) E is the desired margin of error (0.04 in this case)

Let's calculate the sample size:

n = (1.645^2 * 0.5 * (1 - 0.5)) / 0.04^2

n = (2.7056 * 0.25) / 0.0016

n = 0.6764 / 0.0016

n ≈ 422.75

Since we cannot have a fractional sample size, we round up the sample size to ensure the desired margin of error is achieved. Therefore, the required sample size is 423.