Question
(a) What sample size would be needed to estimate the true proportion of American households that own more than one DVD player, with 90 percent confidence and an error of $\pm$ 0.02 ? (b) What sampling method would you recommend? Why?
Step 1
The formula to estimate the sample size \( n \) needed to achieve a certain margin of error \( E \) at a given confidence level for a proportion is: \[ n = \left(\frac{Z^2 \times p \times (1-p)}{E^2}\right) \] where: - \( Z \) is the z-score corresponding to the Show more…
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