Question

A researcher wishes to estimate the mean amount of money spent per month on food by households in a certain neighborhood. She desires a margin of error of $30. Past studies suggest that a population standard deviation of $248 is reasonable. Estimate the minimum sample size needed to estimate the population mean with the stated accuracy.A. 274B. 284C. 264D. 272

Answer 1

Answer:

C. 264

Step-by-step explanation:

Assuming the random variable for amount of money spent per month on food is normal distributed and the researcher wants to estimate the mean with $95\%$ confidence. Then we can use the following formula to determine the minimum sample size:

$n_0 = \bigg(\frac{z_(\frac{\alpha}{2})S_d}{e} \bigg)^2$

Where $S_d$ is the standard deviation and $z_{(\frac{\alpha}{2})}$ is the quantile of the normal distribution with an area of $\frac{\alpha}{2}$.

$n_0 = \big(\frac{z_{(0.025)}\times248}{30} \big)^2 = \big(\frac{1.96\times248}{30} \big)^2 \approx 263$

So we need at least 264 households to sample.