Researchers conducted a survey among 1793 families of a certain city who reported that they participated in charity one or more times in 2016. The sample was selected in a way to be representative for families of a certain city who participated in charity in 2016. For this sample, the mean amount of money spent on charity was $876. Suppose that the sample standard deviation was $514. Find the margin of error and construct a 95% confidence interval for the mean amount of money spent in the certain city on charity in 2016.

Question

Quizplus · Accepted Answer

The margin of error for a 95% confidence interval can be calculated using the formula $ME = z \times \frac{\sigma}{\sqrt{n}}$, where $z$ is the z-score corresponding to the confidence level (approximately 1.96 for 95%), $\sigma$ is the sample standard deviation, and $n$ is the sample size. Plugging in the given values: $ME = 1.96 \times \frac{514}{\sqrt{1793}} \approx 23.79$. The 95% confidence interval is then calculated as $mean \pm ME = 876 \pm 23.79$, which gives us (852.21, 899.79).

Researchers Conducted a Survey Among 1793 Families of a Certain