Find the margin of error at a 95% confidence level.

Question

In a survey, 16 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $46 and standard deviation of $8. Find the margin of error at a 95% confidence level.

Answer 1

You have a small sample and the population variance is unknown, so we're doing a t-test.

Confidence interval = sample mean plus/minus margin of error

That margin of error = t * s / sqrt(n)

Find your test statistic from a table using alpha=0.025 (because 95% confidence = 5% tails = 0.025 on each side) and degrees of freedom = n-1 = 16-1 = 15.

Plug in and solve.