Find the margin of error at a 90% confidence level.

Question

In a survey, 14 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $43 and standard deviation of $7. Find the margin of error at a 90% confidence level.

Give your answer to two decimal places.

Answer 1

Hi Kaitlyn,

Here, we are given a normal (bell-shaped) distribution with known standard deviation (7), so we can use a z confidence interval to get the margin of error. The formula:

CI=x-bar +/- z*SE

Breaking this down:

x-bar=sample mean

z*=z critical value obtained from z-table

SE=standard error

Standard error has its own formula:

SE=sigma/sqrt(n), so let's compute that first.

sigma=7

n=14

SE=7/sqrt(14)

SE=1.87

Now, we need z*, and z* for a 90% two-sided confidence interval is always 1.645--you may want to memorize this value. Now, we have:

CI=x-bar +/- z*SE

x-bar=43

SE=1.87

z*=1.645

We now must address what the question asked for, which is margin of error. Margin of error is equivalent to the back end of the confidence interval. In this case:

MoE=z*SE

but it could also be t*SE if you were working with t confidence intervals. Anyway:

z*=1.645

SE=1.87

MoE=6.16

I hope this helps.