Confidence Intervals

Question

In a survey, 27 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $39 and standard deviation of $7. Find the margin of error at a 95% confidence level.

Give your answer to two decimal places.

Answer 1

95% confidence interval includes the mean plus or minus 2 standard deviations. That's the 68-95-99.7 rule where 68% are within 1 standard deviation, 95%^ within 2 deviations and 99.7 within 3.

That's 39 + or - 2(7) or 39+14, 39-14

The 95% confidence interval is [25, 53] if it's roughly bell shaped.

The margin of error with a 95% confidence interval is 5%, regardless of normality or bell shaped curves.

5% of the time, some parents will spend more than 53 or less than 25 dollars on the child's present, assuming a normal distribution. Some statisticians would likely use t tables for a student's t distribution instead of z tables, if the data has less than 30, but really the t distribution is for when you don't know the variance and have to estimate it. And the t distribution is roughly like the normal distribution, but fatter at the ends.

5% of 27 is 27/20 or slightly more than 1 parent among the 27.

Check the data and see if that's true, if there's 1 or 2 parents outside the interval [25, 53] If 2 or more maybe redo this with the t distribution tables.