In a survey, 27 people were asked how much they spent on their child's last birthday gift.

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 $36 and standard deviation of $7. Construct a confidence interval at a 90% confidence level.

Give your answers to one decimal place. 

Answer 1

The answer to this problem takes the form: x^hat +/- t * ( s / sqrt(n) )

Let's solve for each of these variables:

(1) x^hat = 36, as this is the sample mean from the problem setup

(2) sqrt(n) = sqrt(27), since there are 27 people surveyed according the problem setup, i.e. n = 27

(3) s = 7, as this is the sample standard deviation from the problem setup

(4) t = 1.7056, which requires further explanation below:

First note that we are using a "t score" from the Student T distribution rather than a "z score" from a Standard Normal Distribution. This is because our sample size (n = 27) is less than 30, which is the threshold for the Law of Large Numbers to be applied. If n > 30, then we could have used "z scores" in the confidence interval. Otherwise, we have to use "t scores" due to potential skewness issues.

We calculate the value of t using a T-Distribution table and the following parameters:

(1) Degrees of Freedom = n - 1 = 27 - 1 = 26

(2) Significance Level = 1 - 0.9 = 0.1

(3) "Two-tailed" because our confidence interval is two-sided

==> t = 1.7056

Now that we obtained all of the variables, we can plug them into our solution equation:

==> x^hat +/- t * ( s / sqrt(n) ) = 36 +/- 1.7056 * ( 7 / sqrt(27) ) = 36 +/- 2.3

And that's our solution! 36 +/- 2.3