Question

In a survey, 23 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $34 and sample standard deviation of $14. Construct a confidence interval at a 99% confidence level.Give your answers to one decimal place. < μ <

Answer 1

Step-by-step explanation

The appropriate equation to obtain the confidence interval is the following

`\bar{x}+/-t*s/sqrt\left(n\right)`

Where each value is the following:

x_hat = 34

sqrt(n) = sqrt(23), because there are 23 people surveyed

s = 14 = standard deviation

We need to use a "t score" from the Student T distribution, as follows:

Degrees of Freedom = n - 1 = 23 - 1 = 22

Significance Level = 1 - 0.99 = 0.01

"Two-tailed" because our confidence interval is two-sided

t = 2.819

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

==> xhat +/- t * ( s / sqrt(n) ) = 34 +/- 2.819 * ( 14 / sqrt(23) ) =

34 +/- 8.2292

Therefore, the solution is the following:

34-8.2 < μ < 34 +8.2