Question

in a survey, 13 people were asked how much they spent on their child's last birthday gift. the results were roughly bell-shaped with a mean and standard deviation of $36 and $15, respectively. construct a confidence interval at a 95% confidence level. give your answers to one decimal place. answer should be written in e format.

Answer 1

Sure. Here are the steps on how to construct a confidence interval at a 95% confidence level for the mean amount spent on a child's last birthday gift:

Identify the sample size, n. In this case, n=13.

Identify the sample mean,

x

ˉ

. In this case, $\bar{x} = $36.

Identify the sample standard deviation, s. In this case, $s = $15.

Identify the t-critical value, t. The t-critical value is the value that cuts off 2.5% of the distribution in each tail, leaving 95% of the distribution in the middle. To find the t-critical value, we need to know the degrees of freedom, df. The degrees of freedom are calculated as df=n−1. In this case, df=13−1=12.

We can find the t-critical value using a t-distribution table. The t-distribution table shows the t-critical values for different degrees of freedom and different confidence levels. For a 95% confidence level and 12 degrees of freedom, the t-critical value is 2.179.

Calculate the confidence interval. The confidence interval is calculated as follows:

$\bar{x} \pm t \cdot \frac{s}{\sqrt{n}}$

In this case, the confidence interval is:

$36 \pm 2.179 \cdot \frac{15}{\sqrt{13}} = (27.28, 44.72)$

Therefore, a 95% confidence interval for the mean amount spent on a child's last birthday gift is $27.28 to $44.72.

Explanation: