Answer:
The 95% confidence interval for the mean savings is ($60.54, $81.46).
Explanation:
As there is no information about the population standard deviation of savings and the sample is not large, i.e. n = 20 < 30, we will use a t-confidence interval.
The (1 - α)% confidence interval for population mean (μ) is:

For the data provided compute the sample mean and standard deviation as follows:
![\bar x=(1)/(n)\sum\limits^(n)_(i=1){ X_(i)}=(1)/(20)* [92+34+40+...+53+82]=71](https://img.qammunity.org/2021/formulas/mathematics/college/hilv1zrdorw7xopc510fdfr6hcgluwqq7f.png)

The critical value of t for α = 0.05 and (n - 1) = 19 degrees of freedom is:

*Use a t-table for the value.
Compute the 95% confidence interval for the mean savings as follows:

Thus, the 95% confidence interval for the mean savings is ($60.54, $81.46).