Question

A survey conducted by the american automobile association showed that a family of four spends an average of $215.60 per day while on vacation. suppose a sample of 64 families of four vacationing at niagara falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $70.50.
a. develop a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting niagara falls (to 2 decimals).

Answer 1

The confidence interval goes from 235.18 to 269.72.

We first find the z-score associated with this confidence level:

1-0.95 = 0.05/2 = 0.025 

Using a z-table (http://www.normaltable.com/ztable-righttailed.html) we see that the z-score associated with this is 1.96.

The formula for the confidence interval is:

$x \pm z\times (\frac{s}{\sqrt{n}})$,
where x is the sample mean, s is the sample standard deviation and n is the sample size.  With our data, we have:

$252.45\pm1.96(\frac{70.50}{\sqrt{64}}) \\ \\252.45\pm17.2725$

252.45-17.2725 = 235.1775 ≈ 235.18
252.45+17.2725 = 269.7225 ≈ 269.72