# A producer of a variety of salty snacks would like to estimate the average weight of a bag of BBQ...

## Question:

A producer of a variety of salty snacks would like to estimate the average weight of a bag of BBQ potato chips produced during the filling process at one of its plants. Determine the sample size needed to construct a {eq}90\% {/eq} confidence interval with a margin of error equal to {eq}0.008 {/eq} ounces. Assume the standard deviation for the potato chip filling process is {eq}0.05 {/eq} ounces.

## Confidence Interval

The confidence interval provides the lower and upper limit value for the estimation of a population parameter. The margin of error for estimating population parameters is calculated by half of the confidence width.

Given information

Margin of error: 0.008

Standard deviation: 0.05

The sample size is calculated as follows;

{eq}\begin{align*} n &= \dfrac{{{{\left( {{Z_{\alpha /2}}} \right)}^2} \times {\sigma ^2}}}{{{\varepsilon ^2}}} \ \ \ \ \ \ \ {\rm{ where }}\varepsilon {\rm{ \ is \ margin \ of \ error}}\\ &=\dfrac{{{{1.645}^2} \times {{0.05}^2}}}{{{{0.008}^2}}}\\ &= 105.704 \simeq 106 \end{align*} {/eq}

Therefore, the required sample size is 106.