# Math Help - Choosing an Appropriate Sample Size

1. ## Choosing an Appropriate Sample Size

Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The executives hire a statistical consultant and ask her to determine the mean shopping time, , of customers at the supermarkets. The consultant will collect a random sample of shopping times at the supermarkets and use the mean of these shopping times to estimate . Assuming that the standard deviation of the population of shopping times at the supermarkets is minutes, what is the minimum sample size she must collect in order for her to be confident that her estimate is within minutes of ?

I am having a hard time plugging this data into the formula, Can someone help?????

2. Originally Posted by kendrick02
Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The executives hire a statistical consultant and ask her to determine the mean shopping time, , of customers at the supermarkets. The consultant will collect a random sample of shopping times at the supermarkets and use the mean of these shopping times to estimate . Assuming that the standard deviation of the population of shopping times at the supermarkets is minutes, what is the minimum sample size she must collect in order for her to be confident that her estimate is within minutes of ?

I am having a hard time plugging this data into the formula, Can someone help?????
Since we're told $\sigma$, you want to use the equation $n\geqslant\frac{z_{\alpha/2}^2\sigma^2}{\varepsilon^2}$ where $\varepsilon$ is the error.

We are given $\varepsilon=6$, $\sigma=27$ and $\alpha=.05$

Thus, $n\geqslant\frac{z_{.025}^2\sigma^2}{\varepsilon^2} =\frac{(1.96)^2(27)^2}{6^2}\approx78$

So the sample size needs to be 78 or more in order to get the estimate to be within 6 minutes of $\mu$ (with 95% confidence).

Does this make sense?

3. All these books are incorrect, it's $n\ge$...
That's why they always round up.
The question should be, what's the smallest sample size that will satisfy....
And I see this was asked correctly.

Originally Posted by Chris L T521
Since we're told $\sigma$, you want to use the equation $n\ge\frac{z_{\alpha/2}^2\sigma^2}{\varepsilon^2}$ where $\varepsilon$ is the error.

We are given $\varepsilon=6$, $\sigma=27$ and $\alpha=.05$

Thus, $n\ge\frac{z_{.025}^2\sigma^2}{\varepsilon^2}=\frac {(1.96)^2(27)^2}{6^2}\approx78$

So the sample size needs to be 78 in order to get the estimate to be within 6 minutes of $\mu$ (with 95% confidence).

Does this make sense?