Question

The owner of a shopping mall wants an estimate of the lengthof

time shoppers spend in the mall. The estimate is to be within
5minutes of the true time, with a 98% confidence level. The
estimateof the standard deviation of the length of time spent in
the mallis 20 minutes. How large a sample will be needed?

Answer 1

A sample size of at least 87 will be needed.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.98)/(2) = 0.01`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.01 = 0.99`, so
`z = 2.325`

Now, find M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

`M = 5, \sigma = 20`

So

`M = z*(\sigma)/(√(n))`

`5 = 2.325*(20)/(√(n))`

`5√(n) = 46.5`

`√(n) = 9.3`

`n = 86.49`

A sample size of at least 87 will be needed.