Question

How many observations should be made if she wants to be 86.64 percent confident that the maximum error in the observed time is .5 second? Assume that the standard deviation of the task time is four seconds.

Answer 1

144 observations

Explanation:

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

`\alpha = (1-0.8684)/(2) = 0.0668`

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
`1-0.0668 = 0.9332`, so
`z = 1.5`

Now, find the margin of error 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:

`\sigma = 4, M = 0.5`

We want to find n

`M = z*(\sigma)/(√(n))`

`0.5 = 1.5*(4)/(√(n))`

`0.5√(n) = 6`

`√(n) = (6)/(0.5)`

`√(n) = 12`

`√(n)^(2) = (12)^(2)`

`n = 144`