Question

A National Retail Federation survey found households intend to spend an average of $469 during the December holiday season. Assume that the survey included 600 households and that the sample variance was $375. Calculate the margin of error for a 99% confidence interval.

Answer 1

The margin of error for a 99% confidence interval was of $1.84.

Explanation:

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

`\alpha = (1-0.99)/(2) = 0.005`

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
`1-0.005 = 0.995`, so
`z = 2.325`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation(square root of the variance) and n is the size of the sample. In this question,
`\sigma = √(375), n = 600`

So

`M = z*(\sigma)/(√(n)) = 2.325(√(375))/(√(600)) = 1.84`

The margin of error for a 99% confidence interval was of $1.84.