Question

The Retail Advertising and Marketing Association would like to estimate the average amount of money that a person spends for Mother's Day with a 99% confidence interval and a margin of error within plus or minus $6. Assuming the standard deviation for spending on Mother's Day is $36, the required sample size is ________.

A) 313 B) 284 C) 210 D) 239

Answer 1

D) 239

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 = 2.575`

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.

Required sample size:

Is n

n is found when
`M = 6`

We have that
`\sigma = 36`

So

`M = z*(\sigma)/(√(n))`

`6 = 2.575*(36)/(√(n))`

`6√(n) = 36*2.575`

Simplifying by 6

`√(n) = 6*2.575`

`(√(n))^(2) = (6*2.575)^(2)`

`n = 238.7`

We round up, and the answer is 239.