Question

Five hundred randomly selected adult residents in Sacramento are surveyed to determine whether they believe children should have limited smartphone access. Of the 500 people surveyed, 381 responded yes - they believe children should have limited smartphone access.

You wish to estimate a population mean y with a known population standard devi- ation o = 3.5. If you want the error bound E of a 95% confidence interval to be less than 0.001, how large must the sample size n be?

Answer 1

The sample size must be of 47,059,600.

Explanation:

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

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Z-table as such z has a p-value of
`1 - \alpha`.

That is z with a p-value of
`1 - 0.025 = 0.975`, so Z = 1.96.

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.

Standard deviation:

`\sigma = 3.5`

If you want the error bound E of a 95% confidence interval to be less than 0.001, how large must the sample size n be?

This is n for which M = 0.001. So

`M = z(\sigma)/(√(n))`

`0.001 = 1.96(3.5)/(√(n))`

`0.001√(n) = 1.96*3.5`

`√(n) = (1.96*3.5)/(0.001)`

`(√(n))^2 = ((1.96*3.5)/(0.001))^2`

`n = 47059600`

The sample size must be of 47,059,600.