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

Answer:

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

Step-by-step explanation:

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

$\alpha = \frac{1 - 0.95}{2} = 0.025$

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

$M = z\frac{\sigma}{\sqrt{n}}$

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

$M = z\frac{\sigma}{\sqrt{n}}$

$0.001 = 1.96\frac{3.5}{\sqrt{n}}$

$0.001\sqrt{n} = 1.96*3.5$

$\sqrt{n} = \frac{1.96*3.5}{0.001}$

$(\sqrt{n})^2 = (\frac{1.96*3.5}{0.001})^2$

$n = 47059600$

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