Chapter 10, Problem AP3.22CPT

To determine

To find out the correct options that how would the margin error of a $95\%$ confidence interval be affected if the sample size were increased from $50\text{ to }200$ and the sample proportion of people who have a Facebook page is unchanged.

Answer to Problem AP3.22CPT

Option (d) is the correct option

Explanation of Solution

It is given that:

  $\begin{array}{l}c=0.95\\ n\text{ increases from}:50\text{ to }200\end{array}$

Now, we know that the margin of error is calculated as:

  $MOE=Z_{0.025}\times \sqrt{\frac{p(1-p)}{n}}$

As we know that the sample size has been increased. And the sample size is increased by four times as it was before.

So the margin of error will be now with the new sample size,

  $MOE=Z_{0.025}\times \sqrt{\frac{p(1-p)}{4n}}$

Now, we will take four outside the square root to simplify the formula, then the margin of error will become,

  $MOE=\frac{1}{2}(Z_{0.025}\times \sqrt{\frac{p(1-p)}{n}})$

Thus, from above formula compared to the previous one, we can say that the margin of error is divided by two. Thus, option (d) is the correct option i.e. it is divided by two.