The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the brand. How many adults must he survey in order to be 90 % confident that his estimate is within six percentage points of the true population percentage?

a) Assume that nothing is known about the percentage of adults who have heard of the brand. n=_**____** (Round up to the nearest integer.)

b) Assume that a recent survey suggests that about 76 % of adults have heard of the brand. n=_**__** (Round up to the nearest integer.)

#### Solution

The formula to estimate the sample size required to estimate the proportion is

`$$ \begin{aligned} n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2 \end{aligned} $$`

where $p$ is the proportion of success, $z$ is the $Z_{\alpha/2}$ and $E$ is the margin of error.

(a) Given that margin of error $E =0.06$. The confidence coefficient is $0.9$. If no estimate were available for the proportion of adults who have heard of the brand we assume that the proportion is $p =0.5$.

The critical value of $Z$ is `$Z_{\alpha/2} =Z_{0.05}= 1.645$`

.

The minimum sample size required to estimate the proportion is

`$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.5(1-0.5)\bigg(\frac{1.645}{0.06}\bigg)^2\\ &=187.918\\ &\approx 188. \end{aligned} $$`

Thus, the sample of size $n=188$ will ensure that the $90$% confidence interval for the proportion will have a margin of error $0.06$.

(b) Given that margin of error $E =0.06$. The confidence coefficient is $1-\alpha=0.9$. The sample proportion of adults who have heard of the brand is $p =0.76$.

The critical value of $Z$ is `$Z_{\alpha/2} =Z_{0.05}= 1.645$`

.

The minimum sample size required to estimate the proportion is

`$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.76(1-0.76)\bigg(\frac{1.645}{0.06}\bigg)^2\\ &=137.105\\ &\approx 138. \end{aligned} $$`

Thus, the sample of size $n=138$ will ensure that the $90$% confidence interval for the proportion will have a margin of error $0.06$.