Question

The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of visitors who plan to camp in the state. Current estimates are that 35% of visitors are campers. How many visitors would you sample to estimate the population proportion of campers with a 95% confidence level and an allowable error of 2%?

Answer 1

Answer: 2185

Step-by-step explanation:

Let p be the proportion of visitors are campers.

Given : The Tennessee Tourism Institute (TTI) plans to sample information center visitors entering the state to learn the fraction of visitors who plan to camp in the state.

The prior proportion of visitors are campers : p=0.35

Allowable error : E= 2%= 0.02

We know that the z-value for 95% confidence = $z_c=1.96$

Then by Central Limit Theorem , the required sample size would be :

$n=p(1-p)(\dfrac{z_{c}}{E})^2$

$\Rightarrow\ n=0.35(1-0.35)(\dfrac{1.96}{0.02})^2\\\\ n= 0.35(0.65)(9604)$

Simply , we get

$n=2184.91\approx2185$  [Rounded to the next whole number.]

Hence, the smallest sample size to estimate the population proportion of campers =2185