Question

The automatic ticket machines at a railroad station are programmed to ask every 7th ticket buyer on a given day if they would buy snacks on the train if snacks were available. of the 316 people shown this question, 198 of them answered yes. based on this sample, about how many of the regular 2200 daily passengers would buy snacks on the train?

Answer 1

Calculate the sample proportion (198 yes responses out of 316).
$\hat{p}= \frac{198}{316} =0.6266$

We want to test against a sample size of n = 2200 daily passengers.
In order to use the normal distribution, we should satisfy
$n \hat{p} \ge 10\,\, and\,\, n(1 - \hat{p}) \ge 10$
2200*0.6266 = 1378.5
2200*(1-0.6266) = 821.5
We may use the normal distribution.

Let us use a 95% confidence interval.
The estimate for the population proportion is
$p=\hat{p} \pm z^{*} \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} }$
where z* = 1.96 at the 95% confidence level.

$1.96 \sqrt{ \frac{06266(1-0.6266)}{2200} } =0.0202$
Therefore
p = 0.6266 +/-0.0202 = (0.6064, 0.6468)

Answer:
At the 95% confidence level, about 60% to 64% of regular passengers will buy snacks on the train.