Calculate the sample proportion (198 yes responses out of 316).
[tex]\hat{p}= \frac{198}{316} =0.6266[/tex]
We want to test against a sample size of n = 2200 daily passengers.
In order to use the normal distribution, we should satisfy
[tex]n \hat{p} \ge 10\,\, and\,\, n(1 - \hat{p}) \ge 10[/tex]
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
[tex]p=\hat{p} \pm z^{*} \sqrt{ \frac{\hat{p}(1-\hat{p})}{n} } [/tex]
where z* = 1.96 at the 95% confidence level.
[tex]1.96 \sqrt{ \frac{06266(1-0.6266)}{2200} } =0.0202[/tex]
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.
