Question

A certain transportation system of buses and commuter trains is heavily utilized so that it is not practical to check every traveler's ticket. Rather, only a small, randomly selected group of travelers on any given trip will be asked to show their tickets. Suppose that in a random sample of 621 train travelers is selected and 69 of them admitted they did not buy a ticket. Find the upper bound of a 95% confidence interval for the true proportion of all train travelers who do not buy tickets

Answer 1

Answer:

$0.0864 < p< 0.1358$

We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.  

Step-by-step explanation:

1) Data given and notation  

n=621 represent the random sample taken    

X=69 represent the people admitted they did not buy a ticket

$\hat p=\frac{69}{621}=0.111$ estimated proportion of people admitted they did not buy a ticket

$\alpha=0.05$ represent the significance level (no given, but is assumed)    

z would represent the statistic

p= population proportion of people admitted they did not buy a ticket

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

2) Confidence interval

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing into the confidence interval formula we got:

$0.111 - 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.0864$

$0.111 + 1.96 \sqrt{\frac{0.111(1-0.111)}{621}}=0.1358$

And the 95% confidence interval would be given (0.0864;0.1358).

$0.0864 < p< 0.1358$

We are confident (95%) that the true proportion of people admitted they did not buy a ticket is betwen 0.0864 and 0.1358.