More About MidlineThe equation of the midline of periodic function is the average of the maximum and minimum values of the functionExamples of Midline<span>Figure-1 shows y = sin x and Figure-2 shows y = sin x + 1. The second curve is the first curve shifted vertically up by one unit.
The midline of y = sin x is the x-axis and the midline of y = sin x + 1 is the line y = 1.</span>

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A certain transportation system of buses and commuter trains is heavily utilized so that it is not practical to check every trav

IRINA_888 [86]

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.

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