Suppose that the insurance companies did do a survey. they randomly surveyed 400 drivers and found that 320 claimed they always buckle up. we are interested in the population proportion of drivers who claim they always buckle up. construct a 95% confidence interval for the population proportion who claim they always buckle up. what is the error bound?

The 95% confidence interval would be given (0.761;0.839). The error bound is  

Step-by-step explanation:

1) Data given and notation  

n=400 represent the random sample taken    

X=320 represent the people drivers claimed they always buckle up

estimated proportion of people drivers claimed they always buckle up

represent the significance level (no given, but is assumed)    

Confidence =95% or 0.95

p= population proportion of people drivers claimed they always buckle up

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) Calculating the interval for the proportion

The confidence interval would be given by this formula  

For the 95% confidence interval the value of and , with that value we can find the quantile required for the interval in the normal standard distribution.  

And replacing into the confidence interval formula we got:  

And the 95% confidence interval would be given (0.761;0.839). The error bound is  

We are confident that about 76.1% to 83.9% of people drivers that they always buckle up  at 95% of confidence

if the pizza is divided equally into 2 equal halves then you would get

πr^2/2

where r is the radius of the pizza.

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