Suppose that insurance companies did 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.

Compute the margin of error for a 95% confidence interval and construct this confidence interval for the population proportion who claim they always buckle up.
(i) State the margin of error. (Round your answers to four decimal places.)
(ii) State the confidence interval. (Round your answers to four decimal places.)

i)

And replacing we got:

ii)

Step-by-step explanation:

Notation and definitions

number of people that claimed always buckle up.

random sample taken

estimated proportion of people that claimed always buckle up

true population proportion of people that claimed 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".  

The population proportion have the following distribution

Solution to the problem

Part i

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by and . And the critical value would be given by:

The confidence interval for the mean is given by the following formula:  

The margin of error is given by:

And replacing we got:

Part ii

And the confidence interval would be given by:

the function makes a strait line horizontally with no x intercept but a y intercept at 9.

a

step-by-step explanation: