Suppose 420 randomly selected people are surveyed to determine whether or not they subscribe to cable TV. Of the 420 surveyed, 278 reported subscribing to cable TV. Find the margin of error for the confidence interval for the population proportion with a 95% confidence level.

The estimated proportion is given by:

And replacing we got:

Step-by-step explanation:

Previous concepts

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

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:  

And the margin of error is given by:

The estimated proportion is given by:

And replacing we got:

i believe it's a.

i'm not sure on this one, most likely correct.

step-by-step explanation:

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