When an online news magazine asked viewers to click their agreement or disagreement, 300 out of 1,200 respondents agreed with a statement that the most practical way of becoming a millionaire is winning a lottery. Immediate feedback stated that 25% of the viewers, with a margin of error of /- 2.5%, agreed with the statement. Fine print claimed 95% confidence. What is the proper conclusion?

The estimated proportion on this case is

The margin of error is given by:

So then the claim makes sense for this case since the estimated proportion of 0.25 and the margin of error 0.025 or 2.5%

If we replace the values obtained we got:

The 95% confidence interval would be given by (0.226;0.275)

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:  

The estimated proportion on this case is

The margin of error is given by:

So then the claim makes sense for this case since the estimated proportion of 0.25 and the margin of error 0.025 or 2.5%

If we replace the values obtained we got:

The 95% confidence interval would be given by (0.226;0.275)

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