Assume the population proportion is to be estimated from the sample described. find the approximate margin of error and the 95% confidence interval for the population proportion. sample size = 225, sample proportion = 0.28 the margin of error is 0.0587. (round to four decimal places as needed.) find the 95% confidence interval. (round to the three decimal places as needed.)

We can conclude that the true population proportion at 95% of confidence is between (0.221;0.339)

Step-by-step explanation:

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".  

represent the real population proportion of interest  

represent the estimated proportion for the sample

n=225 is the sample size required  

represent the critical value for the margin of error  

The population proportion have the following distribution  

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.  

The margin of error is given by :

And replacing into the confidence interval formula we got:  

And the 95% confidence interval would be given (0.221;0.339).

We can conclude that the true population proportion at 95% of confidence is between (0.221;0.339)

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could u provide us some answer choices or hmm idk maybe a table for us to anser with ? ?