Amajor news organization selected a simple random sample of 1,600 subscribers from their list of 100,000 subscribers. they asked whether the paper should increase its coverage of entertainment. forty percent of the sample wanted more entertainment. what is the 99% confidence interval for the proportion of subscribers who would like more coverage of entertainment news?

The 99% confidence interval would be given (0.368;0.432).

Step-by-step explanation:

Data given and notation  

n=1600 represent the random sample taken    

X represent the number of subscribers who would like more coverage of entertainment news

estimated proportion of subscribers who would like more coverage of entertainment news

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

Confidence =0.99 or 99%

p= population proportion of subscribers who would like more coverage of entertainment news

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".  

Confidence interval

The population proportion have the following distribution  

The confidence interval would be given by this formula  

For the 99% 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 99% confidence interval would be given (0.368;0.432).

it's both, now can you answer my most recent question?

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step-by-step explanation: