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Clemens Sielaff · Answer 1 · 2020-08-29T21:48:08+0000

Answer:

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

0.368 < p <0.432

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

$\hat p=0.40$ estimated proportion of subscribers who would like more coverage of entertainment news

$\alpha=0.01$ 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

$p \sim N(p,\sqrt{(\hat p(1-\hat p))/(n)})$

The confidence interval would be given by this formula

$\hat p \pm z_(\alpha/2) \sqrt{(\hat p(1-\hat p))/(n)}$

For the 99% confidence interval the value of
$\alpha=1-0.99=0.01$ and
$\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_(\alpha/2)=2.58$

And replacing into the confidence interval formula we got:

$0.40 - 2.58 \sqrt{(0.40(1-0.40))/(1600)}=0.368$

$0.40 + 2.58 \sqrt{(0.40(1-0.40))/(1600)}=0.432$

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

0.368 < p <0.432

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