Solved – the difference between conservative and non-conservative confidence intervals?

I want to answer this question by I can't.

What is the 95% conservative confidence interval for the population of all teens proportion who get from 6 to 8 hours of sleep per night in average ?

I want to know the difference between conservative and non conservative confidence intervals.
How do I answer this type of question?
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I have also uploaded the solution but I want to understand it by equations, and how to reach this answer.
there is also another question which I can't understand about the difference between confidence interval and conservative confidence interval
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The confidence interval for a proportion requires you to calculate the standard error of the proportion. The standard error depends on the population proportion. Since you don't know the population proportion (which is why you're making a confidence interval in the first place), it's ambiguous how to calculate the standard error to use in the confidence interval.

One solution is to use the sample proportion in the formula for the standard error. The formula for this is the following:

$$CI=hat p pm Zsqrt{frac{hat p (1-hat p)}{n}}$$

where $hat p$ is the sample proportion, $Z$ is the critical Z-statistic for $alpha = .05$ for a two-tailed test, and $n$ is the sample size. This is the formula that was used to calculate the bounds for the answer that was marked correct, except that $2$ was used for $Z$ as an approximation to $1.96$, the actual correct value for a 95% confidence interval. Unfortunately, this strategy can lead to confidence intervals that are too narrow. Because this confidence interval has lower than the required coverage (i.e., the true confidence level will be less than 95%), this confidence interval is non-conservative. A conservative confidence interval is a wider one because it excludes fewer potential values of the population parameter.

Another solution is to use the largest standard error possible for this distribution, which is the one that occurs when the population propertion is .5. The largest value that $p(1-p)$ can take is $.25$. If $p$ is anything less than or greater than .5, $p(1-p)$ will be smaller. So, we can replace $hat p(1-hat p)$ in the above formula with $.5(1-.5)$. Because this is the largest standard error possible for a distribution of sample proportions, your confidence interval will be too wide when the population proportion is different from .5 and exactly right when the population proportion is equal to .5. This is what makes it conservative. On the other hand, the previous formula that uses $hat p$ will be systematically too narrow when the population proportion is .5.

The values they used in the formula were

$$frac{460}{1368} pm 2sqrt{frac{frac{460}{1368}(1-frac{460}{1368})}{1368}}$$

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