Question

What does the 99% confidence level in the previous problem tell us? Group of answer choices There is a 99% chance that this particular interval contains the population proportion. The interval will contain 99% of all sample proportions. Of confidence intervals with this margin of error, 99% will contain the population proportion. Of confidence intervals with this margin of error, 99% will contain the sample proportion.

Answer 1

Answer:

The confidence interval for the mean is given by the following formula:  

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

The margin of error is given by:

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

For this case since the confidence is 99% we are confident that the true proportion of interest would be on the interval calculated and the best option for this case is:

Of confidence intervals with this margin of error, 99% will contain the population proportion

Step-by-step 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".  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

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

The margin of error is given by:

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

For this case since the confidence is 99% we are confident that the true proportion of interest would be on the interval calculated and the best option for this case is:

Of confidence intervals with this margin of error, 99% will contain the population proportion