Respuesta :
Answer:
The width of the confidence interval based on the 784 women would be narrower than the confidence interval based on the 327 women.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
From this, we have that the margin of error, and also the width, is inversely proportional to the sample size, that is, a larger sample size leads to a smaller margin of error and a narrower interval.
How would the width of the new confidence interval compare to the width of the confidence interval based on the 327 women in the armed forces?
New interval: 784
Old interval: 327
Sample size increased, so the new interval will be narrower, and the correct answer is:
The width of the confidence interval based on the 784 women would be narrower than the confidence interval based on the 327 women.
