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Sagot :
Using a confidence interval of proportions, it is found that the correct option is:
A. approximately the same estimate but a smaller margin of error.
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]\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]\frac{1+\alpha}{2}[/tex].
The margin of error is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
- From this, it is taken that the margin of error is inversely proportional to the sample size, hence, increasing the sample from 250 to 1000 students, the margin of error will be smaller.
The sample size has no bearing on the estimate, hence, it stays the same, and option A is correct.
A similar problem is given at https://brainly.com/question/15043877
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