Question

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In a random sample of 50 undergraduate students at a college, it was found that 44 students regularly access social networking websites from their college library. What is the margin of error for the true proportion of all undergraduates who access social networking sites from their college library?



0.046



0.038



0.014



0.070



0.092

Answer 1

@precal

Answer 2

sorry, this looks like STATS and I have not done stats is a very long time

Answer 3

The margin of error is the critical value * standard error

When you are doing a confidence interval, you calculate it as:
\(\text{estimate }\pm (\text{critical value})(\text{standard error)}\), or written as:
\(\text{estimate }\pm \text{margin of error}\)

So, the margin of error is found as:
\[ z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] where \(\hat{p}\) is the sample proportion (44/50), and \(n\) is the sample size (50). Here, \(z\) is the critical value of a normal distribution, except that you didn't supply the confidence level in your question.

Answer 4

Usually though they ask for 95% confidence levels, so you could assume that \(z = 1.96\) (or sometimes it's approximated as 2).