Question

question 5 1 pts a survey is administered to a random sample of 320 college students, and 75% of the students in the sample say they check social media right before going to bed at night. if a 95% confidence interval is to be constructed based on this data, what will the approximate margin of error be? 4.7% 5.6% 3.2% 11.5% this question cannot be answered without knowing the population size.

Explanation:

Step1: Identify the formula for margin of error

For a proportion in a large - sample confidence interval, the margin of error $E = z\sqrt{\frac{p(1 - p)}{n}}$, where $z$ is the z - score corresponding to the desired confidence level, $p$ is the sample proportion, and $n$ is the sample size. For a 95% confidence interval, $z\approx1.96$, $p = 0.75$, and $n=320$.

Step2: Calculate $(1 - p)$

$1 - p=1 - 0.75 = 0.25$

Step3: Calculate $\frac{p(1 - p)}{n}$

$\frac{p(1 - p)}{n}=\frac{0.75\times0.25}{320}=\frac{0.1875}{320}\approx0.000586$

Step4: Calculate $\sqrt{\frac{p(1 - p)}{n}}$

$\sqrt{\frac{p(1 - p)}{n}}\approx\sqrt{0.000586}\approx0.0242$

Step5: Calculate the margin of error $E$

$E = 1.96\times0.0242\approx0.0474\approx4.7\%$

Answer:

$4.7\%$