Question

1. 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?

a. 4.7%
b. 5.6%
c. 3.2%
d. 11.5%
e. 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, $p$ is the sample proportion, and $n$ is the sample size. For a 95% confidence interval, the z - score $z\approx1.96$. The sample proportion $p = 0.75$ and the sample size $n=320$.

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 $E$

$E = 1.96\times0.0242\approx0.0474$ or $4.74\%\approx4.7\%$

Answer:

A. 4.7%