select the correct answer.\nin a random sample of 50 undergraduate students at a college, it was found that…

Explanation:

Step1: Calculate sample proportion

Let $n = 50$ (sample size) and $x=44$ (number of successes). The sample proportion $\hat{p}=\frac{x}{n}=\frac{44}{50} = 0.88$.

Step2: Determine z - value for 95% confidence level

For a 95% confidence level, the z - value $z = 1.96$.

Step3: Calculate margin of error formula

The formula for the margin of error $E$ for a proportion is $E=z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$. Substitute $\hat{p}=0.88$, $n = 50$ and $z = 1.96$ into the formula: First, calculate $1-\hat{p}=1 - 0.88=0.12$. Then $\hat{p}(1 - \hat{p})=0.88\times0.12 = 0.1056$. $\frac{\hat{p}(1 - \hat{p})}{n}=\frac{0.1056}{50}=0.002112$. $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}=\sqrt{0.002112}\approx0.04596$. $E = 1.96\times0.04596\approx0.0901\approx0.092$.

Answer:

E. 0.092