Question

In a random survey of 226 college students, 20% reported being "only" children (no siblings). The standard deviation is 2%.

Determine the confidence interval that is likely to contain the percent of all students nationwide who are only children.

Answer 1

Answer:

The confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)

Step-by-step explanation:

Total number of students = n = 226

We are given that 20% reported being "only" children (no siblings).

Number of students being "only" children (no siblings) $= 20\% \times 226=\frac{20}{100} \times 226=45.6\sim46$

So, x = 46

$p=\frac{x}{n}=\frac{46}{226}=0.203$

Z at 95% is 1.96

Formula for confidence interval for sample proportion :

$CI=p \pm Z \times \sqrt{\frac{p(1-p)}{n}}$

Substitute the values in the formula :

$CI=0.203 \pm 1.96 \times \sqrt{\frac{0.203(1-0.203)}{226}}\\CI=(0.1505,0.2554)$

Hence  the confidence interval that is likely to contain the percent of all students nationwide who are only children is (0.1505,0.2554)