Question

question 9 (1 point) based on a random sample of 50, a 95% confidence interval for the population proportion was computed. holding everything else constant, which of the following will reduce the length of the confidence interval by half? quadruple the sample size. decrease the sample proportion by half. none double the sample size.

Explanation:

Step1: Recall confidence - interval formula

The formula for the confidence interval for a population proportion is $\hat{p}\pm z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, and the length of the confidence interval $L = 2z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $z$ is the z - value corresponding to the confidence level, and $n$ is the sample size.

Step2: Analyze the effect of changing $n$

We want to find how to change $n$ to make $L$ half of its original value. Let the original sample size be $n_1$ and the new sample size be $n_2$, and the original length be $L_1$ and the new length be $L_2$. We know $L_1 = 2z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_1}}$ and $L_2=\frac{1}{2}L_1 = 2z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_2}}$.
Substitute $L_2=\frac{1}{2}L_1$ into the equations:
$\frac{1}{2}\times2z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_1}}=2z\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_2}}$
Simplify the equation: $\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_1}} = 2\sqrt{\frac{\hat{p}(1 - \hat{p})}{n_2}}$
Square both sides: $\frac{\hat{p}(1 - \hat{p})}{n_1}=\frac{4\hat{p}(1 - \hat{p})}{n_2}$
Cancel out $\hat{p}(1 - \hat{p})$ (assuming $\hat{p}
eq0$ and $\hat{p}
eq1$), we get $n_2 = 4n_1$.

Answer:

Quadruple the sample size.