Question

Out of 500 yards sampled, 110 had no dandelions. Based on this, construct a 95% confidence interval for the true population proportion of yards with no dandelions.Give your answers as decimals, to three placesI am 95% confident that the proportion of yards without dandelions is between_____ and ______.

Answer 1

Solution:

Given;

$\begin{gathered} x=110,n=500 \\ \\ p=\frac{x}{n} \\ \\ p=\frac{110}{500} \\ \\ p=0.22 \\ \\ \text{ Where }p\text{ is the true proportion} \end{gathered}$

Thus, the confidence interval is;

$\begin{gathered} p\pm z_{\alpha}\sqrt{\frac{p(1-p)}{n}} \\ \\ \Rightarrow0.22\pm1.96\sqrt{\frac{0.22(1-0.22)}{500}} \\ \\ \Rightarrow(0.184,0.256) \end{gathered}$

ANSWER: I am 95% confident that the proportion of yards without dandelions is between 0.184 and 0.256