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

Given;

`\begin{gathered} x=110,n=500 \\ \\ p=(x)/(n) \\ \\ p=(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{(p(1-p))/(n)} \\ \\ \Rightarrow0.22\pm1.96\sqrt{(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