Question

Suppose a researcher is estimating the proportion of trees showing signs of disease in a local forest. They randomly sample 150 trees and find that 25 of them show signs of disease. Calculate the 95% confidence interval for p, the proportion of all trees in the forest showing signs of disease.

Answer 1

Answer:

$0.167 - 1.96 \sqrt{\frac{0.167(1-0.167)}{150}}=0.107$

$0.167 + 1.96 \sqrt{\frac{0.167(1-0.167)}{150}}=0.227$

And the 95% confidence interval would be given (0.107;0.227).

Step-by-step explanation:

Information given

$X= 25$ number of trees with signs of disease

$n= 150$ the sample selected

$\hat p=\frac{25}{150}= 0.167$ the proportion of trees with signs of disease

The confidence interval for the true proportion would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the significance is $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, and the critical value would be

$z_{\alpha/2}=1.96$

And replacing we got:

$0.167 - 1.96 \sqrt{\frac{0.167(1-0.167)}{150}}=0.107$

$0.167 + 1.96 \sqrt{\frac{0.167(1-0.167)}{150}}=0.227$

And the 95% confidence interval would be given (0.107;0.227).