Question

1. Many of the trees in a national forest suffer from a virus that attacks the bark of the tree. Trees with this virus should be removed in order to minimize the risk to nearby trees.   To estimate the proportion of trees that have this virus a random sample of 204 trees was selected. Each selected tree was inspected and it was found that 28% of the trees in the sample had the virus. A 95% confidence interval will be used to estimate the proportion of trees with the virus.
   1. Verify that use of the large-sample z confidence interval is appropriate.
   2. Calculate the 95% confidence interval.
   3. Interpret the confidence interval and the associated confidence level

Thanks for the help!!

Answer 1

Given ; Sample size=n=204

Estimate of the sample proportion=p=28%=0.28

Significance level=

Now , ; From standard normal distribution table

a. Here , np=204*0.28=57.12>10 and n(1-p)=204*(1-0.28)=146.88>10

Therefore , the large sample Z confidence interval is appropriate to use.

b. The 95% confidence interval is ,

c. Interpretation: The 95% confidence that the true population proportion of trees with the virus is within the interval ( 0.2184 , 0.3416 )