Question

A lawn service owner is testing new weed killers. He discovers that a particular weed killer is effective 89% of the time. Suppose that this estimate was based on a random sample of 60 applications. Construct a 90% confidence interval for p, the true proportion of weeds killed by this particular brand. what is the upper confidence limit for p

Answer 1

Answer:

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

He discovers that a particular weed killer is effective 89% of the time. Suppose that this estimate was based on a random sample of 60 applications.

This means that

90% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564

Explanation:

Answer 2

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

He discovers that a particular weed killer is effective 89% of the time. Suppose that this estimate was based on a random sample of 60 applications.

This means that
`\pi = 0.89, n = 60`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.89 - 1.645\sqrt{(0.89*0.11)/(60)} = 0.8236`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.89 + 1.645\sqrt{(0.89*0.11)/(60)} = 0.9564`

The 90% confidence interval for p is (0.8236, 0.9564). The upper confidence limit for p is 0.9564.