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:

Answer:

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

Step-by-step 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

Step-by-step explanation:

Answer 2

Answer:

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

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

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 $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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