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A lawn service owner is testing new weed killers. He discovered that a particular weed killer was effective 89% of time. Suppose that this estimate was based on a random sample of 60 applications. Find the lower confidence limit (LCL) for a 90% confidence interval for p, the true proportion of weeds killed by this particular brand. Round your answer to two decimal places.

User Max Fomin
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4 votes

Answer:

The LCL for a 90% confidence interval for p is 0.82.

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).

A lawn service owner is testing new weed killers. He discovered that a particular weed killer was effective 89% of 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

Rounding to two decimal places, the LCL for a 90% confidence interval for p is 0.82.

User Yorimar
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