Question

Lesson 6.07: In a random sample of 74 homeowners in a city, 22 homeowners said they wouldsupport a ban on nonnatural lawn fertilizers to protect fish in the local waterways. The samplingmethod had a margin of error of +3.1%. SHOW ALL WORK!A) Find the point estimate.B) Find the lower and upper limits and state the interval.

Answer 1

Confidence interval is written in the form,

(point estimate +/- margin of error)

The given scenario involves population proportion

The formula for the point estimate is

p' = x/n

where

p' = estimated proportion of success. p' is a point estimate for p which is the true proportion

x represents the number of success

n represents the number of samples

From the information given,

n = 74

x = 22

p' = 22/74 = 0.297

The formula for finding margin of error is expressed as

`\begin{gathered} \text{margin of error = z}_{(\alpha)/(2)}(\sqrt[]{(p^(\prime)q^(\prime))/(n)} \\ q^(\prime)\text{ = 1 - p'} \\ q^(\prime)\text{ = 1 - 0.297 = 0.703} \end{gathered}`

A) The point estimate is 0.297

B) margin of error = +/-3.1% = 3.1/100 = +/- 0.031

Thus,

the lower limit would be 0.297 - 0.031 = 0.266

Expressing in percentage, it is 0.266 x 100 = 26.6%

the upper limit would be 0.297 + 0.031 = 0.328

Expressing in percentage, it is 0.328 x 100 = 32.8%

Thus, the confidence interval is between 26.6% and 32.8%