Question

A new fertilizer was applied to the soil of 283 bean plants. 21% showed increased growth.

Find the margin of error and 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer. Round all answers to 2 decimal places.

Margin of Error (as a percentage): _______
Confidence Interval: _______% to _______%

Answer 1

TRADITIONAL METHOD
given that,
possible chances (x)=59.43
sample size(n)=283
success rate ( p )= x/n = 0.21
I.
sample proportion = 0.21
standard error = Sqrt ( (0.21*0.79) /283) )
= 0.0242
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
margin of error = 1.96 * 0.0242
= 0.0475
III.
CI = [ p ± margin of error ]
confidence interval = [0.21 ± 0.0475]
= [ 0.1625 , 0.2575]
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DIRECT METHOD
given that,
possible chances (x)=59.43
sample size(n)=283
success rate ( p )= x/n = 0.21
CI = confidence interval
confidence interval = [ 0.21 ± 1.96 * Sqrt ( (0.21*0.79) /283) ) ]
= [0.21 - 1.96 * Sqrt ( (0.21*0.79) /283) , 0.21 + 1.96 * Sqrt ( (0.21*0.79) /283) ]
= [0.1625 , 0.2575]
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interpretations:
1. We are 95% sure that the interval [ 0.1625 , 0.2575] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion
Answers:
margin of error = 4.75%
95% sure that the interval [ 0.1625 , 0.2575] = (16.25,25.75)%