A new fertilizer was applied to the soil of 135 bean plants. $31 \%$ 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. Report answers to at least 2 decimal places.
Margin of Error (as a percentage):
Confidence Interval:
$\%$ to
Answer
\(\boxed{\text{Final Answer: The margin of error is approximately 7.80%. The 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer is approximately 23.20% to 38.80%.}}\)
Step 1 ：First, we calculate the sample proportion (p̂), which is the number of successes (bean plants that showed increased growth) divided by the total number of trials (total bean plants). In this case, the number of successes is 31% of 135 and the total number of trials is 135.
Step 2 ：Next, we calculate the standard error (SE) of the sample proportion. The formula for the standard error of a proportion is \(\sqrt{\frac{{p̂ * (1 - p̂)}}{n}}\), where n is the total number of trials.
Step 3 ：The margin of error (E) for a 95% confidence interval is calculated as E = Z * SE, where Z is the Z-score for a 95% confidence interval. The Z-score for a 95% confidence interval is approximately 1.96.
Step 4 ：Finally, the 95% confidence interval is calculated as (p̂ - E, p̂ + E).
Step 5 ：Using these calculations, we find that the margin of error is approximately 7.80%.
Step 6 ：The 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer is approximately 23.20% to 38.80%.
Step 7 ：\(\boxed{\text{Final Answer: The margin of error is approximately 7.80%. The 95% confidence interval for the percentage of all bean plants which show increased growth after application of the fertilizer is approximately 23.20% to 38.80%.}}\)
