Answer:
Margin of error for a 90% confidence interval for p based on the given sample is 0.028 or 2.8%
Step-by-step explanation:
Total number of residents = n = 850
Number of residents who supported property tax levy = x = 410
Proportion of residents who supported property tax levy = p = [tex]\frac{x}{n}=\frac{410}{850}=0.482[/tex]
Confidence Level= 90%
Since we need to find the confidence interval for the population proportion of 1 sample we will use one sample z-test for population proportion to answer the question.
z value associated with 90% confidence interval as seen from the z table is 1.645.
The formula for Margin of Error for the population proportion is:
[tex]M.E=z \times \sqrt{\frac{p(1-p)}{n} }[/tex]
Substituting the values in this formula, we get:
[tex]M.E=1.645 \times \sqrt{\frac{0.482(1-0.482)}{850} } \\\\ M.E=0.028[/tex]
Therefore, the margin of error for a 90% confidence interval for p based on the given sample is 0.028 or 2.8%
