Question

A local board of education conducted a survey of residents in the community concerning a property tax levy on the coming local ballot. They randomly selected 850 residents in the community and contacted them by telephone. Of the 850 residents surveyed, 410 supported the property tax levy. Let p represent the proportion of all residents in the community that support the property tax levy. What is the margin of error for a 90% confidence interval for p based on this sample?

Answer 1

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 = $\frac{x}{n}=\frac{410}{850}=0.482$

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:

$M.E=z \times \sqrt{\frac{p(1-p)}{n} }$

Substituting the values in this formula, we get:

$M.E=1.645 \times \sqrt{\frac{0.482(1-0.482)}{850} } \\\\ M.E=0.028$

Therefore, the margin of error for a 90% confidence interval for p based on the given sample is 0.028 or 2.8%