Question

in a survey conducted by a firm, a sample of 90 residents in two story homes 45 were found to own their houses. find a 95 % confidence interval for the proportion of residents of two story houses who own their house.

Answer 1

To find a 95% confidence interval for the proportion of residents owning two-story houses, we use the sample proportion of 0.5 (45 out of 90 residents), a z-score of approximately 1.96, and the formula for a confidence interval for a proportion. The resulting confidence interval is approximately (0.397, 0.603), indicating that we are 95% confident that the true proportion lies within this range.

Step-by-step explanation:

To find a 95% confidence interval for the proportion of residents of two-story houses who own their house, we can use the formula for a confidence interval for a proportion:

Confidence Interval (CI) = p ± (z * sqrt(p(1-p)/n))

Where:
p is the sample proportion

• z is the z-score corresponding to the desired confidence level

• n is the sample size

• In this case:

• p = 45/90

• n = 90

• For a 95% confidence interval, z is approximately 1.96 (You can find this value in standard z-tables)

• Calculating the confidence interval:

• First, calculate the sample proportion (p):

• p = 45/90 = 0.5

• Next, plug the values into the CI formula:

• CI = 0.5 ± (1.96 * sqrt(0.5 * (1 - 0.5) / 90))

• Simplify the equation and calculate the margin of error:

• Margin of error = 1.96 * sqrt(0.5 * 0.5 / 90) ≈ 0.103

• Finally, calculate the bounds of the CI:

• Lower bound = 0.5 - 0.103

• = 0.397

• Upper bound = 0.5 + 0.103

• = 0.603

Therefore, the 95% confidence interval for the proportion of two-story house residents who own their houses is approximately (0.397, 0.603).

This interval means that we are 95% confident that the true proportion of residents in two-story homes who own their houses falls between 39.7% and 60.3%.