To calculate the margin of error for the sample proportion at a 95% confidence level, we can follow these steps:
-
Calculate the sample proportion (\( \hat{p} \)): \[ \hat{p} = \frac{x}{n} = \frac{122}{400} = 0.305 \]
-
Calculate the standard error (SE) of the sample proportion: \[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.305 \times (1 - 0.305)}{400}} \] \[ SE = \sqrt{\frac{0.305 \times 0.695}{400}} = \sqrt{\frac{0.212475}{400}} \approx \sqrt{0.0005311875} \approx 0.02305 \]
-
Find the critical value (z) for a 95% confidence level: For a 95% confidence interval, the critical value (\( z \)) is approximately 1.96.
-
Calculate the margin of error (ME): \[ ME = z \cdot SE = 1.96 \cdot 0.02305 \approx 0.0452 \] Converting to a percentage: \[ ME \approx 0.0452 \times 100 \approx 4.52% \]
Now, we can summarize the results:
- The margin of error \( r \) is approximately 4.52%.
- Therefore, the confidence interval for the sample proportion of people who own a house is: \[ \hat{p} \pm ME = 30.5% \pm 4.52% \]
Calculating the bounds:
- Lower bound: \( 30.5% - 4.52% \approx 25.98% \)
- Upper bound: \( 30.5% + 4.52% \approx 35.02% \)
Thus, the interval is approximately between 25.98% and 35.02%.
Final answer:
The margin of error \( r \) is 4.52%. With 95% confidence, the sample proportion of people who own a house is between 25.98% and 35.02%.
Interpretation:
This means that we are 95% confident that the true proportion of people who own a house in the population from which the sample was drawn falls between 25.98% and 35.02%. This range gives us an indication of the variability in our estimation based on the surveyed sample.