A random sample of 5000 people was chosen for a survey. 54% of them said they didn't have children under 18 living at home. Find the 95% confidence interval. Round answer to 4 significant digits and give lowest and highest values of confidence Interval.
I got 200.9 & 339.0 but they seem to be wrong.
To find the 95% confidence interval for the proportion of people who don't have children under 18 living at home, you can use the formula:
Confidence Interval = Sample Proportion ± Margin of Error
To calculate the margin of error, you need to determine the standard error, which is the standard deviation divided by the square root of the sample size. In this case, the sample size is 5000, and the sample proportion is 54%.
Step 1: Calculate the standard error:
Standard Error = sqrt((p * (1 - p)) / n)
p = 54% = 0.54
n = 5000
Standard Error = sqrt((0.54 * (1 - 0.54)) / 5000)
Standard Error ≈ 0.0091
Step 2: Calculate the margin of error:
Margin of Error = Critical Value * Standard Error
For a 95% confidence level, the critical value corresponds to a z-score of approximately 1.96 (since the standard normal distribution is used).
Margin of Error ≈ 1.96 * 0.0091
Margin of Error ≈ 0.0178
Step 3: Calculate the confidence interval:
Confidence Interval = Sample Proportion ± Margin of Error
Sample Proportion = 0.54
Lower Bound = 0.54 - 0.0178
Upper Bound = 0.54 + 0.0178
Lower Bound ≈ 0.5222
Upper Bound ≈ 0.5578
Rounded to four significant digits, the 95% confidence interval for the proportion of people who don't have children under 18 living at home is approximately 0.5222 to 0.5578.
Therefore, the lowest value of the confidence interval is 0.5222, and the highest value is 0.5578.
So, the correct answer is not 200.9 & 339.0.