Question

Out of 300 people sampled, 186 had children under the age of 18 . Based on this, construct a 99% confidence interval for the true population proportion of people with children under the age of 18 . Give your answers as decimals, to three places.

Answer 1

To construct a 99% confidence interval for the true population proportion of people with children under 18, calculate the sample proportion, find the Z-value, calculate the standard error, compute the margin of error, and apply these to the formula. The interval is (0.548, 0.692).

Step-by-step explanation:

To construct a 99% confidence interval for the true population proportion of people with children under the age of 18, we use the formula for a confidence interval for a population proportion:

CI = Π ± Z*√(Π(1-Π)/n)

where Π is the sample proportion, Z* is the Z-value for the 99% confidence level, and n is the sample size.

Step 1: Calculate the sample proportion (Π):
Π = number of successes / sample size
Π = 186/300 = 0.620

Step 2: Find the critical value (Z*) for 99% confidence interval:
The Z-value for a 99% confidence level is approximately 2.576.

Step 3: Calculate the standard error (SE) of the sample proportion:
SE = √(Π(1-Π)/n)
SE = √(0.620(1-0.620)/300) = 0.028

Step 4: Compute the margin of error (ME):
ME = Z* * SE
ME = 2.576 * 0.028 = 0.072

Step 5: Construct the confidence interval:
CI = 0.620 ± 0.072
Lower limit = 0.620 - 0.072 = 0.548
Upper limit = 0.620 + 0.072 = 0.692

Therefore, the 99% confidence interval for the true population proportion is (0.548, 0.692).