The data below is taken from two random, independent samples. Calculate the margin of error for a 99% confidence interval for the difference in population proportions.
$x_1=87$, $x_2=68$
$n_1=120$, $n_2=115$

• A. 0.061
• B. 0.062
• C. 0.158
• D. 0.060

Answer: (C)

Verified step by step guidance

1. Step 1: Begin by calculating the sample proportions for each group. The formula for sample proportion is p̂ = x/n, where x is the number of successes and n is the sample size. For the first sample, calculate p̂₁ = x₁/n₁, and for the second sample, calculate p̂₂ = x₂/n₂.
2. Step 2: Determine the standard error (SE) for the difference in proportions. The formula for SE is: SE = sqrt((p̂₁(1 - p̂₁)/n₁) + (p̂₂(1 - p̂₂)/n₂)). Substitute the values of p̂₁, p̂₂, n₁, and n₂ into this formula.
3. Step 3: Identify the z-value corresponding to the desired confidence level. For a 99% confidence interval, the z-value is approximately 2.576. This value is derived from the standard normal distribution table.
4. Step 4: Calculate the margin of error (ME) using the formula: ME = z * SE. Multiply the z-value by the standard error calculated in Step 2.
5. Step 5: Interpret the margin of error. The margin of error represents the range within which the true difference in population proportions is likely to fall, given the specified confidence level.