Question

1. child care programs a study found that 73% of randomly selected prekindergarten children ages 3 to 5 whose mothers had a bachelor’s degree or higher were enrolled in center - based early childhood care and education programs. how large a sample is needed to estimate the true proportion within 3 percentage points with 95% confidence? how large a sample is needed if you had no prior knowledge of the proportion?

Explanation:

Step1: Recall the formula for sample size when estimating a proportion

The formula for the sample size \( n \) to estimate a population proportion \( p \) is:

$$ n = \frac{z_{\alpha/2}^2 \cdot p \cdot (1 - p)}{E^2} $$

where \( z_{\alpha/2} \) is the z-score corresponding to the desired confidence level, \( p \) is the estimated proportion, and \( E \) is the margin of error.

For a 95% confidence level, the critical value \( z_{\alpha/2} \) is approximately 1.96 (this is a standard value for 95% confidence). The margin of error \( E \) is 0.03 (3 percentage points).

Step2: Case 1: With prior knowledge of \( p = 0.73 \)

Substitute \( z_{\alpha/2} = 1.96 \), \( p = 0.73 \), \( 1 - p = 0.27 \), and \( E = 0.03 \) into the formula:

$$ n = \frac{(1.96)^2 \cdot 0.73 \cdot 0.27}{(0.03)^2} $$

First, calculate the numerator:

$$ (1.96)^2 = 3.8416 $$

$$ 3.8416 \cdot 0.73 \cdot 0.27 = 3.8416 \cdot 0.1971 = 0.7572 $$

Then, calculate the denominator:

$$ (0.03)^2 = 0.0009 $$

Now, divide the numerator by the denominator:

$$ n = \frac{0.7572}{0.0009} \approx 841.33 $$

Since the sample size must be an integer, we round up to the next whole number, so \( n = 842 \).

Step3: Case 2: No prior knowledge of \( p \)

When we have no prior knowledge of \( p \), we use \( p = 0.5 \) (this maximizes the product \( p(1 - p) \), leading to the largest sample size needed). Substitute \( z_{\alpha/2} = 1.96 \), \( p = 0.5 \), \( 1 - p = 0.5 \), and \( E = 0.03 \) into the formula:

$$ n = \frac{(1.96)^2 \cdot 0.5 \cdot 0.5}{(0.03)^2} $$

Calculate the numerator:

$$ (1.96)^2 \cdot 0.5 \cdot 0.5 = 3.8416 \cdot 0.25 = 0.9604 $$

The denominator is still \( 0.0009 \).

$$ n = \frac{0.9604}{0.0009} \approx 1067.11 $$

Round up to the next whole number, so \( n = 1068 \).

Answer:

• With prior knowledge (\( p = 0.73 \)): The required sample size is \(\boldsymbol{842}\).
• Without prior knowledge (\( p = 0.5 \)): The required sample size is \(\boldsymbol{1068}\).