Questions: Question 6 of 10 Estimating Number of Close Confidants More Accurately In a study, we see that the average number of close confidants in a random sample of 2006 US adults is 2.2 with a standard deviation of 1.4. If we want to estimate the number of close confidants with a margin of error within ± 0.08 and with 90% confidence, how large a sample is needed? Round your answer up to the nearest integer. sample size = i

Solution Steps

We are provided with the following information:

• Sample size (\( n \)) = 2006 (not directly used in calculations)
• Sample mean (\( \bar{x} \)) = 2.2 (not directly used in calculations)
• Population standard deviation (\( \sigma \)) = 1.4
• Desired margin of error (\( E \)) = 0.08
• Confidence level = 90%

For a 90% confidence level, the corresponding Z-score is approximately: \[ Z \approx 1.645 \]

Using the formula for the margin of error: \[ E = Z \cdot \frac{\sigma}{\sqrt{n}} \] we can rearrange it to solve for \( n \): \[ n = \left( \frac{Z \cdot \sigma}{E} \right)^2 \]

Substituting the known values: \[ n = \left( \frac{1.645 \cdot 1.4}{0.08} \right)^2 \]

Calculating the numerator: \[ 1.645 \cdot 1.4 = 2.303 \]

Now, substituting back into the equation: \[ n = \left( \frac{2.303}{0.08} \right)^2 \] \[ n = \left( 28.7875 \right)^2 \approx 828.4 \]

Since the sample size must be a whole number, we round up: \[ n = \lceil 828.4 \rceil = 829 \]

Final Answer