Question

Kim wants to determine a 90 percent confidence interval for the true proportion of high school students in the area who attend their home basketball games. How large of a sample must she have to get a margin of error less than 0.04?

Answer 1

Kim would need to survey at least 423 high school students to get a margin of error less than 0.04 for a 90 percent confidence interval. This calculation uses the Z-value for 90% confidence, an estimated proportion of 0.5, and the specific margin of error required.

Step-by-step explanation:

To determine how large of a sample Kim must have to get a margin of error less than 0.04 for a 90 percent confidence interval, we can use the formula for the sample size of a proportion:

n = (Z^2 · p · (1-p)) / E^2

Where:

• n is the sample size,
• Z is the Z-value for the desired confidence level (1.645 for 90% confidence),
• p is the estimated proportion of the population (0.5 is used when the proportion is unknown),
• E is the margin of error (0.04 in this case).

Plugging in the values we get:

n = (1.645^2 · 0.5 · 0.5) / 0.04^2
= (2.708025 · 0.25) / 0.0016
= 0.67700625 / 0.0016
= 423.12890625

Since we can't have a fraction of a person, we would round up to the nearest whole number. Therefore, Kim would need to survey at least 423 high school students to ensure the margin of error is less than 0.04 for the 90 percent confidence interval.