Dylan 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 he have to get a margin of error less than 0.03? [To find n, use the value p* = 1/2 for the sample proportion and the values for z* from Table C.]
n = (1.645/0.03)^2 = 576.5625
Therefore, Dylan must have a sample size of 577 high school students to get a margin of error less than 0.03.
To determine the sample size required to achieve a desired margin of error, we can use the formula:
n = (z*^2 * p*(1-p*)) / (E^2)
Where:
- n is the sample size
- z* is the z-value corresponding to the desired confidence level (90% confidence level corresponds to a z-value of approximately 1.645)
- p* is the estimated sample proportion (we'll use 1/2 or 0.5 in this case, assuming no prior knowledge)
- E is the desired margin of error (0.03 in this case)
Substituting the values into the formula, we get:
n = (1.645^2 * 0.5 * (1 - 0.5)) / (0.03^2)
Simplifying the expression:
n = (2.706025 * 0.5 * 0.5) / 0.0009
n = (0.67650625) / 0.0009
n ≈ 751.6736111
Since we can't have a fraction of a student, we round up the sample size to the next whole number. Therefore, Dylan would need a sample size of at least 752 in order to get a margin of error less than 0.03 with a 90% confidence level.
Note: Table C refers to the z-value table or standard normal distribution table. The z-value for a 90% confidence level is approximately 1.645.
To determine the sample size required to achieve a margin-of-error less than 0.03 for a 90 percent confidence interval, we need to use the formula:
n = (z^2 * p * (1-p)) / E^2
Where:
n is the sample size
z is the z-score corresponding to the desired level of confidence (90 percent confidence corresponds to a z-score of 1.645)
p is the estimated proportion of high school students attending basketball games (p* = 1/2)
E is the desired margin of error (0.03)
Substituting the given values into the formula, we have:
n = (1.645^2 * (1/2) * (1 - 1/2)) / (0.03^2)
Calculating the expression:
n = (2.705025 * 0.5 * 0.5) / 0.0009
n = 1.3525125 / 0.0009
n ≈ 1502
Therefore, Dylan would need a sample size of approximately 1502 to obtain a margin of error less than 0.03 for a 90 percent confidence interval.