Kim wants to determine a 95 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 0.03?
Assuming the worst case of the point estimate of 50% (i.e. 50% participation), and
n=size of sample
Margin of error
= z1-α * standard error of worst case point estimate
= z95% * sqrt(0.5(1-0.5)/n)
=1.96 * 0.5/sqrt(n)
Equate to given margin of error of 0.03, then
0.03=1.96*0.5/sqrt(n)
Solve for n
n=(1.96*0.5/0.03)²
=32.67^2
=1067.11 (round up)
=1068
Well, Kim must do some serious sampling. I hope she has a lot of basketball-loving friends! Now, let's get cracking on this math problem.
To find out the minimum sample size, we need to use the formula:
n = (Z² * p * (1 - p)) / (E²)
Where:
- n is the sample size
- Z is the Z-score that corresponds to the desired level of confidence (in this case, 95% confidence, so Z would be around 1.96)
- p is the estimated proportion
- E is the margin of error
Since we want a margin of error less than 0.03, we plug that value into E. Now, the estimated proportion is what we're really wondering about.
If Kim has no prior estimates, then she could assume a proportion of 0.5 (or 50%), which is the most conservative estimate. However, if she has some insider information, anecdotal evidence, or a crystal ball, then she could use a more specific estimated proportion.
Once Kim figures out the estimated proportion (let's call it p), she can plug in the values and do some mathematical magic to find the necessary sample size. Just be careful not to let the basketball players trip over your numbers — they need to focus on the game!
To determine the required sample size for a given margin of error, we need to use the formula:
n = (Z^2 * p * (1-p)) / E^2
Where:
n = required sample size
Z = Z-value corresponding to the desired level of confidence (95 percent confidence corresponds to a Z-value of approximately 1.96)
p = estimated proportion of high school students attending home basketball games (we can start with 0.5 as it gives the largest sample size)
E = margin of error (0.03)
Plugging in the values:
n = (1.96^2 * 0.5 * (1-0.5)) / 0.03^2
Simplifying:
n = (3.8416 * 0.25) / 0.0009
n = 0.9604 / 0.0009
n ≈ 1067.11
Therefore, Kim would need a sample size of at least 1068 to obtain a margin of error less than 0.03.
To determine the sample size required to get a margin of error less than 0.03 and construct a 95 percent confidence interval for the true proportion, Kim needs to use the formula:
n = (Z^2 * p * (1-p)) / E^2
Here's what each variable in the formula represents:
- n: sample size
- Z: Z-score corresponding to the desired confidence level (in this case, 95 percent)
- p: estimated proportion of high school students attending home basketball games (since there is no initial estimate given, you can assume 0.5 for a conservative estimate)
- E: margin of error
First, find the Z-score associated with the 95 percent confidence level. This can be obtained from a standard normal distribution table or using a calculator or software. For a 95 percent confidence level, the Z-score is approximately 1.96.
Next, substitute the values into the formula:
n = (1.96^2 * 0.5 * (1-0.5)) / 0.03^2
Simplifying this equation will give you the required sample size (n) needed by Kim to obtain a margin of error less than 0.03.