A survey of 500 shoppers showed that 60% were paying cash for their purchases at Walmart. What is the margin of error for this survey?
To calculate the margin of error for this survey, we need to use the formula:
Margin of Error = (Z * √(p * (1-p)) / √n)
Where:
- Z is the z-score corresponding to the desired confidence level
- p is the estimated proportion of shoppers paying cash (60% as per the survey)
- n is the sample size (500)
First, let's determine the z-score. Assuming a 95% confidence level, the z-score is approximately 1.96.
Now, let's plug the values into the formula:
Margin of Error = (1.96 * √(0.6 * (1-0.6)) / √500)
Calculating this expression:
Margin of Error ≈ (1.96 * √(0.6 * 0.4) / √500)
≈ (1.96 * √(0.24) / √500)
≈ (1.96 * 0.489897 / √500)
≈ (0.958255 / √500)
≈ 0.0428
Therefore, the margin of error for this survey is approximately 0.0428 or 4.28%.
To calculate the margin of error for a survey, you need to know the sample size (n) and the percentage (p) of the population that possesses a certain characteristic. In this case, the sample size is 500, and the percentage of shoppers paying cash is 60%.
The margin of error (ME) can be determined using the following formula:
ME = Z * √(p * (1 - p) / n)
where Z represents the standard score associated with the desired level of confidence. The standard score for a 95% confidence level is typically 1.96.
To calculate the margin of error for this survey:
ME = 1.96 * √(0.6 * (1 - 0.6) / 500)
Let's calculate it step by step:
1. Calculate p * (1 - p):
p * (1 - p) = 0.6 * (1 - 0.6) = 0.24
2. Calculate the square root of p * (1 - p):
√(p * (1 - p)) = √0.24 = 0.4899
3. Divide √(p * (1 - p)) by the square root of the sample size (n):
0.4899 / √500 ≈ 0.0347
4. Multiply the result by the Z score (1.96 for a 95% confidence level):
ME = 1.96 * 0.0347 ≈ 0.0681
Therefore, the margin of error for this survey is approximately 0.0681, or 6.81%. This means that if we were to conduct the same survey multiple times with different samples, we would generally expect the true population proportion to fall within 6.81% of the observed proportion (60%).