Assume that a random sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level.
95% confidence the sample size is 2096 of which 25%are successes
The margin of E= (round to four decimal places as needed)
Margin of error = (1.96)[√(pq/n)]
Note: 1.96 represents 95% confidence.
For p in your problem: .25
For q: 1 - p = q
n = 2096
I let you take it from here to calculate.
To find the margin of error (E) that corresponds to the given statistics and confidence level, we will use the formula:
E = Z * √(p̂(1 - p̂) / n)
Where:
E = Margin of error
Z = Z-score corresponding to the desired confidence level
p̂ = Sample proportion
n = Sample size
In this case, the confidence level is 95%, which means we need to find the Z-score for a 95% confidence level. The Z-score can be found using a standard normal distribution table or with the help of a statistical software. For a 95% confidence level, the corresponding Z-score is approximately 1.96.
Given that the sample size (n) is 2096 and the sample proportion (p̂) is 25% (or 0.25), we can substitute these values into the formula:
E = 1.96 * √(0.25(1 - 0.25) / 2096)
Now, let's calculate the margin of error (E):
E = 1.96 * √(0.25 * 0.75 / 2096)
E = 1.96 * √(0.1875 / 2096)
E = 1.96 * √0.000089441
E ≈ 1.96 * 0.009454
E ≈ 0.0185 (rounding to four decimal places)
Therefore, the margin of error (E) that corresponds to the given statistics and a 95% confidence level is approximately 0.0185.