Assume that a sample is used to estimate a population proportion P. Find the margin of error E is that corresponds to the given statistics in confidence level. Round the margin of error to four decimal places95% confidence; the sample size is 10,000, of which 40% are successes
To find the margin of error (E) in estimating a population proportion (P) with a given confidence level, you can use the following formula:
E = Z * sqrt((P * (1-P))/n)
Where:
E = Margin of error
Z = Z-score corresponding to the desired confidence level (in standard deviations)
P = Sample proportion (in decimal form)
n = Sample size
In this case, you need to find the margin of error for a 95% confidence level, given that the sample size is 10,000 and 40% of the sample are successes
To find the margin of error (E) in estimating a population proportion (P) with a given confidence level, you can use the following formula:
E = Z * sqrt((P * (1-P))/n)
Where:
E = Margin of error
Z = Z-score corresponding to the desired confidence level (in standard deviations)
P = Sample proportion (in decimal form)
n = Sample size
In this case, you need to find the margin of error for a 95% confidence level, given that the sample size is 10,000 and 40% of the sample are successes.
First, let’s convert the confidence level to the corresponding Z-score. For a 95% confidence level, we can use a Z-score of 1.96 (which corresponds to the area under the standard normal distribution curve).
Next, we need to convert the sample proportion to decimal form. The sample proportion is 40%, so in decimal form, it is 0.40.
Now, we can plug in the values into the formula:
E = 1.96 * sqrt((0.40 * (1-0.40))/10,000)
Calculating this equation step by step:
E = 1.96 * sqrt((0.40 * 0.60)/10,000)
E = 1.96 * sqrt(0.024/10,000)
E = 1.96 * sqrt(0.0000024)
E ≈ 1.96 * 0.00155
E ≈ 0.0030396 (rounded to four decimal places)
Therefore, the margin of error (E) for estimating the population proportion (P) with a 95% confidence level is approximately 0.0030.
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