Assume that a sample is used to estimate a population proportion p. Find the margin of error M.E. that corresponds to a sample of size 344 with 96 successes at a confidence level of 98%.
Margin of error = (2.33)[√(pq/n)]
...where p = 96/344, q = 1 - p, and n = 344
Note: 2.33 represents 98% confidence interval.
Convert all fractions to decimals.
Plug the values into the formula and calculate.
I let you take it from here.
To find the margin of error (M.E.), we can use the formula:
M.E. = Z * √((p * (1 - p)) / n),
where:
Z is the z-score corresponding to the desired confidence level,
p is the estimated population proportion,
(1 - p) is the complement of the estimated population proportion,
and n is the sample size.
In this case:
Sample size (n) = 344,
Number of successes (96),
Confidence level = 98%.
First, we need to find the z-score that corresponds to a 98% confidence level. The z-score represents the number of standard deviations the sample estimate is from the mean. For a 98% confidence level, the z-score is approximately 2.33.
Next, we calculate the estimated population proportion (p) by dividing the number of successes (96) by the sample size (344):
p = 96 / 344 ≈ 0.2791.
Now, we can substitute these values into the margin of error formula:
M.E. = 2.33 * √((0.2791 * (1 - 0.2791)) / 344).
Calculating this, we get:
M.E. ≈ 2.33 * √(0.1936 / 344) ≈ 0.0641.
Therefore, the margin of error (M.E.) that corresponds to a sample of size 344 with 96 successes at a confidence level of 98% is approximately 0.0641.