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 359 with 273 successes at a confidence level of 99.5%.M.E. = %

Sure, let's go through the steps to find the margin of error (M.E.) for the given sample.

1. Calculate the sample proportion ($\hat{p}$): $\hat{p} = \frac{\text{number of successes}}{\text{sample size}} = \frac{273}{359}$ $\hat{p} \approx 0.7604$

2. Find the critical value (z) for a 99.5% confidence level:

   • The confidence level is 99.5%, so the significance level ($\alpha$) is $1 - 0.995 = 0.005$.
   • Since the confidence level is two-tailed, we need to find the z-value that corresponds to $\alpha/2 = 0.0025$.
   • Using a standard normal distribution table or a calculator, the critical value $z$ for 99.5% confidence level is approximately 2.807.

3. Calculate the standard error (SE) of the sample proportion: $SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.7604 \times (1 - 0.7604)}{359}}$ $SE \approx \sqrt{\frac{0.7604 \times 0.2396}{359}} \approx \sqrt{\frac{0.1821}{359}} \approx \sqrt{0.000507} \approx 0.0225$

4. Calculate the margin of error (M.E.): $M.E. = z \times SE = 2.807 \times 0.0225 \approx 0.0631$

5. Convert the margin of error to a percentage: $M.E. \approx 0.0631 \times 100 \approx 6.31\%$

So, the margin of error (M.E.) that corresponds to a sample of size 359 with 273 successes at a confidence level of 99.5% is approximately 6.31%.