Final answer:
To construct a 98% confidence interval with a margin of error of no more than 7, the required sample size is 245 when rounded up to the nearest whole number.
Step-by-step explanation:
To determine the sample size needed for a 98% confidence interval with a margin of error of no more than 7 to estimate a proportion, we begin with the formula for the margin of error (E) associated with a proportion:
E = z * sqrt[(p_hat * (1 - p_hat)) / n]
Since we don't know the true population proportion, we use p_hat = 0.5 for the maximum variability, which gives the largest sample size. For a 98% confidence interval, we'd use the z-value associated with the middle 98% of a standard normal distribution, which is approximately 2.33 (z-value for a 99% confidence interval is 2.576, but for 98% it's a bit less). Writing the formula in terms of the sample size (n) and solving for n:
n = (z^2 * p_hat * (1 - p_hat)) / E^2
Substitute the values (E = 0.07, z = 2.33, and p_hat = 0.5) into the equation:
n = (2.33^2 * 0.5 * (1 - 0.5)) / 0.07^2
Calculate and round up to the nearest whole number. After calculation, you will find that the required sample size is between 240 and 245. Rounding to the nearest whole number, the answer is in option D: 245.