using the mean standard deviation 66 and confidence level 99

3. Using the mean, standard deviation and confidence level from the previous question, how large a sample size would you need to use to achieve a margin of quesion, how large a sample size would you need to use to achieve a margin of error of 7 dollars? (25 points)

Solution

Necessary Sample Size using

Z- score for 99% CI is 2.576

std. dev = 66

Margin of error = 7 dollars

Step 1: Find z a/2 by dividing the confidence interval by two, and looking that area up in the z-table:
.99/2 = 0.495. The closest z-score for 0.495 is 2.58.

Step 2:Multiply step 1 by the standard deviation.
2.58 * 66 = 170.28

Step 3:Divide Step 2 by the margin of error. Our margin of error (from the question), is 7.
170.28/7 = 24.3257

Step 4:Square Step 3.
24.3257 * 24.3257 = 591.7404 which is approximately 592

Therefore we would need a sample size of 592 using the mean, standard deviation 66 and confidence level 99% to achieve a margin of margin error of 7 dollars