### CFA Practice Question

There are 434 practice questions for this study session.

### CFA Practice Question

To estimate the average cost of a food-shopping event, Delcore Inc. randomly sampled 100 shoppers and found a sample mean of \$72. Assuming a population standard deviation of \$5, a 99% confidence interval for average cost for the food-shopping event is ______.
A. \$70.71 < m < \$73.29
B. \$71.18 < m < \$72.82
C. \$59.12 < m < \$84.88
Explanation: For a 99% confidence interval, we find z(0.005), the cutoff for the top 0.5% of the normal distribution. Looking up 0.995 in the middle of the table, the reading to the row/column values, we get 2.575. Working with the formula for E, we get E = 1.29. So, the 99% confidence interval is \$72 - 1.29 < m < \$72 + 1.29 or \$70.71 < m < \$73.29.

User Comment
danlan Where comes the value 2.575?
danlan From z-table, z(2.575)=0.495, then we get 1.29=2.575*5/sqrt(100)
dimanyc Why do we use 5/sqrt(100) when we the st dev of the population is known and the sample is large?
TonyShen Because we are using the Sample Mean instead of the Pop Mean.
dipu617 I dont see the z-table in the question. Will it be provided in the exam?
floydbite not necessarily for 90, 95, 99% values
tflies51 What is the formula for E?
philerup Memorize 90->1.65, 95->1.96. 99->2.58
mmccoy Where does 1.29 come from?
megageorge Exactly, whats formula for E?
chris21Feb E = 2.58 x 5 / sqrt(100) = 2.58 / 2 = 1.29