Question

In constructing a 90% confidence level estimate of the mean when the population standard deviation (σ) is known what will be your z score used in the formula? HINT: Be sure to review page 236 "Finding Z scores from Known Areas - Special Cases" and Tabel A-2. If you identified it as 1.64 contact me in the DB and I will provide partial credit. Please see my lecture "CIE of the mean z and t examples 2015" as to an explanation as to why the correct answer is 1.645. The Z value is a function of the level of confidence. The level of confidence is determined by whoever is performing the analysis and is often 90%, 95%, or 99%. Once the level of confidence is selected the “Z value” can be found. To find the Z value you divide the level of confidence by 2 and use the normal curve table. For instance if the level of confidence is 95%, you divide .95 by 2 and get .4750 (for our purposes carry the number out to four places because our normal curve table goes out four places). Using the normal curve table you find .4750 in the “areas under the normal curve” section of the table, i.e. the ‘heart’ of the table. Once you find that number on/in the table you read across to the first column (the Z column) simultaneously reading to the top row (the z row), and select the number in the z column (1.9) and the number in the z row (0.06) combining the two thus getting the Z value of 1.96. The Z value corresponding to a level of confidence of 90% is 1.645. Note .90/2 = .4500. In the areas under the normal curve table you see .4495 and .4501, but not .4500. The Z value for .4495 is 1.64. The Z value for .4501 is 1.65. Because .4500 is half way between .4495 and .4501 we use the half-way point between their corresponding Z values of 1.64 and 1.65 which is 1.645.

Answer 1