The mean of a normally distributed group of weekly incomes of a large group of executive is $1,000 and the standard deviation is $100. What is the z-score for an income of $1,100?
Answer(s): E
z = (x-u)/sigma = 1100 - 1000/100 = 1.
A mining firm has purchased a derivative security to partially hedge itself against the losses caused by fluctuations in a base metal price. The security pays a million dollars if the metal price falls below $2 per pound. Otherwise, it pays $100,000. If the expected rate of return on the security is 21% and the security costs $400,000, what is the probability that the metal price will remain above $2 per pound?
Answer(s): C
If p is the probability of the metal price remaining above $2/pound, then the expected total return on the derivative security is (p * 100,000 + (1-p)*1,000,000)/400,000 = (10-9p)/4. Since the expected return is given to be 21%, we get (10-9p)/4 = 1.21. Therefore, p = 57.33%.
Walter Jennings, a quantitative analyst with Smith, Kleen & Beetchnutty Brokerage, has just been informed of an important error in one of his recent statistical endeavors. Specifically, in one hypothesis test, Mr. Jennings failed to reject a null hypothesis that later was proven to be false. Which of the following best describes this type of error in hypothesis testing? Further, if the confidence level of the test were increased, would the probability of this error increase, decrease, or is this probability difficult to determine?
Answer(s): A
In this example, Mr. Jennings has incorrectly failed to reject a null hypothesis. This type of error in hypothesis testing is called a Type II error. In hypothesis testing, the Type I error is given more attention than the Type II error. A Type I error is defined as the act of incorrectly rejecting the null hypothesis. In most hypothesis tests, the probability of a null hypothesis is equal to the significance level of the test. A significance level of 0.01, for example, indicates that a 1% chance exists that the null hypothesis will be rejected when it is indeed true.Another way to think of the probability of a Type I error is to observe the following relationship:{Probability of a Type I error = (1 - confidence level)}.For example, a confidence level of 95% leaves a 5% probability of a Type I error occurring. If this confidence level were to increase to say, 98%, then the probability of a Type I error would reduce to 2%. While a relationship exists between the confidence level for a hypothesis test and the probability of a Type I error, the relationship between confidence levels and Type II errors is not as explicit. The probability of a Type II error is inherently difficult to quantify, and as such, few hypothesis tests call for a determination of the probability of a Type II error. If the confidence level of a hypothesis test to be increased, it would be difficult to determine what effect, if any, this would have on the probability of a Type II error.
If you owe a debt of $1,000 today and also owe $500 in 24 months, what single payment could you make 12 months from today that would pay off both of these debts, if interest is assessed at 8% per year, compounded monthly?
Answer(s): D
To solve this question, set the problem up as the sum of two compound interest calculations. Move the $1,000 from today over to month 12 and add it to the $500 brought back from month 24 to month 12. On the BAII Plus, press 12 N, 8 divide 12 = I/Y, 1000 PV, 0 PMT, CPT FV, which yields $1,083.00. Then press STO 1. Then press 500 FV, CPT PV which yields $461.68. Finally press + RCL 1 = to see the answer. On the HP12C, press 12 n, 8 ENTER 12 divide i, 1000 PV, 0 PMT, FV. Then press STO 1. Then press 500 FV, PV. Finally press RCL 1 + to see the answer.
On a very hot summer day, 5 percent of the production employees at Midland States Steel are absent from work. The production employees are to be selected at random for a special in depth study on absenteeism. What is the probability of selecting 10 production employees at random on a hot summer day and finding that none of them are absent?
This is a binomial probability. The probability of getting r successes out of n trials where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p) is given by: n!(p^r)[q^(n-r)]/r!(n-r)!.Here n = 10, r = 0,p = 0.05 and q = 0.95. Therefore we have 10!(0.05^0)(0.95^10)/0!10! = 0.599.
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