November 13, 2003
Where is theta?
The point of today's lecture is to establish a common structure underlying hypothesis testing and confidence intervals. This structure is found in tests with variance known, variance unknown, sample proportions, and two means.
Can Bob reach Jim? Can Jim reach Bob?
Student's t distribution corrects for the problem that we do not know the population variance of X. We use the sample variance of X and refer to the inferred standard deviation of the mean as the standard error of the mean.
Recall our standard setup to prepare for a "t test" discussion:
How to conduct a "t test":
We concluded with a first pass at applying this structure to the "two means" setting.
Ch. 7, #30
In #29, we are supposed to recognize that p(1-p)/n is largest for p = 0.5. The standard error is then no larger than 0.5 / sqrt(n).
We want a 95% confidence interval of plus or minus 0.03. This means 1.96 x 0.5 / sqrt(n) = 0.03 or sqrt(n) = 1.96 x 0.5 / 0.03 = 32.6667. The required n is 1067.1111 or 1068 observations.
Ch. 7, #14
(a) In other words, we want the 90% confidence interval to be plus or minus 1.0 from the mean. The standard error of the mean is 3.8/sqrt(n). The 5% one-tail critical value is 1.645 from either the standard normal table or the bottom row of the t table. (Please look up both so you understand this point.)
So, we want 1.645 x 3.8/sqrt(n) = 1.0. Or, sqrt(n) = 1.645 x 3.8 = 6.2510. Squaring shows that the required n is 39.075, whick mean n = 40 to strict inequalities
(b) To switch to a 95% confidence interval, we would change 1.645 to 1.96. The required n is found by sqrt(n) = 1.96 x 3.8 = 7.446. This gives us n = 55.47 or 56 observations.
(c) If we relax the requirement to a confidence interval of plus or minus 1.5 we have 1.96 x 3.8/sqrt(n) = 1.5 or sqrt(n) = 1.96 x 3.8 / 1.5. So, sqrt(n) = 4.9653. The required n is 24.65 or 25 observations.
Note that here I did the calculations. I am also not real impressed by the answer to (c) in the back of the book.