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36 36 Discussion A Unit 4. The Basics of Hypothesis Testing A There are many basic types of logical reasoning in mathematics and statistics but implication and probabilistic reasoning are two common types, implication being found in pure mathematics in the form of theorems, and probabilistic reasoning found in statistics in the form of hypothesis testing. Both are connected and we will explore both in this discussion Definition. A sentence which has the property of being either true or false is called a proposition (or a statement). Examples: Today in June 3 (false at the time of this writing). Twenty minus five is fifteen (true). There are 120, 221 birds flying above Merritt Island right now (we don't know whether it is true or false, but this is still a proposition, since it is either true or false, irrespective of our knowledge). Definition. Let P and Q be two propositions. Then P is said to imply Q if it is impossible for Q to be false whenever P is true. An implication is a pair of propositions P and Q, where P implies Q. It is common to use the phrase, "If P then Q" instead of "P implies Q". Notice that if the statement, "If P then Q" is true, this means that P implies Q and hence if P is true, then Q can not be false (by definition). Examples: Let P be, "X is an integer which is divisible by 2", and let Q be 'X ends in a digit in the set {0, 2, 4, 6, 8}". Then P implies Q, since if P really is an integer which is divisible by two, then P is an even number, and even numbers always end in one of the digits in the set {0, 2, 4, 6, 8). Let P be the statement, "X is a number in the interval [0, 10]" and let Q be the statement, "X is not equal to 11". Then P implies Q, since if X really is a number in the closed interval [0, 10], then since 11 is greater than 10, X can not equal 11.
Notice that this type of reasoning has already been used in our class to understand the concept of the degree of freedom. Namely, if there n real numbers whose mean is a given number , then n - 1 of those numbers can be freely chosen to have any value, but since the mean is a given number, the last number can not be freely chosen (and still have the mean equal to ). So there are n - 1 degrees of freedom. From the very definition of implication, we can form a basic principle of reasoning that goes like this: If P then Q Q is false Hence P must be false. Notice that if the first statement is true, and the second statement is true, then the last statement must be true by definition of implication and what it means for the first two statements to be true. Here is a specific example. If X is an even number, then X ends with a digit in the set {0, 2, 4, 6, 8] X does not does not end in a digit in the set {0, 2, 4, 6, 8} Hence it is false that X is an even number. (In this case the statement P is "X is an even number" and the statement Q is "X ends with a digit in the set (0, 2, 4, 6, 8}" The above example leads to this line of abstract reasoning: If P then Q Q is false Hence P must be false. This formula of reasoning is so common that it is often one of the first forms of valid deductive reasoning studied in a logic course (where it is often called modus tollens, the Latin name is a clue that it has been known for quite a long time).
Suppose we weaken implication so that if P is true, then Q is not certain, but highly probable. In this case we do not say that P implies Q, but P gives strong evidence for Q. Example. Let P be the statement, "X is a number chosen randomly from a box of 1000 different even integers", and let Q be the statement, "X ends with a digit in the set {2, 4, 6, 8}" Now in this case, P does not imply Q, since Q could be false and P still be true. For example if X were the number 120, then X is even but does not end with 2, 4, 6, or 8. But we can say that if P is true, then it Q is more likely to be true than a proposition like "X ends with a 0", or put another way, the truth of P gives strong evidence for the truth of Q. This is probabilistic reasoning. In hypothesis testing in statistics, the probabilistic reasoning is used this way. Let some hypothesis about a population parameter (P.P.) be given which contains an equal sign in some part of the statement (e.g. P.P = x, P.P ≤ x, P.P ≥ x). We will call this statement the null hypothesis. If the null hypothesis about a population parameter is true, then a sample statistic derived from a simple random sample taken from population will support the null hypothesis (this follows from the Central Limit Theorem and others). The sample statistic taken from the sample of the population did not support the null hypothesis (within a specified degree of confidence). Hence we reject the null hypothesis as probably being false in favor of another hypothesis (which turns out to be the negation of the null hypothesis.)
Here is an example made up in the spirit of the examples given above. Suppose we have 1000 even integers in a box, and we want to know if they were chosen randomly (or perhaps someone has put more numbers of a certain type in the box, like numbers which are even but end in O). We wish to test whether this is the case. Since even numbers end in one of the digits {0, 2, 4, 6, 8} we expect P(X = d) = 1/5, where d is one of the digits 0, 2, 4, 6, or 8. (In other words, the probability that a randomly chosen integer from the box has a 1/5 chance of being a specific digit in the set {0, 2, 4, 6, 8]. So we perform a test to find out! The test consists of two parts. Taking a random sample of 89 numbers and recording the proportion which end in 0. The second part is mathematical, where we see if the actual outcome of our test supports the null hypothesis. The null hypothesis will be: The proportion of even numbers which end in a 0 in the box is 1/5 (or 0.20) The alternative hypothesis will be: The proportion of even numbers which end in a 0 is not 1/5. We randomly choose a sample of 89 numbers taken from the box, and find that the proportion of even numbers ending in 0 is 0.26 Does that finding support our hypothesis that the proportion of even numbers which end in a 0 in the box is 0.20 or not? Statistics allows us to test it!
(Q1). Find three examples of propositions P and Q such that P implies Q. (Q2). Find three examples of propositions P and Q such that if P is true, then Q is probably true, (but not certain). (Q3). If the proportion of even numbers that end in zero in a sample of 89 were 0.26, do you think that supports the null hypothesis or not? What about 0.29? (For this last question, you can state your intuition. If you have advanced enough with the homework to calculate the actual answer, then tell us the answer -- use a confidence level of 95%, the test is two tailed)