INTRODUCTION TO HYPOTHESIS TESTING From R. B. McCall, Fundamental Statistics for Behavioral Sciences, 5th edition, Harcourt Brace Jovanovich Publishers, New York 1990

OUTLINE Population parameters - sample statistics How to test hypotheses - “Null Hypothesis” H0 2 examples to illustrate - the normal curve General Procedures - Assumptions General Procedures - Hypotheses General Procedures - Significance level General Procedures - Decision rules General Procedures - Reject/not reject H0 General Procedures - Possible errors

POPULATION PARAMETERS Assume that we know that, in the entire population, non-dragged subjects can correctly recall, on average, 7 of 15 learned nouns, with a standard deviation of 2. Thus, population parameters: =7, =2 We also have sample statistics: For a sample of n subjects, we have the mean X and the standard deviation s.

QUESTION & HYPOTHESES Will subjects perform differently if they are given the drug physostigmine? Null Hypothesis H0: Drug will have no effect on performance. Alternative Hypothesis H1: Drug will have some effect on performance.

EXAMPLE 1: SINGLE CASE Assume one subject took drug and correctly recalled 11 nouns. Reject H0 if the subject’s score falls into the most extreme =5% of the distribution of non-drugged subjects. Score could fall in the extreme low /2=2.5% or extreme high 2.5%, so test against /2=2.5% probability.

EXAMPLE 1 (continued) Translate his/her score into a z score (z score is called the standard normal deviate; has mean=0 and standard deviation=1): z=(Xi-)/ = (11-7)/2 = 2.00 In this case, z score is 2 standard deviations above . Look up the Table for Normal Curve to determine if H0 can be rejected.

EXAMPLE 1 (continued) H0 1- /2 Here is a pictorial representation of the situation: H0 /2 1-

Proportions of area under Normal Curve .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 ... … 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 2.9 0.4981 0.4982 0.4983 0.4984 0.4985 0.4986 P(z<2) = 0.5 + 0.4772 = 0.9772; P(z≥2) = 0.0228 < 0.025. Thus, we must reject H0.

EXAMPLE 2: SINGLE GROUP Assume 20 subjects took drug and correctly recalled 8.3 nouns. Reject H0 if the subjects’ score falls into the most extreme =5% of the distribution of non-drugged subjects. Score could fall in the extreme low 2.5% or extreme high 2.5%, so test against 2.5% probability.

EXAMPLE 2 (continued) Translate group’s score into a z score: z=(X-) / X = (X-)/[x/√n] = (8.32-7)/(2/√20) = 2.95 In this case, z score is 2.95 standard deviations above . Look up the Table for Normal Curve to determine if H0 can be rejected.

Proportions of area under Normal Curve .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 0.2 ... … 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817 2.1 2.9 0.4981 0.4982 0.4983 0.4984 0.4985 0.4986 P(z<2.95) = 0.5 + 0.4984 = 0.9984; P(z≥2) = 0.0016 < 0.025. Thus, reject H0.

ASSUMPTIONS Assumptions are statements of circumstances in the population and the samples that the logic of the statistical process requires to be true, but that will not be proved or decided to be true. In Example 2, two assumptions were made: The 20 subjects that the drug was administered to were randomly and independently selected from the non-drugged population. The sampling distribution of the mean is normal in form.

HYPOTHESES Hypotheses are statements of circumstances in the population and the samples that the statistical process will examine and decided their likely truth or validity. Null Hypothesis H0: The observed sample mean is computed on a sample drawn from a population with =7; that is, the drug has no effect. Alternative Hypothesis H1: The observed sample mean is computed on a sample drawn from a population with ≠7; that is, drug has some effect (either positive or negative; two-tailed test).

SIGNIFICANCE LEVEL The significance level (or critical level), symbolized by  (alpha), is the probability value that forms the boundary between rejecting and not rejecting the null hypothesis. Usually, =0.05. If H0 can be rejected, the result is said to be “significant at the 0.05 level”. This is sometimes written “p<0.05”, where p stands for the probability that H0 is true.

DECISION RULES Decision rules are statements, phrased in terms of the statistics to be calculated, that dictate precisely when the null hypothesis H0 will be rejected and when it will not. In our case, we used the following: If the observed sample mean deviated from the population mean to an extent likely to occur in the non-drugged population less than 5% of the time, we reject H0. Notice that we decided on a two-tailed test BEFORE we took a sample of 20 subjects. From the Table, p < 0.05 corresponds to an “extreme” tail area of 0.025, or a cumulative area of (1-0.025 = 0.975) = 0.500+ 0.475, which corresponds to  z ≥ 1.96. .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 0.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.0753 ... … 1.9 0.4713 0.4719 0.4726 0.4732 0.4738 0.4744 0.4750 0.4756 0.4761 0.4767 2.0 0.4772 0.4778 0.4783 0.4788 0.4793 0.4798 0.4803 0.4808 0.4812 0.4817

YOU CAN ONLY REJECT OR NOT REJECT H0 The statistical procedure is designed to test only one hypothesis, H0. Thus, depending on the results, you can only reject, or not reject, the Null Hypothesis H0. If p>, the only thing you can do is not reject H0. You cannot accept the Null Hypothesis H0. In our examples, the drug may still have an effect, but the effect may be very small. You may need a larger sample to observe an effect. If p≤, the only thing you can do is reject H0. You cannot accept the Alternative Hypothesis H1. The decision to reject H0 is not equivalent to accepting H1.

Correct non-rejection POSSIBLE ERRORS H1  H0 /2 1- 1- ACTUAL --> SITUATION DECISION H0 IS TRUE H0 IS FALSE REJECT H0 Type I error p =  Correct rejection P = 1 -  (POWER) DO NOT REJECT H0 Correct non-rejection P = 1 -  Type II error p = 

SUMMARY Adopt appropriate decision rules. State, examine and justify the assumptions. State the Null Hypothesis H0 and the Alternative Hypothesis H1. Translate the raw scores into z scores. Look up the normal distribution Tables to reject or not reject H0.