SUPPOSE HYPOTHESIS TESTING the we think the mean value of a population of SAT scores is 455 with 100 four steps Introduction to Statistics in Psychology PSY 201 Professor Greg Francis Lecture 19 Hypothesis testing of the mean we take a sample of 144 from the population and calculate the sample mean of SAT scores X 535 1 State the hypothesis 2 Set the criterion for rejecting H0 3 Compute the test statistic 4 Decide whether to reject H0 Why clinical studies use thousands of subjects 2 3 RECAP OF LAST TIME RECAP OF LAST TIME DIFFERENT MEAN 1 State the hypotheses 3 Compute the test statistic suppose we had the same situation as before but we had instead found H0 455 z Ha 455 0 05 2 Set the criterion Find the z scores that identify the 5 of scores farthest from zcv 1 96 z X X 535 455 9 60 8 33 X 465 1 State the hypotheses H0 455 4 Decide Ha 455 z zcv so reject H0 0 05 the found sample mean would be very rare if H0 were true 2 Set the criterion Find the z scores that identify the 5 of scores farthest from zcv 1 96 4 5 6 DIFFERENT MEAN SAMPLE SIZE SAMPLE SIZE 3 Compute the test statistic suppose we had the same situation as before but we had instead found 3 Compute the test statistic z X X 465 455 z 1 20 8 33 4 Decide z zcv so do not reject H0 the found sample mean would not be very rare if H0 were true z X 465 with a sample size of n 500 1 State the hypotheses H0 455 Ha 455 X X we need to recompute X 100 X 4 47 n 500 465 455 z 2 24 4 47 0 05 0 05 4 Decide 2 Set the criterion z zcv so do reject H0 the probability that X 465 would be found by random sampling is greater than 05 Find the z scores that identify the 5 of scores farthest from zcv 1 96 sampling is less than 05 7 8 9 CLINICAL TRIALS COMMENTS DIRECTIONAL HYPOTHESIS often hear about medical studies that track thousands of patients several things are worth noting why do they need so many people a larger sample makes for less variation in the sampling distribution thus if the null hypothesis really is false you are more likely to reject it with a larger sample the found sample mean would be very rare if H0 were true the probability that X 465 would be found by random 1 Even when we reject H0 there is always a chance that it is true 2 Even when we do not reject H0 there is always a chance that it is false 3 The statement p 0 05 is about the statistic given the hypothesis not about the hypothesis We never conclude that H0 is false with probability 0 95 we choose a significance level indicates probability of Type I error earlier we split this error across the two tails of the sampling distribution 0 04 0 03 4 Technically we have done all of this before 0 02 if the null hypothesis is really true you are not more likely to reject it no extra mistakes with a larger sample size 5 No inclusion of knowledge about the direction of difference 0 01 0 6 We have to know population standard deviation 400 420 440 7 These techniques are quantifiable 10 11 460 Sample Mean 12 480 500 DIRECTIONAL HYPOTHESIS REGION OF REJECTION EXAMPLE if we only have to worry about one tail the region of rejection in that tail is larger we know that the sampling distribution is suppose we know or strongly suspect that if the sample mean X is different from the population mean it will be greater last 5 starts with a z score of 1 645 then we don t need to worry about the left side tail we can reject H0 when the difference between X and is smaller H0 455 0 4 Ha 455 1 Normal 2 Has a mean of 455 3 Has a standard error of the mean 100 X 8 33 n 144 0 3 0 2 0 04 0 03 0 1 0 02 0 4 2 0 2 4 X 0 01 0 400 420 440 460 480 500 Sample Mean 13 14 15 REGION OF REJECTION REGION OF REJECTION EXAMPLE area under the curve represents the probability of getting the corresponding sample means given that is where specified we convert sample mean to z score suppose everything was the same except we had hypotheses the extreme right tail of the sampling distribution corresponds to what should be very rare sample means 535 455 z 9 60 8 33 critical z score value is 1 645 z X X H0 455 Ha 455 then we would shift the region of rejection to the left tail greater than critical value 9 60 1 645 0 4 0 04 reject H0 0 03 0 4 0 3 0 02 0 3 0 2 0 01 0 2 0 1 0 400 420 440 460 Sample Mean 0 1 0 4 2 0 2 4 X 0 4 2 0 2 4 X 16 17 18 480 500 EXAMPLE TABLE the critical z score value becomes 1 645 NOTE you only need to know a few critical value z scores with our sample mean of X 535 and z 9 60 we cannot reject H0 0 4 0 3 0 2 0 1 0 4 2 0 2 4 Level of Level of Critical Significance Significance Value of Two tailed test One tailed test Test Statistic 0 20 0 10 1 282 0 10 0 05 1 645 0 05 0 025 1 960 0 02 0 01 2 326 0 01 0 005 3 291 0 001 0 0005 3 291 X 19 20 CONCLUSIONS NEXT TIME hypothesis testing when is unknown sample size t distribution directional test t test significance Do nuclear plants make you sick 22 23 SIGNIFICANCE vs IMPORTANCE something may make a big difference in the value of X but not be statistically significant something can be statistically significant without being practically important will you change your diet to avoid cancer depends on the diet and how much of a difference it makes 21