1. 1 Test a hypothesis about a mean Formulate hypothesis about mean, e.g., mean starting income for graduates from WSU is $25,000. Get random sample, say n=400 Calculate mean, sd, and se: say mean = $22,352, sd = $13,873, se = $693.65 Check to see if sample mean is in critical region

2. 2 Quick review Mean Standard Deviation Standard Error of the Mean

3. 3 Test hypothesis mean is $25,000 $25,000 Mean: $23,352 sd: $13,873 n=400 se: $693.65 $25,000 + 2* $693.65 = $26,387 $25,000 - 2* $693.65 = $23,613 Sample Mean REJECT HYPOTHESIS Critical Region

4. 4 Create a 95% confidence interval We want to know what is the mean starting income for graduates from WSU. Get random sample, say n=400 Calculate mean, standard deviation: say mean = $23,352, sd = $13,873 Calculate the standard error: $13,873 / 20 = $693.65

5. 5 Create 95% confidence interval Use formula: Low end = $23,352-2*$693.65=$21,964 High end= $23,352+2*$693.65=$24,739 Interval is: $21,964 - $24,739 note this does not include $25,000

6. 6 Always a mean? Proportions are a form of mean, but if the population proportion is known or estimated, its variance is given. Mean Where X is either 0 or 1 sd se

7. 7 Always a mean? Sometime we wish to test a difference between two means. We consider paired data now and independent samples later. Paired data would exist if we sampled pairs of cases, e.g., husband and wife.

8. 8 Paired difference of means Calculate difference for each pair, for example age of husband and age of wife. This difference is a new variable Mean sd Standard error

9. 9 Hypothesis of no difference For paired data, this amounts to a hypothesis that d-bar is zero. 0 Plus 2 standard errors Minus 2 standard errors

10. 10 Some examples (husband - wife) Var | Mean Std. Err. t P>|t| age | 1.677316.2001067 8.38211 0.0000 income | 6.193548.8836383 7.00914 0.0000 love | 2.852716.6497145 4.39072 0.0000 depress | -2.055539.7851498 -2.61802 0.0093 esteem |.659164.3386991 1.94616 0.0525 efficacy | -.4057508.2331804 -1.74007 0.0828 What decisions do we make in each case?

11. 11 Decisions Reject hypothesis Correctly --- ??? Power of the test Error --- 5% Type I or alpha error Fail to reject hypothesis Correctly --- 95% Error --- Type II or beta error

12. 12 One more test Sign test for median A nonparametric test -- not based on a measurement Like a one sample t test for the mean Especially good for small samples Based on binomial distribution

13. 13 Sign test for median Hypothesize a particular median value Count number of observations above and below the hypothesized value If hypothesis is true, there should be approximately equal numbers of observations above and below Probability of being above is.5 if the hypothesis is true

14. 14 Consider Hypothesize median age: 23 Obs: 24 25 27 23 23 22 18 35 19 26 20 28 24 26 21 (14 cases) Count: above: 8 below: 4 at: 2

15. 15 Find binomial probability Find the probability of getting a split of 8 to 4 (ignore the two cases at the hypothesized median) Use the binomial Find p(8) or p(9) or p(10) or p(11) or p(12), assuming p of being above median is.5

16. 16 Calculate.1208.0537.0161.0029.0002.1937 Fail to reject hypot.

17. 17 Large samples What happens when sample size increases? Consider sample sizes of 4, 25, 100, 400. Assume: s = 5 se = 2.5, 1,.5,.25 for n = 4 25 100 400

18. 18 How big a sample do we need? To show a 1 year difference in ages is significant at.05 level for husbands and wives. Assume sd of age is 5 years Want se of.5 so that 2 se is 1 years

19. 19 Sample size? 01 Difference in years.5 = 5 / sqrt(n) Square both sides.25 = 25 / n n = 100