Hypothesis test flow chart frequency data Measurement scale number of variables 1 basic χ 2 test (19.5) Table I χ 2 test for independence (19.9) Table I 2 correlation (r) number of correlations 1 Test H 0 :  =0 (17.2) Table G 2 Test H 0 :   =   (17.4) Tables H and A number of means Means Do you know  ? 1 Yes z -test (13.1) Table A t -test (13.14) Table D 2 independent samples? Yes Test H 0 :   =   (15.6) Table D No Test H 0 : D=0 (16.4) Table D More than 2 number of factors 1 1-way ANOVA Ch 20 Table E 2 2-way ANOVA Ch 21 Table E No START HERE

Chapter 14: Testing Hypotheses about the Difference between Two Independent Groups Example: The mothers of 10 left handed students in this class have a mean height of 64.1 inches with SS X = 60.9, and the mothers of the 86 right handed students in this class have a mean height of inches with SS Y = Are the left hander’s mothers significantly different than the right-hander’s mothers, using  =.05? This chapter is all about conducting a t-test on problems like this, where we’re testing for differences in the means from two independent samples. Setting up the hypotheses: let X be heights of left-handed student’s mothers and Y be heights of the right handed student’s mothers. Null Hypothesis, H 0 :  X =  Y Alternative Hypothesis H A :  X ≠  Y

To calculate the standard error of the estimate, we need to find the ‘average’ standard deviation from the two samples. This is done by calculating the ‘pooled’ standard deviation: The standard error of the estimate is calculated from the pooled variance by: And then calculate our value for the t-distribution with n X +n Y -2 degrees of freedom. This thing is zero when we don’t expect a difference of means under H 0

H 0 :  X =  Y H A :  X ≠  Y Back to our example: Looking up Table D for df = (10-1)+(86-1) = 94 (we’ll use 95), two-tailed,  =.05, tcrit = Decision: Our observed value of t does not fall in the critical region, so we fail to reject H 0

To apply a t-test on the difference between two independent means, we must assume: 1)Each sample is drawn at random from its respective population. This is the basis rule of random sampling, but for two populations. 2)The two random samples must be independently selected. i.e. there should be no expected correlation between X and Y. 3)Samples are drawn with replacement This is almost never done, but if the population is large it doesn’t matter much. 4) The sampling distribution of the difference between means is normal. This where the central limit theorem comes to the rescue. The larger the sample size, the more normally distributed the difference of means will become. Other names for this test are ‘Unpaired two-sample t-test’ and ‘Independent two- sample” t-test.

If you are given the standard deviations s X and s Y instead of SS X and SS Y, then you can use the fact that (note the lower case s for estimating population standard deviation): Or To calculate: and

Example: The heights of the 45 students in our class with fathers above 70 inches has a mean of 66.8 inches and a standard deviation of 4.14 inches. The heights of the remaining 51 students has a mean of 64.9 and a standard deviation of 3.62 inches. Is there a significant difference between the heights of these two groups of students using  =.05? Answer: We will use a two-tailed independent samples t-test with H O :  X =  Y, and H A :  X ≠  Y Since you’ve been given just means, standard deviations and sample sizes, you should use this estimate for the standard error of the difference between means:

Calculating t with n X +n Y -2 = = 94 degrees of freedom: Look up t crit in table D with in one tail: t crit = So our rejection region will be t Since our observed t = , we reject H 0. Using APA format: There is a significant difference between the heights of students with tall fathers (M = 66.8, SD = 4.14) and students with less tall fathers (M = 64.9, SD = 3.62), t(94) = 2.4, p<.05. For fun at home, would we still reject H 0 using  =.01?

Plotting bar graphs with ‘error bars’ It is useful to compare means by plotting them as bar graphs with error bars representing plus and minus one standard error of the mean. Example (again): The heights of the 45 students in our class with fathers above 70 inches has a mean of 66.8 inches and a standard deviation of 4.14 inches. The heights of the remaining 51 students has a mean of 64.9 and a standard deviation of 3.62 inches.

Rule of thumb: If the error bars representing standard errors of the mean overlap, a one-tailed t-test will probably fail to reject H O with  =.05. In general, the bars need a gap to reach statistical significance.

In the pursuit of science, you measure the anger of 10 fluffy and 10 noisy athletes and obtain for fluffy athletes a mean anger of and a standard deviation of 19.21, and for noisy athletes a mean of 92.8 and a standard deviation of Using an alpha value of 0.05, is the mean anger of fluffy athletes significantly greater than for the noisy athletes? We fail to reject H 0. (t(18) = 1.08, t crit = ) The anger of fluffy athletes is not significantly greater than the anger of noisy athletes. p =

Your Psych 315 professor asks you to measure the money of 10 laughable and 10 flimsy candy bars and obtain for laughable candy bars a mean money of 106 and a standard deviation of 12.25, and for flimsy candy bars a mean of and a standard deviation of Using an alpha value of 0.05, is the mean money of laughable candy bars significantly greater than for the flimsy candy bars? We fail to reject H 0. (t(18) = -0.18, t crit = ) The money of laughable candy bars is not significantly greater than the money of flimsy candy bars. p = laughableflimsy money candy bars

Tomorrow you measure the clothing of 10 kind and 10 red geeks and obtain for kind geeks a mean clothing of and a standard deviation of 14.91, and for red geeks a mean of 87.4 and a standard deviation of Using an alpha value of 0.05, is the mean clothing of kind geeks significantly greater than for the red geeks? We reject H 0. (t(18) = 3.62, t crit = ) The clothing of kind geeks is significantly greater than the clothing of red geeks. p = kindred clothing geeks

Just for fun, you measure the conductivity of 14 straight and 12 strange teams and obtain for straight teams a mean conductivity of 19.4 and a standard deviation of 11.74, and for strange teams a mean of 13.2 and a standard deviation of Using an alpha value of 0.05, is the mean conductivity of straight teams significantly greater than for the strange teams? We fail to reject H 0. (t(24) = 1.34, t crit = ) The conductivity of straight teams is not significantly greater than the conductivity of strange teams. p = straightstrange conductivity teams

On a dare, you measure the age of 20 capricious and 19 enthusiastic Asian food and obtain for capricious Asian food a mean age of 22.8 and a standard deviation of 12.05, and for enthusiastic Asian food a mean of 12.5 and a standard deviation of Using an alpha value of 0.05, is the mean age of capricious Asian food significantly greater than for the enthusiastic Asian food? We reject H 0. (t(37) = 2.39, t crit = ) The age of capricious Asian food is significantly greater than the age of enthusiastic Asian food. p = capriciousenthusiastic age Asian food

I measure the conduct of 17 aware and 19 easy beers and obtain for aware beers a mean conduct of 26.5 and a standard deviation of 14.9, and for easy beers a mean of 17.6 and a standard deviation of Using an alpha value of 0.05, is the mean conduct of aware beers significantly greater than for the easy beers? We reject H 0. (t(34) = 1.78, t crit = ) The conduct of aware beers is significantly greater than the conduct of easy beers. p = awareeasy conduct beers

We decide to measure the visual acuity of 12 husky and 16 wrathful Seattleites and obtain for husky Seattleites a mean visual acuity of 50.1 and a standard deviation of 11.45, and for wrathful Seattleites a mean of 42.4 and a standard deviation of Using an alpha value of 0.05, is the mean visual acuity of husky Seattleites significantly greater than for the wrathful Seattleites? We fail to reject H 0. (t(26) = 1.47, t crit = ) The visual acuity of husky Seattleites is not significantly greater than the visual acuity of wrathful Seattleites. p = huskywrathful visual acuity Seattleites

I go and measure the water of 13 super and 20 tan brains and obtain for super brains a mean water of 41.5 and a standard deviation of 14.32, and for tan brains a mean of 31.9 and a standard deviation of Using an alpha value of 0.05, is the mean water of super brains significantly greater than for the tan brains? We reject H 0. (t(31) = 1.77, t crit = ) The water of super brains is significantly greater than the water of tan brains. p =

Effect size and power for the independent samples t-test Effect size is the difference between means divided by the estimate of the standard deviation. For tests of the difference between means our estimate of the standard deviation is the pooled standard deviation: Where (as before) or Again, by convention, effect sizes of 0.2 are small, 0.5 are medium and 0.8 are large.

Example (again): The mothers of 10 left handed students in this class have a mean height of 64.1 inches with SS X = 60.9, and the mothers of the 86 right handed students in this class have a mean height of inches with SS Y = What is the effect size? Answer: these are our parameters: This is a very small effect size.

Example (again): The heights of the 45 students in our class with fathers above 70 inches has a mean of 66.8 inches and a standard deviation of 4.14 inches. The heights of the remaining 51 students has a mean of 64.9 and a standard deviation of 3.62 inches. What is the effect size? Answer: This time, we have been given standard deviations, so we’ll use: This is a medium effect size. Remember it was a ‘significant’ t-test.

Power curves for independent samples t-test: Remember, the power of a test is the probability of correctly rejecting H 0 when it is false. Power depends on , n X, n Y, and the effect size.

Example: Suppose you randomly chose two groups of 25 subjects from the general population and gave a new drug designed to enhance IQ to one group and a placebo to the other. Suppose you later give all subjects an IQ test and find that the group that got the new drug had a mean IQ of 110 and a standard deviation of 12, and the placebo group had a mean of 102 and a standard deviation of 16. 1)Conduct an independent sample t-test with  =.05 to determine if the drug had a significant effect on IQ Answer: We’ll test the null hypothesis that  X -  Y = 0, where X is the drug group and Y is the placebo group. Since we’re given standard deviations, we’ll use:

Calculating t with n X +n Y -2 = = 48 degrees of freedom: Look up t crit in table D with in one tail (using df = 50) t crit = So our rejection region will be t Since our observed t = 2.0, we fail to reject H 0, so we cannot conclude that there is a significant difference between the heights of these two groups. We just missed the  =.05 cutoff for rejecting H 0

Example contd. 2) What was the effect size in this example? Answer: We use the pooled standard deviation: and then effect size:

Example contd. 3) Estimate the power value (to the nearest tenth) from this test given this effect size? We have a power of about 0.5

Example contd. 4) How large of a sample size would we need to obtain a power level of 0.8? We’d need a sample size of about 50 people in each group.

Example: 37 of the 96 students in the class were born first in their family. The GPA’s of these first-born students has a mean of 3.41 and a standard deviation of The GPA’s of the remaining 59 students has a mean of 3.33 and a standard deviation of Using an a value of 0.05, is there a significant difference in the GPA’s of first-born students and the rest of the class?