 # Testing means, part III The two-sample t-test. Sample Null hypothesis The population mean is equal to  o One-sample t-test Test statistic Null distribution.

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Testing means, part III The two-sample t-test

Sample Null hypothesis The population mean is equal to  o One-sample t-test Test statistic Null distribution t with n-1 df compare How unusual is this test statistic? P < 0.05 P > 0.05 Reject H o Fail to reject H o

Sample Null hypothesis The mean difference is equal to  o Paired t-test Test statistic Null distribution t with n-1 df *n is the number of pairs compare How unusual is this test statistic? P < 0.05 P > 0.05 Reject H o Fail to reject H o

4 Comparing means Tests with one categorical and one numerical variable Goal: to compare the mean of a numerical variable for different groups.

5 Paired vs. 2 sample comparisons

6 2 Sample Design Each of the two samples is a random sample from its population

7 2 Sample Design Each of the two samples is a random sample from its population The data cannot be paired

8 2 Sample Design - assumptions Each of the two samples is a random sample In each population, the numerical variable being studied is normally distributed The standard deviation of the numerical variable in the first population is equal to the standard deviation in the second population

9 Estimation: Difference between two means Normal distribution Standard deviation s 1 =s 2 =s Since both Y 1 and Y 2 are normally distributed, their difference will also follow a normal distribution

10 Estimation: Difference between two means Confidence interval:

11 Standard error of difference in means = pooled sample variance = size of sample 1 = size of sample 2

12 Standard error of difference in means Pooled variance:

13 Standard error of difference in means df 1 = degrees of freedom for sample 1 = n 1 -1 df 2 = degrees of freedom for sample 2 = n 2 -1 s 1 2 = sample variance of sample 1 s 2 2 = sample variance of sample 2 Pooled variance:

14 Estimation: Difference between two means Confidence interval:

15 Estimation: Difference between two means Confidence interval: df = df 1 + df 2 = n 1 +n 2 -2

16 Costs of resistance to disease 2 genotypes of lettuce: Susceptible and Resistant Do these differ in fitness in the absence of disease?

17 Data, summarized Both distributions are approximately normal.

18 Calculating the standard error df 1 =15 -1=14; df 2 = 16-1=15

19 Calculating the standard error df 1 =15 -1=14; df 2 = 16-1=15

20 Calculating the standard error df 1 =15 -1=14; df 2 = 16-1=15

21 Finding t df = df 1 + df 2 = n 1 +n 2 -2 = 15+16-2 =29

22 Finding t df = df 1 + df 2 = n 1 +n 2 -2 = 15+16-2 =29

23 The 95% confidence interval of the difference in the means

24 Testing hypotheses about the difference in two means 2-sample t-test

25 2-sample t-test Test statistic:

26 Hypotheses

27 Null distribution df = df 1 + df 2 = n 1 +n 2 -2

28 Calculating t

29 Drawing conclusions... t 0.05(2),29 =2.05 t <2.05, so we cannot reject the null hypothesis. These data are not sufficient to say that there is a cost of resistance. Critical value:

30 Assumptions of two-sample t - tests Both samples are random samples. Both populations have normal distributions The variance of both populations is equal.

Sample Null hypothesis The two populations have the same mean  1  2 Two-sample t-test Test statistic Null distribution t with n 1 +n 2 -2 df compare How unusual is this test statistic? P < 0.05 P > 0.05 Reject H o Fail to reject H o

Quick reference summary: Two-sample t-test What is it for? Tests whether two groups have the same mean What does it assume? Both samples are random samples. The numerical variable is normally distributed within both populations. The variance of the distribution is the same in the two populations Test statistic: t Distribution under H o : t-distribution with n 1 +n 2 -2 degrees of freedom. Formulae:

33 Comparing means when variances are not equal Welch’s t test

34 Burrowing owls and dung traps

35 Dung beetles

36 Experimental design 20 randomly chosen burrowing owl nests Randomly divided into two groups of 10 nests One group was given extra dung; the other not Measured the number of dung beetles on the owls’ diets

38 Hypotheses H 0 : Owls catch the same number of dung beetles with or without extra dung (  1 =  2 ) H A : Owls do not catch the same number of dung beetles with or without extra dung (  1   2 )

39 Welch’s t Round down df to nearest integer

40 Owls and dung beetles

41 Degrees of freedom Which we round down to df= 10

42 Reaching a conclusion t 0.05(2), 10 = 2.23 t=4.01 > 2.23 So we can reject the null hypothesis with P<0.05. Extra dung near burrowing owl nests increases the number of dung beetles eaten.

Quick reference summary: Welch’s approximate t-test What is it for? Testing the difference between means of two groups when the standard deviations are unequal What does it assume? Both samples are random samples. The numerical variable is normally distributed within both populations Test statistic: t Distribution under H o : t-distribution with adjusted degrees of freedom Formulae:

44 The wrong way to make a comparison of two groups “Group 1 is significantly different from a constant, but Group 2 is not. Therefore Group 1 and Group 2 are different from each other.”

45 A more extreme case...

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