Testing means, part III

The two-sample t-test

Sample Null hypothesisThe population mean

is equal to o

One-sample t-test

Test statistic Null distributiont with n-1 dfcompare

How unusual is this test statistic?

P < 0.05 P > 0.05

Reject Ho Fail to reject Ho

€

t = Y − μo

s / n

Sample Null hypothesisThe mean difference

is equal to o

Paired t-test

Test statistic Null distributiont with n-1 df

*n is the number of pairscompare

How unusual is this test statistic?

P < 0.05 P > 0.05

Reject Ho Fail to reject Ho

€

t = d − μdo

SEd

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

€

Y 1 −Y 2

Normal distributionStandard deviation s1=s2=s

Since both Y1 and Y2 are normally distributed, their difference will also follow a normal distribution

10

Estimation: Difference between two

means

€

Y 1 −Y 2

Confidence interval:

€

Y 1 −Y 2( ) ± SEY 1 −Y 2tα 2( ),df

11

Standard error of difference in means

€

SEY 1 −Y 2= sp

2 1n1

+ 1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

= pooled sample variance= size of sample 1= size of sample 2

€

sp2

n1

n2

12

Standard error of difference in means

€

SEY 1 −Y 2= sp

2 1n1

+ 1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

sp2 =

df1s12 + df2s2

2

df1 + df2

Pooled variance:

13

Standard error of difference in means

€

sp2 =

df1s12 + df2s2

2

df1 + df2

df1 = degrees of freedom for sample 1 = n1 -1df2 = degrees of freedom for sample 2 = n2-1s1

2 = sample variance of sample 1s2

2 = sample variance of sample 2

Pooled variance:

14

Estimation: Difference between two

means

€

Y 1 −Y 2

Confidence interval:

€

Y 1 −Y 2( ) ± SEY 1 −Y 2tα 2( ),df

15

Estimation: Difference between two

means

€

Y 1 −Y 2

Confidence interval:

€

Y 1 −Y 2( ) ± SEY 1 −Y 2tα 2( ),df

df = df1 + df2 = n1+n2-2

16

Costs of resistance to disease

2 genotypes of lettuce: Susceptible and Resistant

Do these differ in fitness in the absence of disease?

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.QuickTime™ and a

TIFF (Uncompressed) decompressorare needed to see this picture.

17

Data, summarizedSusceptible Resistant

Mean numberof buds 720 582

SD of numberof buds 223.6 277.3

Sample size 15 16

Both distributions are approximately normal.

18

Calculating the standard errordf1 =15 -1=14; df2 = 16-1=15

19

Calculating the standard errordf1 =15 -1=14; df2 = 16-1=15

€

sp2 = df1s1

2 + df2s22

df1 + df2

=14 223.6( )

2 +15 277.3( )2

14 +15= 63909.9

20

Calculating the standard errordf1 =15 -1=14; df2 = 16-1=15

€

sp2 = df1s1

2 + df2s22

df1 + df2

=14 223.6( )

2 +15 277.3( )2

14 +15= 63909.9

€

SEx 1 −x 2= sp

2 1n1

+ 1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟= 63909.9 1

15+ 1

16 ⎛ ⎝ ⎜

⎞ ⎠ ⎟= 90.86

21

Finding t

df = df1 + df2= n1+n2-2

= 15+16-2=29

22

Finding t

df = df1 + df2= n1+n2-2

= 15+16-2=29

€

t0.05 2( ),29 = 2.05

23

The 95% confidence interval of the

difference in the means

€

Y 1 −Y 2( ) ± sY 1 −Y 2tα 2( ),df = 720 − 582( ) ± 90.86 2.05( )

=138 ±186

24

Testing hypotheses about the difference

in two means2-sample t-test

The two sample t-test compares the means of a numerical

variable between two populations.

25

2-sample t-test

€

t = Y 1 −Y 2SE

Y 1−Y 2

Test statistic:

€

SEY 1 −Y 2= sp

2 1n1

+ 1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

sp2 =

df1s12 + df2s2

2

df1 + df2

26

Hypotheses

H0: There is no difference between the number ofbuds in the susceptible and resistant plants.

(1 = 2)

HA: The resistant and the susceptible plants differ intheir ean nu ber of buds. (1 ≠ 2)

27

Null distribution

€

tα 2( ),df

df = df1 + df2 = n1+n2-2

28

Calculating t

€

t =x 1 − x 2( )SEx 1 −x 2

=720 − 582( )

90.86=1.52

29

Drawing conclusions...

t0.05(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.

SampleNull hypothesis

The two populations have the same mean

1=2

Two-sample t-test

Test statistic Null distributiont with n1+n2-2 dfcompare

How unusual is this test statistic?

P < 0.05 P > 0.05

Reject Ho Fail to reject Ho

€

t = Y 1 −Y 2SE

Y 1−Y 2

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 Ho: t-distribution with n1+n2-2 degrees of freedom.

• Formulae:

€

t = Y 1 −Y 2SE

Y 1−Y 2€

SEY 1 −Y 2= sp

2 1n1

+ 1n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟

€

sp2 =

df1s12 + df2s2

2

df1 + df2

33

Comparing means when variances are not

equalWelch’s t test

Welch's approximate t-test compares the means of two

normally distributed populations that have unequal

variances.

34

Burrowing owls and dung traps

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

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

37

Number of beetles caught

• Dung added:

• No dung added:

€

Y = 4.8s = 3.26

€

Y = 0.51s = 0.89

38

Hypotheses

H0: Owls catch the same number of dung beetles with or without extra dung (1 = 2)

HA: Owls do not catch the same number of dung beetles with or without extra dung (1 2)

39

Welch’s t

€

t =Y 1 − Y 2s1

2

n1

+ s22

n2

€

df =

s12

n1

+s2

2

n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

s12 n1( )

2

n1 −1+

s22 n2( )

2

n2 −1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Round down df to nearest integer

40

Owls and dung beetles

€

t = Y 1 −Y 2s1

2

n1

+ s22

n2

= 4.8 − 0.513.262

10+ 0.892

10

= 4.01

41

Degrees of freedom

€

df =

s12

n1

+ s22

n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

s12 n1( )

2

n1 −1+

s22 n2( )

2

n2 −1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

=

3.262

10+ 0.892

10

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

3.262 10( )2

10 −1+

0.892 10( )2

10 −1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

=10.33

Which we round down to df= 10

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Reaching a conclusion

t0.05(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 Ho: t-distribution with adjusted degrees of freedom

• Formulae:

€

t =Y 1 − Y 2s1

2

n1

+ s22

n2

€

df =

s12

n1

+ s22

n2

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

s12 n1( )

2

n1 −1+

s22 n2( )

2

n2 −1

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

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...