testing means, part iii the two-sample t-test

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

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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.

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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.

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

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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

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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

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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.”

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A more extreme case...