Hypothesis testing applied to means

Characteristics of the Sampling Distribution of the mean

The sampling distribution of means will have the same mean as the population

:0 = :

Characteristics of the Sampling Distribution of the mean

The sampling distribution of means has a smaller variance.

x

N2

2

x N

x = standard deviation of the mean

= standard error

This is because the means of samples are less likely to be extreme compared to individual scores.

Characteristics of the Sampling Distribution of the mean

The shape of the sampling distribution approximates a normal curve if either:

the population of individual cases is normally distributed

the sample size being considered is 30 or more

IQ Scores - Ranked

71, 76, 76, 77, 79, 80, 81, 82, 83, 83, 84, 84, 84, 85, 85, 86, 87, 88, 88, 88, 89, 90, 90, 91, 91, 91, 92, 92, 92, 93, 93, 93, 93, 93, 93, 94, 94, 94, 94, 95, 95, 95, 96, 96, 97, 97, 97, 97, 97, 97, 97, 98, 99, 99, 99, 99, 100, 100, 100, 100, 101, 101, 101, 102, 102, 102, 102, 103, 103, 103, 103, 103, 103, 103, 104, 104, 104, 105, 105, 106, 106, 107, 107, 107, 107, 107, 108, 108, 108, 108, 109, 109, 110, 111, 111, 112, 112, 112, 113, 113, 113, 113, 114, 114, 115, 115, 115, 117, 118, 120, 121, 121, 121, 123, 123, 125, 125, 126, 131, 136

Sample Size lowest extreme score highest extreme score

N =1 71 71 136 136

N=2 71+76 73.5 131+136 133.5

N=3 71+76+76 74.3 126+131+136 131.0

N=4 71+76+76+77 75.0 125+126+131+136 127.0

zX

115 10015

10.

Probability is approximately .16

Example

Jill now has to choose 25 intelligent people. 0 = 106

IQ test: := 100 F = 15.

Hypotheses:

H1: :1 … :2

H0: :1 = :2 = 100

Sampling Distribution

:1 = :2 = 100

x N

1525

1553

zX

x

zX

x

106 1003

63

200.

Look up area in the tail of the curve

= .0228 (one tailed) or .0456 (two tailed)

If signficance level (") = .05, reject the Null Hypothesis

t Distribution

$What happens if we do not have the population standard deviation?<We can use the sample standard deviation and an estimate.

$Problem<Cannot use z and the normal distribution to estimate probability.

BSample variance tends to underestimate the population variance$Have to use a slightly different distribution - Student=s t

Distribution

zX

x

tXsx

df = n - 1

Different t distribution for each degree of freedom

Degrees of freedom

Sample: N = 56 0 = 104.3 S = 12.58

Norms: : = 100

Hypotheses:

H1: :1 … :2

H0: :1 = :2 = 100

tXsx

XsN

10413 100125856

..

41251682..

t245. df = 56 - 1 = 55

tcrit or t.025 = " 2.009

tobs > tcrit Reject H0

Percentage Points of the t Distribution

See Howell page 247

Matched-Sample t-test

Use related samples In SPSS called the >Paired-Samples t-test'

Data: Howell p 193

H0: D 1 2 0

tDs

Ds

DsN

D D D 0 0

DsND

0 429 0160431

429288

149..

..

.

df = N - 1 (N is number of pairs of observations)

df = 31 - 1 = 30

t.025(30) = " 2.042

tobs < tcrit

Fail to reject H0

Two Independent Samples

Distribution of Differences between Means

Variance of the Distributions of the Means

12

1

22

2N N&

Standard Error of the Distributions of the Means

1

1

2

2N N&

Variance of the Distribution of Mean Difference

X X N N1 2

2 12

1

22

2

Standard Error of the Distribution of Mean Difference

X X N N1 2

12

1

22

2

Mean of the Distribution of Mean Differences

1 2 0

t - test for two independent samples

tXsx

tX X

sX X

( ) ( )1 2 1 2

1 2

( ) ( )X X

sN

sN

1 2 1 2

12

1

22

2

( )X X

sN

sN

1 2

12

1

22

2

Pooled Variance estimate sN s N s

N Np2 1 1

22 2

2

1 2

1 12

( ) ( )

tX X

sN

sN

( )1 2

12

1

22

2

( )X X

s

N

s

Np p

1 22

1

2

2

tX X

sN Np

( )1 2

2

1 2

1 1

Degrees of Freedom:

df = (N1 - 1) + (N2 - 1) = N1 + N2 -2