One-sample

In the previous cases we had one sample and were comparing its mean to a hypothesized population mean

However in many situations we will use the t-test to compare two samples/groups of data.

Two samples

Previously compared sample mean to a known population meanNow we want to compare two samplesNull hypothesis: the mean of the population of scores from which one set of data is drawn is equal to the mean of the population of the second data set H0: 1=2 or 1 - 2 = 0

Independent samples

Consider the original case

Now want to consider not just 1 mean but the difference between 2 means H0: 1 - 2 = 0

t X

sX

Which leads to...

Now statistic of interest is:

Mean of the sampling distribution is:

Sampling Distribution of the Difference Between Means

X1 X 2

1 2

Variability

Standard error of the difference between means

Since two independent variables, variance of the difference between means equals sum of their variances (sqrt for standard error)

X 1 X 2

X1

2 X 2

2 1

2

n1

2

2

n2

Note for the first timers:

Does not involve subtraction. The is simply a designation for the fact that we are dealing with the difference between two means (e.g. rather than standard error for one mean)

1 2X X

1 2X X

Same problem, same solution

Usually we do not know population variance (standard deviation)

Again use sample to estimate it

Result is distributed as t (rather than z)

Formula

All of which leads to:

t X1 X 2 1 2

sX 1 X 2

X1 X 2 1 2

s12

n1

s22

n2

But...

Recall before we stated that our null hypothesis is

In other words, we are testing the value for no difference (most of the time)So, if the null hypothesis is true…

1 2 0

t test

t X1 X 2 s

X1 X 2

X1 X 2 s1

2

n1

s22

n2

t X1 X 2 1 2

sX 1 X 2

X1 X 2 1 2

s12

n1

s22

n2

1 2

1 2 1 2

2 21 2

1 2

0 0

X X

X X X Xt

s s sn n

Original

Simplified

Degrees of freedom

Across the 2 samples we have

(n1-1) and (n2-1)

df = (n1-1) + (n2-1) = n1 + n2 – 2

Steps for hypothesis test are still the same

1) State the research question2) State the null hypothesis, which simply gives us a

value to test, and alternative hypothesis3) Construct a sampling distribution based on the null

hypothesis and locate region of rejection (i.e. find the critical value on your table)

How large a difference between means is required for statistical significance?

4) Calculate the test statistic (t) and see where it falls along the distribution in relation to the critical value (tcv)

5) Reach a decision and state your conclusion

Assumptions of a t-test

In order for the t-test to be a valid analysis, certain assumptions must be met (or close enough) or we will have to worry about generalizability (external validity) and consistency issues (internal validity)

Assumptions of a parametric* t-test

Dealing with normal populations

Variances of groups in question are equal (i.e. the standard deviations are not markedly different from one another)

Samples are randomly drawn from their respective populations and are independent of one another

*A non-parametric t-test is ‘distribution-free’ and typically used for small sample sizes in which normality cannot be assumed. It is calculated differently.

Example t-test

Compare two therapies for a particular affliction (e.g. depression) on a scale of 1-50 where 50 would be totally cured of all sadness forever after the therapy.

Mean Therapy 1 = 30 s = 5 N = 25

Mean Therapy 2 = 26 s = 4 N = 25

Ho:

H1: not Ho, or

Non-directional test (two-tailed)

1 2 0

1 2 0

Plug into formula

= 30 s = 5 N = 25

= 26 s = 4 N = 25

1 2

1 2 1 2

X X

X Xt

s

2 2

30 26 0 43.12

1.645 425 25

t

1X

2X

Example Continued

t = 3.12

Two-tailed test

Critical value:t.05(25 + 25 – 2) = t.05(48) = ~2.01

Our t statistic is beyond the t critical value we’ve decided upon

Reject H0 There is a statistical difference between the means Treatment 1 is more effective. Dole out the happy pills!

Unequal sample sizes

Assumption: independent samples t test requires samples come from populations with equal variancesTwo estimates of variance (one from each sample)Generate an overall estimate that reflects the fact that bigger samples offer better estimatesIf our two samples are quite different in size, the original formula for standard error will not be as good an estimate as it normally would be

Weighted average

Calculate the pooled variance:

Final result is:

spooled2

(n1 1)s1

2 (n2 1)s22

df1 df2

spooled2

(n1 1)s1

2 (n2 1)s22

n1 n2 2

Pooled variance estimate

Before we had

With unequal sample sizes we’ll use our pooled variance estimate

If sample sizes were equal we’d get the same result as our previous formula for std. error

1 2

1 2 1 2

2 21 2

1 2

X X

X X X Xt

s s sn n

1 2

1 2 1 2

2 2

1 2

X X p p

X X X Xt

s s s

n n

Note the sp

2

Example

New drug MemoPlus is supposed to aid memory. 5 people take the drug and we compare them to a control group of 10 people who take a placebo.Look at their free recall of 20 items on a list of related words.Drug: mean = 14, SD = 4.0Control: mean = 10, SD = 6.0

Start by calculating the standard error

Using

Get:

[note pooled variance nearer to group’s with larger n]

spooled2

(n1 1)s1

2 (n2 1)s22

n1 n2 2

spooled2

(5 1)(42 ) (10 1)(62 )

5 10 229.85

sX1 X 2

sp2 1

n1

1

n2

29.85

1

5

1

10

2.99

Calculate t

Enter the values:

Critical value for a two-tailed test

t.05(13) = 2.16

1 2

1 2 14 10.0 41.33

2.99 2.99X X

X Xt

s

tcv

t = 1.33

Make a decision and draw your conclusion

Our t-statistic does not meet our criterion for rejection of the null hypothesis

MemoPlus isn’t effective