Quantitative MethodsQuantitative Methods

Part 3T- Statistics

Standard DeviationStandard Deviation

Measures the spread of scores within the data set◦Population standard deviation is

used when you are only interested in your own data

◦Sample standard deviation is used when you want to generalise for the rest of the population

Z - ScoresZ - ScoresA specific method for describing a specific location within a distribution

◦Used to determine precise location of an in individual score◦Used to compare relative positions of 2 or more scores

Normally Distributed (Bell Normally Distributed (Bell shaped) shaped)

Distribution of the MeansDistribution of the Means

X X X X

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• Frequency Distribution of 4 scores (2, 4, 6,8)

• It is flat and not bell shaped• Mean of population is (2+4+6+8)/4 = 5

Distribution of the MeansDistribution of the Means

• Take all possible samples of pairs of scores (2,4,6,8)• Use random sampling and replace each individual sample into data set• Calculate average of all sample pairs

2+2 /2 = 22+4 /2 = 34+2 /2 = 3

4+4 /2 = 44+6 /2 = 56+4 /2 = 5

2+6 /2 = 46+2 /2 = 42+8 /2 = 58+2 /2 = 5

0 1 2 3 4 5 6 7 8 9

X XX

XXX

XXXX

X X XXX

X

“For any population with a mean μ and standard deviation σ , the distribution of sample means for sample size n will have a mean of μ and standard deviation of σ/√n and will approach a normal distribution as n gets very large.”

How big should the sample size be? n=30X

X X X

X X X X X

X X X X X X X

0 1 2 3 4 5 6 7 8 9

Central Limit TheoremCentral Limit Theorem

σ/√n is used to calculate the Standard Error of the sample mean

Sample data = X The mean of each sample = Then the standard error becomes It identifies how much the observed sample

mean differs from the un-measurable population mean μ.

So to be more confident that our sample mean is a good measure of the population mean, then the standard error should be small. One way we can ensure this is to take large samples.

Standard ErrorStandard Error

The population of SATs scores is normal with μ= 500, σ =100. What is the chance that a sample of n=25 students has a mean score =540? Since the distribution is normal, we can use the z-score

First calculate Standard Error ◦ 100/5 = 20

Then Z-Score◦ 540-500/20 =2

z-value is 2, therefore around 98% of the sample means are below this and only 2% are above. So we conclude that the chance of getting a sample mean of 540 is 2%, so we are 98% confident that this sample mean, if recorded in an experiment is false.

Example Example Z = - Z = - μμ// σ σ

T - StatisticsT - StatisticsSo far we’ve looked at mean and

sd of populations and our calculations have had parameters

But how do we deduce something about the population beyond our sample?

We can use T-Statistic

T - StatisticsT - StatisticsRemember SD from last week?

Great for population of N but not for sample of n

Why n -1? Because we can only freely

choose n-1 (Degree of freedom = df)

T - StatisticsT - StatisticsStandard ErrorZ-Score redone to show above =

To T-Statistic, we substitute σ (SD of population) with s (SD of sample)

But what about μ ?

Hypothesis TestingHypothesis TestingSample of computer game

players n =16Intervention = inclusion of rich

graphical elementsLevel has 2 rooms

◦Room A = lots of visuals◦Room B = very bland

Put them in level 60 minutesRecord how long they spend in B

ResultsResults

Average time spent in B = 39 minutes

Observed “sum of squares” for the sample is SS = 540.

A B

Stage1: Formulation of Stage1: Formulation of HypothesisHypothesis

: “null hypothesis”, that the visuals have no effect on the behaviour.

: “alternate hypothesis”, that the visuals do have an effect on the players’ behaviour.

If visuals have no effect, how long on average should they be in room B?

Null hypothesis is crucial; here we can infer that μ = 30 and get rid of the population one

Stage 2: Locate the critical Stage 2: Locate the critical regionregion

We use the T-table to help us locate this, enabling us to reject or accept the null hypothesis. To get we need:◦Number of degree of freedom (df) 16

-1 =15◦Level of significance of confidence◦Locate in T-table (2tails)= critical

value of t=-2.131, t=2.131

Stage 3: Calculate Stage 3: Calculate statisticsstatisticsCalculate sample sd = 6

Sample Standard Error = 6 / 4 =1.5

T-Statistic = 6The μ 30 came from the null

hypothesis if visuals had no effect, then the player would spend 30 minutes in both rooms A and B.

Stage 4: DecisionStage 4: DecisionCan we reject the , that the visuals

have no effect on the behaviour?◦T = 6 which is well beyond the value of

2.313 which indicates where chance kicks in.

So yes we can safely reject it and say it does affect behaviour

Which room do they prefer?◦They spent on average 39 minutes in

Room B which is bland

WorkshopWorkshopWork on Workshop 6 activitiesYour journal (Homework)Your Literature Review

(Complete/update)

ReferencesReferences Dr C. Price’s notes 2010 Gravetter, F. and Wallnau, L. (2003) Statistics for the

Behavioral Sciences, New York: West Publishing Company