Psych 230 Psychological Measurement and Statistics Pedro Wolf October 28, 2009

Psych 230

Psychological Measurement and Statistics

Pedro Wolf

October 28, 2009

Last Time….

• Hypothesis testing

• Statistical Errors

• Z-test

This Time….

• T-Test

• Confidence intervals

• Practice problems

Hypothesis Testing

• Experimental hypotheses describe the predicted outcome we may or may not find in an experiment– As scientists, we try to be conservative– we assume no relationship

• The Null Hypothesis (H0)– there is no relationship between the variables

• The Alternative Hypothesis (H1)– there is a real relationship between the variables

Steps to Hypothesis testing

1. State the hypotheses2. Design the experiment 3. Collect the data4. Create the statistical hypotheses5. Select the appropriate statistical test6. Decide the size of the rejection region

(value of )7. Calculate the obtained and critical values8. Make our conclusion

Statistical tests (so far)

• The statistical tests we have used so are concentrate on finding whether a sample is representative of a known population

• Two characteristics of these tests:– one sample is drawn– we know the population mean

• Z-test– we also know the population variance

• T-test (one sample)– we do not know the population variance

Statistical Testing

1. Decide which test to use 2. State the hypotheses (H0 and H1)

3. Calculate the obtained value4. Calculate the critical value (size of )5. Make our conclusion

One sample T-test

One-sample T-test

• We use the one sample T-test when we do not know the population variance

• Only differences from before:

• Tobt uses a slightly different formula

• Tcrit comes from a different distribution (the T-distribution), and so we need different tables to get this value

The T-test - summary

1. Create H0 and H1

2. Compute tobt

1. Compute X and s2x

2. Compute sx

3. Compute tobt

3. Find tcrit by using the T-tables with df = N - 1

4. Compare tobt to tcrit

The T-value of our sample (Tobt)

• Calculating Tobt

Tobt = X - µ sx

sx= √(s2x / N) : estimated standard error of the mean

In General:Test statistic = Observed - Expected

Standard Error

The T-distribution (Tcrit)

• When using z-scores, we always looked at the same distribution (the Z-distribution)

• The T-distribution is actually a family of curves, all which look slightly different depending on how many samples were used to create them– Therefore, as N changes, the exact curve we will use

will change

• For small samples (a small N) the curve is only roughly similar to the standard normal curve

• Large samples (a big N) look very close to the standard normal curve


• Two different T-distributions


• We choose the curve to sample from based on not N exactly, but rather the quantity N-1

• This is termed the degrees of freedom (df)– Degrees of freedom: the number of observations

in a set of data that are variable

• The larger the df, the closer the t-distribution resembles a standard normal curve– When df > 120, the t-distribution is virtually

identical to the standard curve, and in fact tcrit = zcrit


• To decide whether the observed value (Tobs) is in the region of rejection, we need to know Tcrit

• Tcrit is defined as the value that marks the most extreme 5% (usually) of the distribution– 5% when = 0.05

• Different distributions are different shapes and so will have different critical values for the extreme 5% of scores

• So, when performing a t-test, we use one specific curve (and one set of critical values) depending on the value of df (or, N-1)


• Example: Assume the experiment had N=22 and = 0.05, and we want a two-tailed test

• df = N-1 = 22-1 = 21• Look up t-tables (page 551 of book)• df = 0.05 = 0.01• 1 12.706 63.657• 2 4.303 9.925• 3 3.182 5.841• 21 2.080 2.831


• Practice: What are the Tcrit values for each of the following scenarios

• N=16; = 0.05; Two-tailed• N=31; = 0.05; Two-tailed• N=28; = 0.01; Two-tailed• N=9; = 0.05; Two-tailed• N=25; = 0.05; One-tailed• N=15; = 0.01; One-tailed


• Practice: What are the Tcrit values for each of the following scenarios

• N=16; = 0.05; Two-tailed ±2.131• N=31; = 0.05; Two-tailed ±2.042• N=28; = 0.01; Two-tailed ±2.771• N=9; = 0.05; Two-tailed ±2.306• N=25; = 0.05; One-tailed 1.711• N=15; = 0.01; One-tailed 2.624

Problem 1

Your instructor thinks that men and women have different levels of enthusiasm about statistics classes. When asked for their ratings of how much they were looking forward to a stats class, the for women is 5.23. A sample of 7 male students gave the following scores for how excited they were about this class:5, 7, 5, 7, 8, 6, 5

1. What is the appropriate statistical test?2. Is this a one-tailed or two-tailed test? Why?3. What are H0 and HA?4. Compute Tobt

5. With =0.05, what is Tcrit?6. What conclusion should we draw?

Problem 1


1. What is the appropriate statistical test?

Problem 1



We are comparing a sample of scores to a population mean, therefore we will use a one-sample test.

As we do not know the population variance, we must estimate it and use a one-sample T-test

Problem 1


1. Is this a one-tailed or two-tailed test? Why?

Problem 1



Two-tailedWe are interested in whether men differ from women

Problem 1


2. What are H0 and H1?

Problem 1


2. What are H0 and H1?H0 : Men and women are equally enthusiasticH0 : men = 5.23H1 : Men and women differ in enthusiasmH1 : men 5.23

Problem 1


3. Compute Tobt

Problem 1


3. Compute Tobt Tobt = (X - µ) / sx

sx= √(s2x / N)

=5.23; N=7

X= ??sX= ?? s2X= ??

Problem 1


sx= √(s2x / N)

3. Compute Tobt

X=(43/7)=6.14s2

X = [273-(1849/7)] / [7-1]=(273-264.14)/6=1.48sX= √(1.48/7) = √(0.21) = 0.46

1

)( 22

2

NNX

XsX

Problem 1


3. Compute Tobt Tobt = (X - µ) / sx

=5.23; N=7; X=6.14; sX= 0.46Tobt = (6.14 - 5.23) / (0.46) Tobt = 1.97

Problem 1


4. With =0.05, what is Tcrit?

Problem 1


4. With =0.05, what is Tcrit?

=0.05, two-taileddf=N-1=7-1=6

Tcrit = 2.447

Tcrit and Tobt

a a a

a

Tcrit= -2.447

Tobt=+1.97

Tcrit= +2.447

Problem 1


5. What conclusion should we draw?

Problem 1


5. What conclusion should we draw?As Tobs = Tcrit , we retain H0

Men do not differ significantly from women on how enthusiastic they are about this statistics class

Problem 2

A researcher predicts that smoking cigarettes decreases a person’s sense of smell. On a test of olfactory sensitivity, the for nonsmokers is 18.4. People who smoke a pack a day produced the following scores:16, 14, 19, 17, 16, 18, 17, 15, 18, 19, 12, 14

1. What is the appropriate statistical test? Is this a one-tailed or two-tailed test? Why?

2. What are H0 and HA?3. Compute the obtained value4. With =0.05, what is the critical value?5. What conclusion should we draw from this study?

Problem 2



Problem 2



We are comparing a sample of scores to a population mean, therefore we will use a one-sample test.

As we do not know the population variance, we must estimate it and use a one-sample T-test

Problem 2



Problem 2



It will be a one-tailed test, as we are predicting the direction that the scores will change. That is, we are specifically asking whether smoking leads to a decreased sense of smell

Problem 2


2. What are H0 and H1?

Problem 2


2. What are H0 and H1?H0 : Smoking is not associated with decreased sense of smellH0 : smokers >= 18.4H1 : Smoking is associated with a decreased sense of smellH1 : smokers < 18.4

Problem 2


3. Compute the obtained value

Problem 2


3. Compute the obtained value Tobt = (X - µ) / sx

sx= √(s2x / N)

=18.4; N=12

X= ??sX= ??s2

X= ??

1

)( 22

2

NNX

XsX

Problem 2


3. Compute the obtained value

X=(195/12)=16.25

s2X = [3221-(38025/12)] /

[12-1]=(3221-3168.75)/11=4.75

sX= √(4.75/12) = √(0.396) = 0.629

1

)( 22

2

NNX

XsX

Problem 2


3. Compute the obtained value Tobt = (X - µ) / sx

=18.4; N=12; X=16.25; sX= 0.629

Tobt = (16.25 - 18.4) / (0.629)

Tobt = -3.42

Problem 2


4. With =0.05, what is the critical value?

Problem 2


4. With =0.05, what is the critical value?

=0.05, one-taileddf=N-1=12-1=11

Tcrit = -1.796

Tcrit and Tobt

a a a

a

Tcrit=-1.796

Tobt=-3.42

Problem 2


5. What conclusion should we draw from this study?

Problem 2


5. What conclusion should we draw from this study?

As Tobs < Tcrit , we reject H0 and accept H1.

People who smoke have a significantly decreased sense of smell.

Problem 3

A candidate running for local sheriff claims that she will reduce the average speed of emergency response to less than the current average, which is 30 minutes. Thanks to this campaign, she is elected to office and records are kept. The response times (in minutes) for the first month are:26, 30, 28, 29, 25, 28, 32, 35, 24, 23Using =0.05, did she keep her promise?

Problem 3


1. Decide which test to use 2. State the hypotheses (H0 and H1)3. Calculate the obtained value4. Calculate the critical value (size of )5. Make our conclusion

Problem 3


1. Decide which test to use

Problem 3


1. Decide which test to use We are comparing a sample of scores to a population mean, therefore we will use a one-sample test.As we do not know the population variance, we must estimate it and use a one-sample T-test

Problem 3


1. One or Two-tailed? It will be a one-tailed test, as we are predicting the direction that the scores will change. That is, we are specifically asking whether the new sheriff has decreased response times.

Problem 3


2. State the hypotheses (H0 and H1)

Problem 3


2. State the hypotheses (H0 and H1)H0 : The new response time is not faster then the old oneH0 : new >= 30H1 : The new response time is faster than the old one H1 : new < 30

Problem 3


3. Calculate the obtained value

Problem 3


3. Calculate the obtained value Tobt = (X - µ) / sx

sx= √(s2x / N)

=30; N=10X= ??sX= ?? s2

X= ??

Problem 3


3. Calculate the obtained value X= (280/10) = 28s2

X= [7964-(78400/10)] / [10-1]=(7964-7840)/9=13.78sX= √(s2

x / N) = √(13.78 / 10) = 1.17

1

)( 22

2

NNX

XsX

Problem 3


3. Calculate the obtained value Tobt = (X - µ) / sx

=30; N=10; X=28; sX= 1.17

Tobt = (28 - 30) / (1.17) Tobt = -1.71

Problem 3


4. Calculate the critical value (size of )

Problem 3


4. Calculate the critical value (size of )=0.05, one-taileddf=N-1=10-1=9 Tcrit = -1.833

Tcrit and Tobt

a a a

a

Tcrit=-1.833

Tobt=-1.71

Problem 3


5. Make our conclusion

Problem 3


5. Make our conclusionAs Tobs = Tcrit , we retain H0

The new sheriff has not significantly reduced response times.

Reporting t

• In scientific papers, we report the result of our statistical test like this:

• t(df) = tobs, p < 0.05 [if significant]; p > 0.05 [if not significant]

• For stats enthusiasm study (problem 1)

• t(6) = 1.97, p > 0.05

• For smoking study (problem 2)

• t(11) = -3.42, p < 0.05

Reporting t

• For break-up study

• t(9) = 4.09, p < 0.05

• But, the tcrit at a 0.01 level is 3.25

• So, the result we found is actually significant at a higher level

• t(9) = 4.09, p < 0.01

• Statistical programs will give the exact level of significance - the exact probability that the sample mean would occur if H0 is true

Estimating

• There are two ways to estimate the population mean

• Point estimation in which we describe a point on the variable at which the population mean is expected to fall

• Interval estimation in which we specify an interval (or range of values) within which we expect the population mean to fall

Estimating

• We perform interval estimation by creating a confidence interval

• The confidence interval for a particular describes an interval containing values of

sx= √(s2x / N)

XtsXts XX ))(())(( critcrit

Confidence Intervals

• I pick 5 exam scores at random. How can I estimate the average score of the class on the this exam? The scores are: 75, 63, 68, 86, 73

• I can use the sample mean to estimate the population mean

• X = (365/5) = 73

• For more precision, I can also construct a confidence interval for the population mean

• A 95% confidence interval means I am 95% sure that the true population mean lies within the range of values I find


• I pick 5 exam scores at random. How can I estimate the average score of the class on the this exam? The scores are: 75, 63, 68, 86, 73

(3.86)(-2.776) + 73 <= <= (3.86)(+2.776) + 73

62.29 <= <= 83.71

XtsXts XX ))(())(( critcrit


A researcher predicts that smoking cigarettes decreases a person’s sense of smell. People who smoke a pack a day produced the following scores:16, 14, 19, 17, 16, 18, 17, 15, 18, 19, 12, 14

1. Compute the 95% confidence interval for the of smokers



1. Compute the 95% confidence interval for the of smokersCI = [(sX)(- Tcrit) + X] <= <= [(sX)(+ Tcrit) + X]



1. Compute the 95% confidence interval for the of smokersCI = [(sX)(- Tcrit) + X] <= <= [(sX)(+ Tcrit) + X]

[(0.629)(-2.201) + 16.25] <= <= [(0.629)(+2.201) + 16.25]

14.87 <= <= 17.63

Next Time….

• What do we do when we don’t know the population mean?

• Variants of the T-Test– Independent samples T-test– Related samples T-test

Homework

• Exercises 14, 15, and 16 for chapter 12

• Syllabus change- Read Chapter 13 for next week.

Documents

Psych 230 Psychological Measurement and Statistics Pedro Wolf October 28, 2009