> >bjbj ANN6###8[DT#E "+++pDrDrDrDrDrDrD$G3JPDD++E###++pD#pD##RAC+Et#oB\DE0E*BJ J<CJCV\#PDD#EJ : Hypothesis testing
Making decisions about the null
You can either accept the null (there is no effect of your treatment)
You can reject the nullthere is an effect
You can also accept the null when there is an effect
You can also reject the null when there is no effect
Accept NullReject NullNull is trueCorrectType IAlternate is trueType IICorrect
Potential Hypothesis-Testing Pitfalls and Ethical Issues
When planning a study to test a hypothesis several questions need to be asked to ensure we are using the right methodology:
What is the goal of the experiment or research? Can we describe it in terms of the null and alternative hypothesis?
Is this going to be a one-tailed or two-tailed test?
Can we draw a random sample?
Is the data numerical or categorical?
What significance should we be using? (We want to avoid a Type I error.)
Is the sample large enough?
What statistical test procedure is to be used and why?
What kind of conclusions and interpretations can be drawn from the results of the hypothesis test?
In addition we need to be concerned with ethical issues of our methodology, including:
Making sure our data is collected in randomized fashion. We need to eliminate potential bias in the sample. This can be especially difficult when sampling people, since the researchers will be tempted to take the responses they can get, rather than working for a real randomized sample selection. People cannot be allowed to self-select for the study.
Obtaining informed consent from human subjects being "treated." Any time people are used in a study we need to obtain their consent.
The type of testone tail or two. One approach to testing is to use a two-tailed test. Researchers may be tempted to test for the single tail to prove their point, disregarding opposite data that would change the results of their study.
Our choice of level of significance. Select the level of significance ahead of the study, not after it is done.
Data snooping. Dont review the data and then draw up your hypothesis.
Cleansing and discarding of data. When cleansing the data, look for outliers and bad data input errors, or missing data items that will affect the results of your analysis. You have to look carefully at any decision to remove data from a study.
Reporting of findings. Report ALL of your findings. Dont be selective about what data to report and what not to report.
Unethical behavior occurs when a researcher willfully causes a selection bias in data collection, manipulates the treatment of human subjects without informed consent, uses data snooping to select the type of test and/or level of significance to his or her advantage, hides the facts by discarding observations that do not support a stated hypothesis, or fails to report pertinent findings.
F TestFor students
Named after Sir Ronald Fisher, one of the founders of modern statistics
Used when we want to see if two samples are from two normal populations with equal variances
Used also when want to compare several population means (from normal populations) simultaneously (ANOVA)
Properties of F distribution
Like the t distribution, there is a family of F distributions based on the df. There are two dfs for F teststhe df for the numerator and the df for the denominator
The F distribution is continuous
It can not be negative
It is positively skewed
It is asymptotic (as the value of X increases, it approaches, but never touches, the X axis)
Comparing two population variances
The F distribution can be used to test if the variance of one normal population equals the variance of another normal population
Remember: the standard deviation is the square root of the variance, so to find the variance, you square the standard deviation.
The formula for the F ratio is:
2
F=S
1
_____
2
S
2
F= Variance of sample 1/Variance of sample 2
Example:
If the standard deviation of sample 1 is 6, then the variance is 36
If the standard deviation of sample 2 is 5, the variance is 25
F= 36/25=1.44
Null hypothesis= Variance 1=Variance 2
Alternate hypothesis=Variance 1 `" Variance 2
The closer F is to 1, the more alike the two variances are.
How do you calculate an F test?
(From Koosis, D. J. (1997). A Self-Teaching Guide: Statistics, Fourth Edition. John Wiley and Sons, Inc. New York. Chapter 6.)
A researcher believes stress will increase the variability on test scores, so he tells one group that the test they are taking will be an important factor in deciding if theyre admitted to the college. The other group he tells that the test is for research purposes only.
Null hypothesis: Variance 1=Variance 2
Alternate hypothesis=Variance 1 `" Variance 2
Significance level=.01
Critical region F>2.66
Each condition has 25 students, so the df=24 for the numerator and 24 for the denominator (n-1)
Group 1
StressGroup 2
No stressn=25
n=25
X=120X=110s=20s=10s2=400s2=100
What is F?
F= 400/100=4
Exceeds the critical value, so there is a difference in the variances of the 2 samples at the .01 level.
Complete the problems:
The following hypotheses are given:
Ho=(2 group 1= (2 group 2
Ha=(2 group 1`"(2 group 2
A random sample of eight observations from the first population resulted in a standard deviation of 12. A random sample of seven observations in the second population resulted in a standard deviation of 7. At the .02 significance level, is there a difference in the variation of the two populations?
Group 2=Exercise 3 on page 419
The following hypotheses are given:
Ho=(2 group 1= (2 group 2
Ha=(2 group 1`"(2 group 2
A random sample of eight observations from the first population resulted in a standard deviation of 12. A random sample of seven observations in the second population resulted in a standard deviation of 7. At the .01 significance level, is there a difference in the variation of the two populations?
ANOVA (Analysis of Variance)
This test is used to compare three or more population means to determine whether they could be equal. (From, Lind, D.A., Marchal, W.G., and Mason, R.D. (2002). Statistical Techniques in Business and Economics, Eleventh Edition. McGraw-Hill Irwin, Boston. Chapter 12, p. 419.)
The assumptions for ANOVA are:
The populations are normally distributed
The populations have equal standard deviations
The samples are selected independently (Ibid)
Why couldnt you just run a bunch of Z or T tests to test the differences between the means to see if theyre equal?
Well, because you increase the probability of a Type I error.
For example, if you had 4 population means, youd have to run 6 tests to compare them all:
A vs B
A vs C
A vs D
B vs C
B vs D
C vs D
If you set the significance level for each test at .05, the probability of a correct decision on each test=.95 (1-.05=.95).
For all 6 tests, the probability we do not make a correct decision due to sampling error is:
(.95) (.95) (.95) (.95) (.95) (.95)=.735
To find the probability of at least one error due to sampling, subtract 1-.735=.265. Your probability of a Type 1 error if you run all 6 tests now equals .265, not .05.
So, you want to run 1 test, not 6, and ANOVA allows you to do that.
Now back to Koosis, D. J. (1997). A Self-Teaching Guide: Statistics, Fourth Edition. John Wiley and Sons, Inc. New York. Chapter 6. (Most of the following is direct quoting)
In the analysis of variance, we use the difference between sample means to estimate the variance in the population. We also make a separate estimate of the population variance based only on the differences among individuals within each sample. If the samples all come from populations with the same mean, the difference between sample means should be (large/small).
Answer: Small
If the differences between sample means are relatively large, we will conclude that the mean of the population (is/is not) the same.
Answer: is not
The estimate of variance based on the differences between the means of groups is called between-groups variance. The estimate of the variance based only on the differences between individuals is called within groups variance or error variance. To reject the null hypothesis, between groups variance must be large compared to___________.
Answer: Within groups variance or error variance.
Estimating Between Groups and Within Groups Variance
A researcher believes that the color of a toy will affect how long, in a 10 minute observation period, children will play with the toy. She obtains 4 sample groups of 10 children each. She gives them the same toy to play with, but each group gets a different color toy.
Ho= (1=(2=(3=(4
H1=(1`"(2`"(3`"(4
Here is the data:
Group 1
Red toyGroup 2
Yellow toy
Group 3
Green toyGroup 4
Blue toyMinutes children played with toy in 10 minute observation period1325224356217312622118332744251246334821n1=10n2=10n3=10n4=10X1=3.4X2=5.0X3=2.4X4=10s2=4.5s2=5.6s2=1.2s2=1.8
Remember: s2= " (X-x)2/n-1
EMBED Equation.3 (Only square the top!)
All groups combined:
N=40
X= 3.3
s2=4.1
The between variance is based on the differences between group means.
Consider the distribution of the four group means. This distribution is:
A population distribution
A sample distribution
A sampling distribution?
Answer: C
At the moment, we are interested in computing s2, an estimate of the variance of the population. We can find it by using a formula based on the formula for standard deviation. We can say that s (standard deviation of the means) = s/(n. then s2x (Variance of the means)=s2/n and s2=ns2x.
What is the size of the samples that make up our sampling distribution? n=_____
Answer=10
Total variation=The sum of the squared differences between each observation and the total mean. For the example we are working on, the total mean = 3.3
Group 1
Red toyGroup 2
Yellow toy
Group 3
Green toyGroup 4
Blue toyMinutes children played with toy in 10 minute observation period
Total Variation(1-3.3)2(3-3.3)2(2-3.3)2(5-3.3)2(2-3.3)2(2-3.3)2(4-3.3)2(3-3.3)2(5-3.3)2(6-3.3)2(2-3.3)2(1-3.3)2(7-3.3)2(3-3.3)2(1-3.3)2(2-3.3)2(6-3.3)2(2-3.3)2(2-3.3)2(1-3.3)2(1-3.3)2(8-3.3)2(3-3.3)2(3-3.3)2(2-3.3)2(7-3.3)2(4-3.3)2(4-3.3)2(2-3.3)2(5-3.3)2(1-3.3)2(2-3.3)2(4-3.3)2(6-3.3)2(3-3.3)2(3-3.3)2(4-3.3)2(8-3.3)2(2-3.3)2(1-3.3)2n1=10n2=10n3=10n4=10X1=3.4X2=5.0X3=2.4X4=2.5s2=4.5s2=5.6s2=1.2s2=1.8
Lets let Excel do this.
Total variation=160.8
Treatment variation=The sum of the squared difference between each treatment mean and the overall mean= s2x (variance of the mean) is simply the variance of the following numbers: 3.4, 5.0, 2.4 and 2.5.
What are these numbers?
Answer: the means of the 4 samples
The mean for the red toy=3.4
The mean for the yellow toy=5.0
The mean of the green toy=2.4
The mean of the blue toy=2.5
10(3.4-3.3)2 + 10(5.0-3.3)2 + 10 (2.4-3.3)2 + 10 (2.5-3.3)2=
Again lets let Excel do the math.
Treatment variation=43.41
Random variation=The sum of the squared differences between each observation and its treatment mean
Group 1
Red toyGroup 2
Yellow toy
Group 3
Green toyGroup 4
Blue toyMinutes children played with toy in 10 minute observation period
Total Variation(1-3.4)2(3-5.0)2(2-2.4)2(5-2.5)2(2-3.4)2(2-5.0)2(4-2.4)2(3-2.5)2(5-3.4)2(6-5.0)2(2-2.4)2(1-2.5)2(7-3.4)2(3-5.0)2(1-2.4)2(2-2.5)2(6-3.4)2(2-5.0)2(2-2.4)2(1-2.5)2(1-3.4)2(8-5.0)2(3-2.4)2(3-2.5)2(2-3.4)2(7-5.0)2(4-3.3)2(4-2.5)2(2-3.4)2(5-5.0)2(1-2.4)2(2-2.5)2(4-3.4)2(6-5.0)2(3-2.4)2(3-2.5)2(4-3.4)2(8-5.0)2(2-2.4)2(1-2.5)2n1=10n2=10n3=10n4=10X1=3.4X2=5.0X3=2.4X4=2.5s2=4.5s2=5.6s2=1.2s2=1.8
Again, well let Excel do this.
Random variation comes out to 117.3
Treatment variation + Random variation= Total variation
In this case: 43.41+ 117.3 =160.71 (We calculated 160.8===rounding thing!)
Our calculations are correct
But what is the F for this?
F= Estimate of the population variance based on the differences among the sample means
_____________________________________________________________________
Estimate of the population variances based on the variation within the samples
F= 43.41/3 (3=number of groups=4-1) / 117.3/ 36 (Total of 40 observations-4 treatments)=
F=14.47/3.26=4.44
Look in F table for a .10 significance (two way) and 3 df in numerator and 36 df in denominator= 3 and 40 is 2.84, so I think we can reject the null.
But, we can get Excel to give us the critical value and the p level for our result
To find a probability: =FDIST (F, dfnum, dfdenom) so we would enter (4.44,3,36)
So lets let Excel do its =0.009374
Critical value=FINV (p, dfnum, dfdenom)=2.42
Yep, we can reject the null and say the differences between the treatments would only occur by chance about 9/1000 times or less.
To present these results, we need to set up an ANOVA table:
Anova TableSource of variationSum of SquaresDegrees of freedomMean squareFTreatmentsSSTk-1SST/k-1=MSTMST/MSEErrorSSEn-kSSE(n-k)=MSETotalSSTotaln-1
Anova TableSource of variationSum of SquaresDegrees of freedomMean squareFTreatments43.41343.41/3 =14.4714.47/3.26=4.44Error117.336117.3/36=3.27Total160.7139
There it is! We still cant tell where the differences are, but we know we have a difference! Were going to stop there, because I want you all to take the next hour and do a problem each.
Complete the problems
The following is sample information. Test the hypotheses that the treatment means are equal. Use the .05 significance level.
Treatment 1Treatment 2Treatment 38336241045934
State the null hypothesis and alternative hypothesis.
What is the decision rule?
Compute the SST, SSDE and SS Total.
Complete an ANOVA table.
State your decision regarding the null hypothesis.
Give a real life example of when you could use ANOVA.
The other two groups will do p 428, # 8
8) The following is sample information. Test the hypothesis at the .05 significance level that the treatment means are equal.
Treatment 1Treatment 2Treatment 391310720911141591314121510
State the null hypothesis and alternative hypothesis.
What is the decision rule?
Compute the SST, SSDE and SS Total.
Complete an ANOVA table.
State your decision regarding the null hypothesis.
Give a real life example of when you could use ANOVA.
How do you find out where the differences are?
You can either construct confidence intervals for each mean. If the CIs do not overlap, then thats where a difference is. Also, you can run a t-test for a difference between 2 of the means. That formula is:
(X1-X2)( t (MSE(1/n1 + 1/n2) (Square root of all of that is after the +/-)
So, lets look at problem 7. Lets do CIs for that one. Look at the data first. The mean for group 1 is 8.25. The mean for group 2 is 3 and the mean for group 3 is 4. I would bet there isnt going to be a difference between 3 and 4, but there might be a difference between 1 and 2 and/or 1 and 3.
Lets do a 95%CI. The df are n-k, so the df is 9 and the t value is 2.262 (2 tailed). The formula for the CI is:
X ( t (s/(n)
Mean 1=8.25
8.25 ( 2.62 (1.707/2)=
8.25 ( 2.62(.8535)=
8.25 ( 2.24=
(10.49, 6.01)
Mean 2=3
3 ( 2.62 (1.707/2)=
3 ( 2.24= (5.24, .757)
The CIs do not overlap, so you have a significant difference between Group 1 and Group 2 means
Mean = 4
4 ( 2.62 (1.707/2)=
4 ( 2.24= (6.24, 1.76)
There is no difference between Group 3 and Group 1 and theres no difference between Group 3 and Group 2, because the CIs overlap.
Now lets do Problem 8 using the other formula.
(X1-X2) ( t (MSE(1/n1 + 1/n2) (Square root of all of that)
Our t value is df 12 for two way .05 value=2.179
Mean for Group 1=9.67
Mean for Group 2=15
Mean for Group 3=12
You may have a difference between Group 1 and 2 and Group 2 and 3.
If the calculations have a zero in the CI, then there is no differences. If the Interval does not contain 0, then there is a difference.
Group 1 and Group 2
(9.67-15) ( 2.179 (6.88/1/6+1/4=
-5.33 ( 2.179 (6.88/ .17+.25=
-5.33 ( 2.179 (6.88/.42=
-5.33 ( 2.179 (4.05)=
-5.33 ( 8.79=
(3.46, -14.12)
This interval contains zero (0), so there is not a significant difference!
So, there s no point in checking the others, probably.
Chi Square Chapter 15 in Lind
Chapter 8 in Koosis
2 distribution is a theoretical sampling distribution that allows you to test the assumption that a sample was drawn from a population with a given distribution. It allows you to compare a sample distribution with a population distribution derived from a theory or null hypothesis and decide whether the sample could reasonable be a random sample from that population. (Koosis, p. 210).
2 is based on a comparison of the expected frequencies of an occurrence and the actual frequencies in your sample. It is used for categorical data.
So, let s say (still Koosis, p 210) that, on the basis of genetic theory, you expect the population of gui5 m
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40% spotted
20% white
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Brown
Spotted
White
BrownSpottedWhiteExpected number20 (50 * .40)20 (50 * .40)10 (50 * .20)
Lets say we have a sample of guinea pigs, and this is what you actually observe:
BrownSpottedWhiteObserved number30155
So, your full table would look like this:
BrownSpottedWhiteExpected number20 20 10 Observed number30155
Now you want to find out if your sample comes from the population of the expected frequencies or if you need to change your expectations, based on the sample you drew.
You can do this using 2:
2 = " (f-F2) /F
In the formula, F is the predicted frequency and f is the observed frequency
Let s expand our table by using Excel.
So, 2 = 8.75.
If the differences between the predicted and expected values is large, then 2 will also tend to be large.
What would a smaller 2 mean? That the observed and predicted frequencies are closer.
Is a negative value of 2 possible? No. We are working with squared differences.
You can look up this value of 2 of 8.75, but first you need the df.
The df for 2 is the number of categories minus 1. There are 3 categories of guinea pig colors: white, brown and spotted, so the df=2.
Lets look up 8.75 in the table. This value falls between .02 (7.824) and .01 (9.210), so we have a significant difference between the expected and observed frequencies of guinea pig colors.
The characteristics of Chi Square are:
Chi Square is never negative, which we already said.
Like the t and F distributions, it is a family of distributions based on the df.
It is positively skewed.
Complete the problems
In a particular Chi Square goodness of fit test there are 4 categories and 200 observations. Use the .05 significance level.
a) How many degrees of freedom are there?
b) What is the critical value of Chi Square (you will need to look at the table).
The null hypothesis and the alternate are:
Ho: The cell categories are equal.
Ha: The cell categories are not equal.
CategoryfoA10B20C30
a) State the decision rule, using the .05 significance level.
b) Compute the value of Chi Square.
c) What is your decision regarding Ho?
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2) In a particular Chi Square good ness of fit test, there are 6 categories and 500 observations. Use the .01 significance level.
a) How many degrees of freedom are there?
b) What is the critical value of Chi Square (you will need to look at the table).
4) The null hypothesis and alternate are:
Ho: The cell categories are equal.
Ha: The cell categories are not equal.
CategoryfoA10B20C30D20
a) State the decision rule, using the .05 significance level.
b) Compute the value of Chi Square.
c) What is your decision regarding Ho?
d) Give a real life example of when youd use Chi Square.
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