Formula: ANOR
November 25, 2008 by
Filed under Formulas
SSregression = SStotal * rsquared
SSerror = SStotal * (1- rsquared)
SStotal = SSy
df(regression) = k – 1
df(error) = N – k
df(total) = N-1
F = mean squares regression divided by means squares error
NOW YOU CHOOSE:
Day 7: Probability
Bit More About Probability
Even More About Probability
Even More About ANOR
Calculate ANOR
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Quiz 7
Summary
More Practice Problem ANSWERS: 1-Way ANOVA
November 25, 2008 by
Filed under Calc Solutions
Item 3
Which is the best coffee (most cups ordered):
Blue-Label Green-Label Red-Label
13 1 5
4 1 2
10 2 2
13 2 2
11 2 6
3 4 4
Sum 54 12 21 87
Sum X2 584 30 89 703
N 6 6 6 18
SS 98 6 15.50 119.50
SS df ms
Between 163.00 2 81.50
Within 119.50 15 7.97
Total 282.50 17 16.62
F = 10.23
Critical value of F = 3.68. F is significant.
Which is the best coffee. You’ll have to do t-tests to find out. But looking at the numbers above, you can guess that Blue Label is most likely to be the big winner. But you’ll know more after you do the post tests.
Item 4
Which is candle lasts more days?
Non-Sented Low-Sented High-Sented
3 5 8
9 1 2
5 2 6
11 5 4
Sum 28 13 20 61
Sum X2 236 55 120 411
N 4 4 4 12
SS 40 12.75 20 72.75
SS df ms
Between 27.17 2 14.08
Within 72.75 9 8.08
Total 100.92 11 9.17
F = 1.74
Critical value of F = 4.26. F is not significant.
Which is candle lasts more days? No significant difference between them. Choose any.
NOW YOU CHOOSE
Day9: 1-Way ANOVA
Bit More About 1-Way ANOVA
Even More About 1-Way ANOVA
Calculate 1-Way ANOVA
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Quiz 9
Summary
More Practice Problems: 1-Way ANOVA
November 24, 2008 by
Filed under ANOVA, Practice Problems
13 1 5
4 1 2
10 2 2
13 2 2
11 2 6
3 4 4
Between
Within
Total
Critical value of F =
3 5 8
9 1 2
5 2 6
11 5 4
Between
Within
Total
Critical value of F = .
Which is candle lasts more days?
Day9: 1-Way ANOVA
Bit More About 1-Way ANOVA
Even More About 1-Way ANOVA
Calculate 1-Way ANOVA
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Quiz 9
Summary
Practice Problem ANSWERS: 1-Way ANOVA
November 24, 2008 by
Filed under Calc Solutions
Item 1
Year in school and car accidents.
10th 11th 12th
2 13 4
9 17 8
3 14 2
1 9 1
7 1 4
SS df ms
Between 150.53 2 75.27
Within 228.80 12 19.07
Total 379.33 14 27.10
F = 3.95
Which grade has the most car accidents: The critical value is 3.74, so F is significant. You can now do multiple t-tests to discover which means are significantly different from each other.
Item 2
House color and people’s stay (in years).
Blue Green Peach
8 11 4
7 9 8
3 7 9
1 18 2
9 12 4
SS df ms
Between 116.13 2 58.07
Within 151.60 12 12.63
Total 267.73 14 19.12
F = 4.60
Which color of house is lived in the longest (in years)? The critical value is 3.74, so F is significant. You can now do multiple t-tests to discover which means are significantly different from each other.
NOW YOU CHOOSE
Day9: 1-Way ANOVA
Bit More About 1-Way ANOVA
Even More About 1-Way ANOVA
Calculate 1-Way ANOVA
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Quiz 9
Summary
Practice Problems: 1-Way ANOVA
November 24, 2008 by
Filed under ANOVA, Practice Problems
Item 1
Year in school and car accidents.
10th 11th 12th
2 13 4
9 17 8
3 14 2
1 9 1
7 1 4
SS df ms
Between ____ ___ ____
Within ____ ___ ____
Total ____ ___ ____
F =
Which grade has the most car accidents:
Item 2
House color and people’s stay (in years).
Blue Green Peach
8 11 4
7 9 8
3 7 9
1 18 2
9 12 4
SS df ms
Between ____ ___ ____
Within ____ ___ ____
Total ____ ___ ____
F =
Which color of house is lived in the longest (in years)?
NOW YOU CHOOSE
Day9: 1-Way ANOVA
Bit More About 1-Way ANOVA
Even More About 1-Way ANOVA
Calculate 1-Way ANOVA
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Quiz 9
Summary
Word Problem ANSWERS: Regression #1
November 23, 2008 by
Filed under Word Prob Sol.
As a director of counseling, you are interested in how well your therapy works. In particular, you are curious whether the number of counseling sessions is a good predictor of happiness (smiles per hour). Here is the sample you collected.
Sessions Happiness
12 4
3 7
9 2
4 11
7 9
5 5
6 9
2 7
2 9
What is the SS for Counseling: 76
What is the standard deviation of Counseling: 3.08
What is the range of Happiness: 9
What is the SS for Happiness: 66
What is the SSxy: - 37
Since you are interested in how well one variable acts as a linear predictor of another, which of the following tests should you perform:
a. t-test
b. ANOVA
c. correlation
d. regression
e. multiple regression
Perform the comparison you selected in the item above. Select only the appropriate one(s). What was the result of your calculations?
a = 9.92
b = -.49
r =
t =
F =
If a client has 10 sessions with you, what would you predict their happiness to be? 5.05
Within what boundaries is this estimate accurate? Plus or minus 2.62.
Note: If happiness predicted sessions, than a = 9.92, b = -.52 and standard error of estimate = 2.81.
Day 6: Regression
Bit More About Regression
Even More About Regression
Calculate Regression
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Quiz 6
Summary
Even More About z-Scores
November 22, 2008 by
Filed under Even More
There are 5 primary applications of z-scores.
First, z-scores can be used for describing the location of an individual score. If your score is at the mean, your z score equals zero. If z = 1, you are one standard deviation above the mean. If z = -2, you are two standard deviations below the mean. If z = 1.27, you score is a bit more than one and 1/4th standard deviations above the mean.
What is the z-score for a raw score of 104, a mean of 110 and a standard deviation of 12? 104-110 equals -6; -6 divided by 12 equals -.5. The raw score of 104 is one-half a standard deviation below the mean.
Second, raw scores can be evaluated in relation to some set z-score standard; a cutoff score. For example, all of the scores above a cutoff z-score of 1.65 could be accepted. In this case, z-scores provide a convenient way of describing a frequency distribution regardless of what variable is being measured.
Each z score’s location in a distribution is associated with an area under the curve. A z of 0 is at the 50th percentile and indicates that 50% of the scores are below that point. A z score of 2 is associated with the 98th percentile. If we wanted to select the top 2% of the individuals taking a musical ability test, we would want those who had a z score of 2 or higher. Z scores allow us to compare an individual to a standard regardless of whether the test had a mean of 60 or 124.
Most statistics textbooks have a table that shows the percentage of scores at any given point of a normal distribution. You can begin with a z score and find an area or begin with an area and find the corresponding z score. Areas are listed as decimals: .5000 instead of 50%. In order to save space, tables only list positive values are shown. The tables also assume you know that 50% of the scores fall below the mean and 50% above the mean. The table usually has 3 columns: the z score, the area between the mean and z, and the area beyond z.
The area between the mean and z is the percentage of scores located between z and the mean. A z of 0 has an area between the mean and z of 0 and the area beyond (the area toward the end of the distribution) as .5000. Although there are no negatives, notice that a z score of -0 would also have an area beyond (toward the low end of the distribution) of .5000.
A z score of .1, for example, has an area between the mean and z of .0398. That is, 3.98% of the scores fall within this area. And the third column shows that the area beyond (toward the positive end of the distribution) is .4602. If the z has -.1, the area from the mean down to that point would account for 3.98% of the scores and the area beyond (toward the negative end of the distribution) would be .4602.
Areas under the curve can be combined. For example, to calculate the percentile of a z of .1, the area between the mean and z (.0398) is added to the area below z (which you know to be .5000). So the total percentage of scores below a z of .1 is 53.98 (that is, .0398 plus .5000). A z score of .1 is at the 53.98th percentile.
Third, an entire variable can be converted to z-scores. This process of converting raw scores to z-scores is called standardizing and the resulting distribution of z-scores is a normalized or standardized distribution. A standardized test, then, is one whose scores have been converted from raw scores to z-scores. The resultant distribution always has a mean of 0 and a standard deviation of 1.
Standardizing a distribution gets rid of the rough edges of reality. If you’ve created a nifty new test of artistic sensitivity, the mean might be 123.73 and the standard deviation might be 23.2391. Interpreting these results and communicating them to others would be easier if the distribution was smooth and conformed exactly to the shape of a normal distribution. Converting each score on your artistic sensitivity test to a z score, converts the raw distribution’s bumps and nicks into a smooth normal distribution with a mean of 0 and a standard deviation of 1. Z scores make life prettier.
Fourth, once converted to a standardized distribution, the variable can be linearly transformed to have any mean and standard deviation desired. By reversing the process, z-scores are converted back to raw score by multiplying each by the desired standard deviation and add the desired mean. Most intelligence tests have a mean of 100 and a standard deviation of 15 or 16. But these numbers didn’t magically appear. The original data looked as wobbly as your test of artistic sensitivity. The original distribution was converted to z scores and then the entire distribution was shifted.
To change a normal distribution (a distribution of z scores) to a new distribution, simply multiply by the standard deviation you want and add the mean you want. It’s easy to take a normalized distribution and convert it to a distribution with a mean of 100 and a standard deviation of 20. Begin with the z scores and multiply by 20. A z of 0 (at the mean) is still 0, a z of 1 is 20 and a z of -1 is -20. Now add 100 to each, and the mean becomes 100 and the z of 1 is now 120. The z of -1 becomes 80, because 100 plus -20 equals 80. The resulting distribution will have a mean of 100 and a standard deviation of 20.
Fifth, two distributions with different means and standard deviations can be converted to z-scores and compared. Comparing distributions is possible after each distribution is converted into z’s. The conversion process allows previously incomparable variables to be compared. If a child comes to your school but she old school used a different math ability test, you can estimate her score on your school’s test by converting both to z scores.
If her score was 65 on a test with a mean of 50 and a standard deviation of 10, her z score was 1.5 on the old test (65-50 divided by 10 equals 1.5). If your school’s test has a mean of 80 and a standard deviation of 20, you can estimate her score on your test as being 1.5 standard deviations above the mean; a score of 110 on your test.
NOW YOU CHOOSE:
Day 4: z-Score
A Bit More About z-Scores
Even More About z-Scores
How To Calculate z-Scores
Practice Problems
Basic Facts About z-Scores
Vocabulary
Formulas For z-Scores
Quiz 4
Summary
Practice Problem ANSWERS: Dispersion
November 22, 2008 by
Filed under Calc Solutions
Item 1
Calculate the range of the following scores:
19
15
8
5
5
2
High Score 19
Low Score 2
Range 17
Item 2
Calculate the range and SS of the following scores:
11
9
8
7
5
7
High Score 11
Low Score 5
SumX 47
SumX2 389
(SumX)2 2209
N 6
SS 20.83
variance 3.47 (if a population) or 4.17 (if a sample)
Item 3
Calculate the variance of the following population:
18
9
7
5
4
2
SumX 45
SumX2 499
(SumX)2 2025
N 6
SS 161.50
variance 26.92
Item 4
Calculate the SS, variance and stdev of the following population:
3
18
9
6
10
6
4
SumX 56
SumX2 602
(SumX)2 3136
N 7
SS 154
variance 22
stdev 4.69
Item 5
Calculate the range, SS, variance and stdev of the following sample:
12
3
9
5
5
5
7
11
14
- 1
Range 15
N 10
SS 186
variance 20.67 (it is a sample, so it is Sum of Squares divided by N-1)
The standard deviation would be 4.55
NOW YOU CHOOSE:
Day 3: Dispersion
A Bit More About Dispersion
Even More About Dispersion
Range
MAD
Sum of Squares
Variance
Standard Deviation
How To Calculate
Range
MAD
Sum of Squares
Variance
Standard Deviation
Formulas For Dispersion
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Basic Facts About Dispersion
Vocabulary
Quiz 3
Summary
Calculate: Range
November 22, 2008 by
Filed under How To Calculate, Range
Range is the high score minus the low score.
It’s easy to calculate, and very helpful for identifying typing mistakes. If your 5-point rating scale has a range of 72, you know something is wrong. It should only have values from 1 to 5. So the range should only be 4.
NOW YOU CHOOSE:
Day 3: Dispersion
A Bit More About Dispersion
Even More About Dispersion
Range
MAD
Sum of Squares
Variance
Standard Deviation
How To Calculate
Range
MAD
Sum of Squares
Variance
Standard Deviation
Formulas For Dispersion
Practice Problems
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Basic Facts About Dispersion
Vocabulary
Quiz 3
Summary
Even More About Standard Deviation
November 22, 2008 by
Filed under Even More
Like the other four measures of dispersion, the standard deviation gets smaller as the scores get more homogeneous, and larger the more heterogeneous they become. A small standard deviation indicates the scores are quite similar to the mean (closer to the peak); a large standard deviation says the score vary from the mean.
This measure of dispersion is calculated by taking the square-root of variance. Regardless of whether you used N or N-1 to calculate variance, standard deviation is the square-root of variance. If variance is 7.22, the standard deviation is 2.69. If variance is 8.67, the standard deviation equals 2.94.
Technically, the square-root of a population variance is called sigma and the square-root of a sample variance is called the standard deviation. As a general rule, population measures use Greek symbols and sample parameters use English letters. Since we tend to use large samples, we’ll focus on the standard deviation.
STANDARD
The standard deviation is “standard” in the sense that it takes steps of equal distance from the mean. Think of it as standing at the mean and taking 3 steps in one direction. If doesn’t matter if you step toward the high end or the low end. It only takes three steps to get from the mean to the end of a distribution. If you start at the mean and go toward the positive end, you’re there in 3 steps; and it’s 3 steps from the mean to the lowest end of the distribution. So the entire distribution is comprised of 6 steps (3 positive steps and 3 negative steps).
STEPS (Deviations From The Mean)
Each of these steps is the equal in distance but accounts for a different amount of people. The normal curve is like a mountain. If you’re standing on top of the mountain, your first step is always your largest. In a frequency distribution of a normally distributed variable, your first step accounts for the most people. Because most scores are close to the mean, most scores fall within plus or minus one standard deviation from the mean.
In fact, that’s our definition of normal. Normal is being close to the mean. Normal musical ability is scoring at the mean plus or minus one standard deviation. Normal basketball throwing is at the mean, plus or minus one standard deviation.
In a normally distributed variable, the percentages are consistent, regardless of what is being measured. Starting from the mean, the first step accounts for just over 34% of the scores. The next steps has 14% and the last step has 2%. Since normal frequency distributions are symmetrical, the percentages work on either side of the man. So the entire distribution looks like this: 
NOW YOU CHOOSE:
Day 3: Dispersion
A Bit More About Dispersion
Even More About Dispersion
Range
MAD
Sum of Squares
Variance
Standard Deviation
How To Calculate
Range
MAD
Sum of Squares
Variance
Standard Deviation
Formulas For Dispersion
Practice Problems
More Practice Problems
Basic Facts About Dispersion
Vocabulary
Quiz 3
Summary
