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Inquiry 1 proposals were/are due in lab Th 9/9 or M 9/13 Homework #1 in lab Today: Mean, Median, Mode, T-test xkcd.com
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Most statistical work can be done, and more easily done, by computer using programs such as: MS Excel Open Office SPSS SAS
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The Basics: mean, median, and mode
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Mean- aka the average. Sum of all numbers divided by the number of data points. (14+17+7+6+4+11+8)/7 = 9.57
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Median- the middle number of a group of ordered numbers 1 17 7 6 4 11 8 4 6 7 8 11 14 17 median is 8
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Median- the middle number of a group of ordered numbers 1 17 7 6 4 11 8 26 7 8 11 14 17 median is 8 What about 4 6 7 11 14 17?
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Median- the middle number of a group of ordered numbers 1 17 7 6 4 11 8 26 7 8 11 14 17 median is 8 What about 4 6 7 11 14 17? Median is 9.
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Mode- the most common value in a group. 9, 8, 3, 4, 5, 2, 4, 5, 2, 3, 6, 1, 6, 2, 3, 9, 2, 6 Mode is 2
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Why are there 3 ways to analyze a group of numbers?
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The mean is the most common form of analysis.
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Why are there 3 ways to analyze a group of numbers? The mean is the most common form of analysis. 2, 3, 2, 4, 2, 7, 2, 5, 3, 2, 5, 4, 3, 5, 6, 121, 130 Mean = 18
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Why are there 3 ways to analyze a group of numbers? 2, 3, 2, 4, 2, 7, 2, 5, 3, 2, 5, 4, 3, 5, 6, 121, 130 Mean = 18 Is this an accurate representation of these numbers?
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Why are there 3 ways to analyze a group of numbers? 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7, 121, 130 Median = 4 Mean = 18 Median can be more accurate when there are a few especially large or small numbers.
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What is your favorite color?
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Mode can be used with non-numerical data.
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Is there a numerical way to determine the accuracy of our analysis? 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7, 121, 130 Median = 4 Mean = 18
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Is there a numerical way to determine the accuracy of our analysis? 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7, 121, 130 Median = 4 Mean = 18 Standard Deviation is a measure of variability.
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Standard deviation is a measure of variability. The standard deviation is the root mean square (RMS) deviation of the values from their arithmetic mean. where ∑ = Sum of X = Individual score M = Mean of all scores N = Sample size (Number of scores)
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Is there a numerical way to determine the accuracy of our analysis? 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7, 121, 130 Mean = 18 Standard deviation = 40.5 Standard deviation is a measure of variability.
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Is there a numerical way to determine the accuracy of our analysis? 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 Standard deviation = 1.6 Standard deviation is a measure of variability.
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Is there a numerical way to determine the accuracy of our analysis? 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 (121, 130) Mean = 3.67 Median was 4
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Is there a numerical way to determine the accuracy of our analysis? 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 Standard deviation is a measure of variability.
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Percent of data at 1, 2, or 3 standard deviations from the mean http://www.westgard.com/lesson34.htm 3.7 2.15.3 -1.1 0.56.9 8.5
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http://www.westgard.com/lesson34.htm 3.7 2.15.3 -1.1 0.56.9 8.5 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 8/15=53% 14/15=93% 15/15=100%
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Standard deviation can also be used to determine if 2 data sets are statistically different.
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Standard deviation can also be used to determine if 2 data sets are statistically different… Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 And Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 9, 5 Mean = 7.25 ± 1.48
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Do the standard deviations overlap? Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 range = 2.07 to 5.27 And Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 9, 5 Mean = 7.25 ± 1.48 range = 5.77 to 8.73 No overlap, might be different
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Do the standard deviations overlap? Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 range = 2.07 to 5.27 And Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 4, 5 Mean = 6.83 ± 1.64 range = 5.19 to 8.47 Overlap, different means, but might not be a statistically significant difference
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How significant of a difference is this? Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 range = 2.07 to 5.27 And Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 9, 5 Mean = 7.25 ± 1.48 range = 5.77 to 8.73 No overlap, might be different
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The ‘Students’ T-test is a method to assign a numerical value of statistical difference.
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X 1 –X 2 T= n1n1 Sx 1 22 + n2n2 Sx 2
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The ‘Students’ T-test is a method to assign a numerical value of statistical difference. X 1 –X 2 T= n1n1 Sx 1 22 + n2n2 Sx 2 (Difference between means) (variance) (sample size)
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The ‘Students’ T-test is a method to assign a numerical value of statistical difference. X 1 –X 2 T= n1n1 Sx 1 22 + n2n2 Sx 2 T is then used to look up the P-value from a table. Also need ‘degrees of freedom’ = (n 1 +n 2 )-1.
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P-value Df0.050.020.01 112.7131.8263.66 24.3036.9659.925 33.1824.5415.841 42.7763.7474.604 52.5713.3654.032 62.4473.1433.707 72.3652.9983.499 82.3062.8963.355 92.2622.8213.250 102.2282.7643.169 T Partial table for determining P from T
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How significant of a difference is this? Using a speadsheet to get a P value = 3.44x10 -6. Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 And Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 9, 5 Mean = 7.25 ± 1.48
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How significant of a difference is this? P value = 3.44x10 -6. So the chance that these 2 sets of data are not significantly different is 3.44x10 -6 Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 And Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 9, 5 Mean = 7.25 ± 1.48
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How significant of a difference is this? P value = 3.44x10 -6. So the chance that these 2 sets of data are significantly different is 1 - 3.44x10 -6 or 0.999996559 We can be 99.9996559% certain that the difference is statistically significant. Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 9, 5 Mean = 7.25 ± 1.48
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Overlap, different means, but might not be a statistically significant difference Set 1= 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5, 5, 6, 7 Mean = 3.67 ±1.6 range = 2.07 to 5.27 Set 2= 8, 6, 7, 8, 9, 5, 6, 7, 9, 8, 4, 5 Mean = 6.83 ± 1.64 range = 5.19 to 8.47 P-value = 4.41 x 10 -5
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Generally a P-value of 0.05 or less is considered a statistically significant difference. 20% random difference : 80% confidence 10% random difference : 90% confidence 5% random difference : 95% confidence 1% random difference : 99% confidence 0.1% random difference : 99.9% confidence
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Standard deviation is NOT a valid method for determining statistical signifigance. T-test is one valid and accurate method for determining if 2 means have a statistically significant difference, or if the difference is merely by chance.
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Inquiry 1 proposals were/are due in lab Th 9/9 or M 9/13 Homework #1 in lab Next: More Stats and Outliers... xkcd.com
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