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Published byDarlene Floyd Modified over 9 years ago
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Research Methods: 2 M.Sc. Physiotherapy/Podiatry/Pain Frequency/Probability Polygons, and the Normal Distribution
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Part one: Frequency Tables Un-grouped Tally observations Frequency table Histogram Polygon Grouped Set class limits Tally number in class Frequency table Histogram Polygon
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Ungrouped Frequency Tables; Data from n = 25, rating 1-5 of RM2 teaching
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Ungrouped Frequency Tables; Frequency Table
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Ungrouped Frequency Tables; Data from n = 25, rating 1-5 of RM2 teaching
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Grouped Frequency Tables; Data of weights (kg) n = 12
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Grouped Frequency Tables; Setting class limits Find range Choose number of classes (5 20) Classes equal size (Outliers?) Choose limits at level of measurement precision Tally
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Grouped Frequency Tables Class boundaries Half way between classes One more decimal place than limits Class intervals Distance between boundaries Midpoints Half way between boundaries Mid point of interval
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Grouped Frequency Tables
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Histograms Present information from Frequency tables Show distribution of the data set Columns start and end at class boundaries Midpoints are marked Join midpoints = Frequency/Probability Polygon Area represent frequency/ probability; total area under curve; p = 1.00
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Histograms; Frequency
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Histograms; Probability
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Frequency/Probability Polygons
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Part two: The Normal Distribution A type of (family) of distributions Most important of all known distributions Natural parameters in populations Symmetrical bell shaped curve
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Normal Distribution Or Frequency Probability SD or
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68.2% ± 1SD 95.4% ± 2SD 99.7% ± 3SD
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p = 0.682 ± 1SD p = 0.954 ± 2SD p = 0.997 ± 3SD
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p if not exact multiple of SD away from mean ?
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Z scores Data point of interest = x Mean = Standard deviation = Z score is number of multiples of SD the data point is away from mean ; z = x -
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Z scores Look up the Z score in Tables to find; Probability associated with values below x and vice versa. Why ???
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Graph of number of visits to Physiotherapist for Sports rehabilitation; 16 z = (16 - 10) /4 z = 1.5 p = 0.9332 p = 1 - 0.9332 p = 0.067
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95% of data p = 0.95 p < 0.05
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