Terms & Characteristics

Transcription

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2 NORMAL CURVE Knowledge that a variable is distributed normally can be helpful in drawing inferences as to how frequently certain observations are likely to occur.

3 NORMAL CURVE A Normal distribution: Distribution with specific characteristics symmetrical, bell-shaped mean, median, and mode are the same this is an assumption for many statistical techniques and should be checked 68% w/i + 1SD, 95% w/i + 2SD, 99% w/in + 3SD

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5 Terms & Characteristics Outliers - scores that differ so markedly from the main body of data as to raise questions about their accuracy need to double check outliers, do not delete without a good reason.

6 Terms & Characteristics Skewness - refers to the asymmetry of a distribution Positively skewed or right skewed means that extreme scores lie to the right. (Hint: order of appearance on axis is reverse alphabetic: mode, median, mean)

7 Terms & Characteristics Negatively skewed or left skewed means that extreme scores lie to the left. (Hint: order of appearance on axis is alphabetic: mean, median, mode)

8 Terms & Characteristics Kurtosis - peakedness or flatness, or somewhere in between. Platykurtic: flat (broad hump and thick tails) Mesokurtic: peakedness like the normal distribution. Leptokurtic: peaked (narrow and thin tails)

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10 Quantiles maximum 100.0% % % % quartile 75.0% median 50.0% quartile 25.0% % % % minimum 0.0%

11 Moments Mean Std Dev Std Error Mean Upper 95% Mean Lower 95% Mean N Sum Weights Test for Normality Shapiro-Wilk W Test W Prob<W

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13 The standard normal is a very useful distribution. It has a mean (μ) of 0 and a standard deviation (σ) of 1. It expresses scores in terms of how many sd (+or-) from the mean. z = (x - μ)/σ Let s look at Table A

14 Example Determine the area under the normal curve falling to the right of the given Z: Z = 1.84

15 Example Find the value (1.84) in the table and locate the associated area between the value and the mean 1.84

16 Example Find the value (1.84) in the table and locate the associated area between the value and the mean

17 Example Subtract that area from 0.5 Area = =

18 Example Interpretation: 3.29 % of values fall above 1.84 standard deviations

19 Exercise #1 Serum cholesterol levels were taken from a population of college students. The results were normally distributed. Males: μ = 195, σ = 10 Females: μ = 185, σ = 12

20 What %age of females would have a cholesterol level greater than 200? What kind of picture should you draw?

21 What is the z-score associated with cholesterol of 200? Recall: z = (x - μ)/σ In this case: μ = 185 & σ = 12 z = (200 - )/ z =

22 What percentage of (female) scores are greater than z = 1.25? Table A value Area of interest z =

23 Interpretation: What %age of females would have cholesterol level greater than 200? % of females have cholesterol > 200.

24 What %age of males would have cholesterol level less than 180? What is the z-score associated with cholesterol of 180?

25 What is the z-score associated with cholesterol of 180? Recall: z = (x - μ)/σ In this case: μ = 195 & σ = 10 z = (180 - )/ z =

26 z = 0

27 What %age of males would have a cholesterol level less than 180? 6.68 % of males have cholesterol < 180.

28 Exercise #2 We have two weights on a female patient. Reading A is 175 lbs. and Reading B is 74.8 kg. Has there been a change in the patient between readings? Given: A: μ = 175, σ = 10 B: μ = 68, σ = 4.5

29 Solution Exercise #2 z a = ( ) / 10 = 2.5 area between mean & 2.5 is.4938 z b = ( )/ 4.5 = 1.5 area between mean & 1.5 is.4332

30 What is the 95th percentile in pounds? Kilograms? (Given the above means and standard deviations.) You are looking for the x-value (ie, pounds) that corresponds to the 95th percentile.

31 Solution Solve the equation for x (Wt in Pounds). Plug in the values that you know. Z= x - μ = x σ (10) = x = x = x = weight in lbs

32 Solution Solve the equation for x. Plug in the values that you know. Z= x - μ = x -68 σ (4.5) = x = x = x 77 = weight in kg.

33 Central Limit Theorem states: If random samples are selected from a population with mean μ and finite standard deviation σ, as the sample size n increases, the distribution of X approaches a normal distribution with mean, μ and standard deviation, σ/ n.

34 Why is this so-o-o-o cool? Regardless of the shape of the sampled population, the means of sufficiently large samples will be nearly normally distributed. Therefore, we can work with sample means (provided n is large enough), and meet the assumption of normality!!!

35 A simple example. We have a uniform distribution with four data points. They are: 1,2,3,4 The mean and standard deviation are: μ = 2.5 and σ = 1.118

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37 Sample Mean Sample Mean 1,1 1 2, , , , ,4 3 1,3 2 4,2 3 3,1 2 3,3 3 1, , , , ,2 2 4,4 4

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39 So what is μ and σ for the means? μ = μ (from original distribution) σ = σ / n μ = 2.5 σ = / 2 =.790 which is what we get when we calculate directly from 16 means.

40 From JMP mean Mean Std Dev

41 How Smart are you? Standardized IQ test: μ = 100 σ = 15 If you score 130, what percentile?

42 How Smart are you? Standardized exam: μ = 1000 σ = 200 If you score 1390, what is your IQ?

43 How Smart are WE? Standardized IQ test: μ = 100 σ = 15 If WE (n = 9) have an average score = 125, are we the same as the general population?

44 Are WE different? How different is our average score = 125 than the general population mean = 100? Z = ( )/ (15/ 9) Z = 25 / 5 = 5

45 Are WE different? Table A value Area of interest z = 0 5