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COURSE: JUST 3900 TIPS FOR APLIA Chapters 4-6: Test Review Developed By: Ethan Cooper (Lead Tutor) John Lohman Michael Mattocks Aubrey Urwick Chapter 4: Key Concepts Know that this chapter is on variability, which measures the differences between scores and describes the degree to which the scores are spread out or clustered together. Variability also measures how well an individual score represents the entire distribution. The 3 measures of variability: the range, standard deviation, and the variance. Chapter 4: The Range The range is the distance covered by the scores in a distribution, from the smallest to the largest score. There are 2 formulas for the range that you should be aware of for the test: range = URL for Xmax – LRL for Xmin range = Xmax – Xmin Remember: URL stands for upper real limit, and LRL stands for lower real limit. This is the definition used by many computer programs. Chapter 4: The Range Find the range for the following set of scores using both formulas: 8, 1, 5, 1, 5 Chapter 4: The Range range = URL for Xmax – LRL for Xmin range = 8.5 – 0.5 = 8 range = Xmax – Xmin range = 8 – 1 = 7 Chapter 4: Sum of Squares Sum of Squares (SS) is the sum of the squared deviations. There are 2 formulas you need to know to compute SS: Definitional Formula (Population): = − Definitional Formula (Sample): = − 2 2 Computational Formula (Population): = Computational Formula (Sample): = 2 − 2 − Use when mean is a whole number. 2 2 Use for fractional means. Chapter 4: Variance Variance is the mean squared deviation, or the average squared distance from the mean. 2 = Population Variance: 2 Sample Variance: = The variance will play a bigger role as we move into inferential statistics in the coming chapters. For now, just know that it is a necessary step to finding the standard deviation. −1 Chapter 4: Standard Deviation Standard deviation provides a measure of the standard, or average, distance from the mean. A large standard deviation tells us that our scores are widely distributed. A small standard deviation tells us that they are clustered closely around the mean. It is calculated by taking the square root of the variance. Population Standard Deviation: = Sample Standard Deviation: = 2 or = 2 or = −1 Chapter 4: Standard Deviation Find the standard deviation for the following population of N =7 scores: 8, 1, 4, 3, 5, 3, 4 Chapter 4: Standard Deviation Step 1: Find the mean = = 8+1+4+3+5+3+4 7 = 28 7 =4 Step 2: Find SS = − 2 Because the mean is a whole number. X (X - µ) (X - µ)2 8 8–4=0 (4)2 = 16 1 1 – 4 = -3 (-3)2 = 9 4 4–4=0 (0)2 = 0 3 3 – 4 = -1 (-1)2 = 1 5 5–4=1 (1)2 = 1 3 3 – 4 = -1 (-1)2 = 1 4 4–4=0 (0)2 = 0 SS = 8 + 9 + 0 + 1 + 1 + 1 + 0 = 28 Chapter 4: Standard Deviation Step 3: Calculate the variance 2 = = 28 7 =4 Step 4: Find the standard deviation = = 28 7 = 4=2 Chapter 4: Sample Standard Deviation A sample statistic is unbiased if the average value of the statistic is equal to the population parameter. A sample statistic is biased if the average value of the statistic either overestimates or underestimates the corresponding population parameter. To avoid bias when calculating s2 or s, we use n – 1, instead of n. n – 1 is referred to as degrees of freedom, or df. Chapter 4: Sample Standard Deviation Find the standard deviation for the following sample of n = 4 scores: 7, 4, 2, 1 Chapter 4: Sample Standard Deviation Step 1: Find the mean = = 7+4+2+1 4 = 14 4 = 3.5 Step 2: Find SS = 2 − 2 Because the mean contains decimals. X X2 7 (7)2 = 49 4 (4)2 = 16 2 (2)2 = 4 1 (1)2 = 1 = 14 2 = 70 Chapter 4: Sample Standard Deviation Step 2: Find SS 2 − = 70 − 14 2 4 = 70 − 196 4 = 70 − 49 = 21 Step 3: Calculate the variance = 2 2 = −1 = 21 4−1 = 21 3 =7 Use n-1 in the denominator because we’re working with a sample. Step 4: Find the standard deviation = −1 = 21 3 = 7 = 2.65 Chapter 4: More About Standard Deviation It should be noted that adding a constant to each score in a distribution does not change the standard deviation. Remember that standard deviation is a measure of variability, or the distance between scores. If the same number were added to every score, the entire distribution would shift to the right, but the distance between each score would remain the same. Chapter 4: More About Standard Deviation However, multiplying every score by a constant would cause the standard deviation to be multiplied by the same constant. Imagine a distribution that included scores of X = 10 and X = 11. The distance between these scores is only 1 point. Now imagine we multiply every score in this distribution by 2 points. Our new scores would be X = 20 and X = 22, 2 points apart. Because the distance between scores changes when we multiply by a constant, the standard deviation also changes by the same constant. Chapter 5: Key Concepts z-Scores specify the precise location of each X value within a distribution. The sign of the z-score (+ or -) signifies whether the score is above the mean (positive) or below the mean (negative). The numerical value of the z-score describes the distance from the mean by counting the number of standard deviations between X and µ. The z-score distribution will always have a mean of µ = 0 and a standard deviation of σ = 1. Chapter 5: z-Scores The formulas for calculating the z-score: − − Population z-score: = Sample z-score: = Sometimes it’s necessary to calculate the X-value using the z-score, mean, and standard deviation. = + = + Chapter 5: z-Scores For a population with µ = 50 and σ = 8, find the z-score for each of the following X values: X = 54 X = 42 X = 62 X = 48 Chapter 5: z-Scores X = 54 = 54−50 8 = 4 8 − = 42−50 8 = −8 8 = −1.00 − = 62−50 8 = 12 8 = 1.50 − = 48−50 8 = −2 8 = −0.25 = 0.50 = X = 62 − X = 42 = = X = 48 = Chapter 5: Find X Find the X value that corresponds with the following zscores: (µ = 50 and σ = 8) z = -0.50 z = 0.75 z = -1.50 z = 0.25 Chapter 5: Find X z = -0.50 z = 0.75 = + = 50 + (0.75)(8) = 50 + 6 = 56 z = -1.50 = + = 50 + (−.50)(8) = 50 − 4 = 46 = + = 50 + (−1.50)(8) = 50 − 12 = 38 z = 0.25 = + = 50 + (0.25)(8) = 50 + 2 = 52 Chapter 5: Standardized Distributions Standardized distributions are composed of scores that have been transformed to create predetermined values for µ and σ. Standardized distributions are used to make dissimilar distributions comparable. A z-score distribution is an example of a standardized distribution. Chapter 5: Standardized Distributions A distribution with μ = 62 and σ = 8 is transformed into a standardized distribution with μ = 100 and σ = 20. Find the new, standardized score for each of the following X values: X = 60 X = 54 X = 72 X = 66 Chapter 5: Standardized Distributions Step 1: Find the z-scores X = 60 = − = 60−62 8 = −2 8 = −0.25 Step 2: Find the X value for the new distribution = + = 100 + (−0.25)(20) = 100 − 5 = 95 Chapter 5: Standardized Distributions Step 1: Find the z-scores X = 54 − = 54−62 8 = −8 8 = −1.00 − = 72−62 8 = 10 8 = 1.25 − = 66−62 8 = 4 8 X = 72 = = X = 66 = = 0.50 Chapter 5: Standardized Distributions Step 2: Find the X values for the new distribution X = 54, z = -1.00 X = 72, z = 1.25 = + = 100 + (−1.00)(20) = 100 − 20 = 80 = + = 100 + (1.25)(20) = 100 + 25 = 125 X = 66, z = 0.50 = + = 100 + (0.50)(20) = 100 + 10 = 110 Chapter 6: Key Concepts For a situation in which several different outcomes are possible, the probability for any specific outcome is defined as a fraction of all the possible outcomes. Probability can be represented as a fraction, decimal, or percent. Probability of A = p = 0.25 = ¼ = 25% A random sample requires that each individual in the population has an equal chance of being selected. An independent random sample requires that each individual has an equal chance of being selected and that the probability of being selected remains constant from one selection to the next. Chapter 6: Random Sampling Sampling with replacement requires that selected individuals be returned to the population before the next selection is made. This ensures that the probability of selection remains constant from one selection to the next. Unless otherwise specified, random sampling assumes replacement. Probability does not remain constant when sampling without replacement. Chapter 6: Random Sampling A psychology class consists of 14 males and 36 females. If the professor selects names from the class list using random sampling, a) b) What is the probability that the first student selected will be a female? If a random sample of n = 3 students is selected and the first two are both females, what is the probability that the third student selected will be a male? Chapter 6: Random Sampling a) b) p = 36/50 = 0.72 p = 14/50 = 0.28 Because this is a random sample, replacement is assumed. Therefore, the probability of selection remains constant. Chapter 6: The Normal Distribution When dealing with probability in this chapter, we are dealing with normal distributions. 34.13% Normal distributions are symmetrical. 13.59% Percentage of the population located between z = 0 and Z = 1. 2.28% The unit normal table lists proportions of the normal distribution for a full range of possible z-score values. Chapter 6: The Unit Normal Table The first column (A) in the table lists z-scores corresponding to different positions in a normal distribution. Column B presents the proportion in the body. Column C presents the proportion in the tail. The body is always the larger part of the distribution. The tail is always the smaller part of the distribution. Column D identifies the proportion located between the mean and the z-score. Note: The normal distribution is symmetrical, so the proportions on the right are the same as the proportions on the left. And although the z-score values change signs (+ and -), the proportions are always positive. Chapter 6: The Unit Normal Table Find each of the probabilities for a normal distribution: p(z > 0.25) p(z > -0.75) p(z < 1.20) p(-0.25 < z < 0.25) p(-1.25 < z < 0.25) Chapter 6: The Unit Normal Table p(z > 0.25) p = 0.4013 p(z > -0.75) p = 0.7734 p(z < 1.20) p = 0.8849 p(-0.25 < z < 0.25) p(-1.25 < z < 0.25) Find the proportion in column D for both z-scores and add them together. p = 0.0987 p = 0.4931 Chapter 6: Binomial Distributions When a variable is measured on a scale consisting of exactly two categories, the resulting data are called binomial. The normal distribution can be used to compute probabilities with binomial data. The two categories are defined as A and B. The probabilities associated with each category are: p = p(A) = the probability of A q = p(B) = the probability of B p + q = 1.00 The number of individuals or observations in the sample is identified by n. The variable X refers to the number of times category A occurs in the sample. Chapter 6: Binomial Distributions Binomial distributions tend to approximate a normal distribution when two conditions are met: pn ≥ 10 qn ≥ 10 Under these circumstances, the binomial distribution has the following parameters: Mean: µ = pn Standard deviation: σ = z= − = − Chapter 6: Binomial Distributions Notice that each X value is represented by a bar in the histogram. This means a score of X = 8 spans the interval of 7.5 to 8.5. Chapter 6: Binomial Distributions A multiple choice test has 48 questions, each with four response choices. If a student is simply guessing at the answers, a) b) c) d) What is the probability of guessing correctly for any question? On average, how many questions would a student get correct for the entire test? What is the probability that a student would get more than15 answers correct simply by guessing? What is the probability that a student would get 15 or more answers correct simply by guessing? Chapter 6: Binomial Distributions a) b) p = ¼ = 0.25 pn = 0.25(48) = 12 Chapter 6: Binomial Distributions c) Step 1: Find pn and qn Both are greater than 10, so our distribution approximates a normal distribution. Step 2: Find μ and σ µ = pn = 12 σ = = 48(0.25)(0.75) = 9 = 3 URL because we are excluding the score X = 15. Step 3: Find Z pn = (0.25)(48) = 12 qn = 0.75(48) = 36 z= − = − = 15.5 −12 3 = 3.5 3 Step 4: Find the proportion p(z > 1.17) = 0.1210 = 1.17 Chapter 6: Binomial Distributions d) Step 1: Find pn and qn Step 2: Find μ and σ µ = pn = 12 σ = = 48(0.25)(0.75) = 9 = 3 LRL because we are including the score X = 15. Step 3: Find Z pn = (0.25)(48) = 12 qn = 0.75(48) = 36 z= − = − = 14.5 −12 3 = 2.5 3 Step 4: Find the proportion p(z > 0.83) = 0.2033 = 0.83