Statistics, Measures of Central Tendency I We are considering a - - PowerPoint PPT Presentation

Statistics, Measures of Central Tendency I

We are considering a random variable X with a probability distribution which has some parameters. We want to get an idea what these parameters are. We perfom an experiment n times and record the

• utcome. This means we have X1, . . . , Xn i.i.d. random variables, with

probability distribution same as X. We want to use the outcome to infer what the parameters are. Mean The outcomes are x1, . . . , xn. The Sample Mean is x := x1+···+xn

n

. Also sometimes called the average. The expected value of X, EX, is also called the mean of X. Often denoted by µ. Sometimes called population mean. Median The number so that half the values are below, half above. If the sample is of even size, you take the average of the middle terms. Mode The number that occurs most frequently. There could be several modes, or no mode.

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Statistics, Measures of Central Tendency II

Example

You have a coin for which you know that P(H) = p and P(T) = 1 − p. You would like to estimate p. You toss it n times. You count the number

• f heads. The sample mean should be an estimate of p.

EX = p, and E(X1 + · · · + Xn) = np. So E X1 + · · · + Xn n

• = p.

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Descriptive Statistics I

Frequency Distribution Divide into a number of equal disjoint intervals. For each interval count the number of elements in the sample occuring. Histogram see the next slide Grouped Data Mean Essentially calculate the mean of the frequency

• distribution. Intervals are used, rather than single values. It

is assumed that all these values are located at the midpoint

• f the interval. The letter xM is used to represent the

midpoints and f represents the frequencies: fixM,i n Frequency Polygon Connect the middles of the tops of each interval.

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Histogram

A histogram is a graphical representation of the distribution of numerical

• data. It is a kind of bar graph. To construct a histogram, the first step is

to ”bin” the range of values, that is, divide the entire range of values into a series of intervals, and then count how many values fall into each

• interval. The bins are usually specified as consecutive, non-overlapping

intervals of a variable. The bins (intervals) must be adjacent, and are

• ften (but are not required to be) of equal size.

Bin Count −3.5 − 2.51 9 −2.5 − 1.51 32 −1.5 − 0.51 109 −0.5 − 0.49 180 0.5 − 1.49 132 1.5 − 2.49 34 2.5 − 3.49 4 Mean:

(−3)·9+(−2)·32+(−1)·109+·(0)180+1·132+2·34+3·4 500

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Example

The table on the next page gives the number of days in June and July of recent years in which the temperature reached 90 degrees or higher in New Yorks Central Park. Source: The New York Times and Accuweather.com.

• a. Prepare a frequency distribution with a column for intervals and
• frequencies. Use seven intervals, starting with [0 4].
• b. Sketch a histogram and a frequency polygon, using the intervals in part

a.

• c. Find the mean for the original data.
• d. Find the mean using the grouped data from part a.
• e. Explain why your answers to parts c and d are different.
• f. Find the median and the mode for the original data.

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Temperature Data

fi 1972 11 1985 4 1998 5 1973 8 1986 8 1999 24 1974 11 1987 14 2000 3 1975 3 1988 21 2001 4 1976 8 1989 10 2002 13 1977 11 1990 6 2003 11 1978 5 1991 21 2004 1 1979 7 1992 4 2005 12 1980 12 1993 25 2006 5 1981 12 1994 16 2007 4 1982 11 1995 14 2008 10 1983 20 1996 2009 1984 7 1997 10 2010 20 Year Days Year Days Year Days

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Measures of Variation

Summary of Section 9.2 Range The difference Largest Data - Smallest Data in a Sample. Deviation from the Mean

1 Variance σ2 = s2 =

x2

i −nx2

n−1

=

(xi−x)2 n−1

2 Standard Deviation σ = s =

√ s2 These are random variables called Sample Variance and Sample Standard Deviation. For a random variable X, µ = E(X) is called the mean. The variance Var(X) is σ2 = Var(X) = E((X − µ)2). Main Property/ Explanation for dividing by n − 1: If Xi are i.i.d with distribution X, then if you set S2 =

(Xi−X)2 n−1

, its expected value is E(S2) = σ2. This is not true for the standard deviation, E(S) = σ. Grouped Data s = fix2

M,i − nx2

n − 1 .

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Examples I

Example (Range)

Data 15, −3, 4, 7, 18. The smallest is −3, the largest 18 so Range = 18 − (−3) = 21. Always a nonnegative number.

Example (Deviation from the Mean)

In the previous example, x = 15−3+4+7+18

5

= 8.2. So 15 − 8.2 = 6.8, −3 − 8.2 = −11.2, 4 − 8.2 = −3.8, 7 − 8.2 = −1.2, 18 − 8.2 = 9.8.

Example (Variance and Standard Deviation)

s2 = 6.82+11.22+3.82+1.22+9.82

4

= 152+32+42+72+182−5·8.22

4

s = √ s2.

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Examples II

Example (Binomial Distribution)

P(X = 1) = p, P(X = 0) = 1 − p. Then µ = E(X) = p, and σ2 = E((X − p)2) = (1 − p)2p + (0 − p)2(1 − p) = p(1 − p). This is the same as E(X 2 − p2) = (1 − p2)p + (−p2)(1 − p) = (1 − p)p. Remark: Note that the formula for variance and standard deviation only holds for n > 2. Otherwise, for n = 1, you would be dividing by 0. For one random variable, the variance is defined as Var(X) = E((X − E(X))2). For X1, X2,, two independent random variables, Var(X1 + X2) = Var(X1) + Var(X2). Suppose X is a random variable. We can write a table X a1 a2 . . . an P(X) p1 p2 . . . pn

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Examples III

For the expected value µ = E(X), you multiply the two terms in each column, and add

• i

ai × pn = a1p1 + · · · + anpn. In a spreadsheet program, the data would be in columns and you would add over the products from the rows. You use a command like sumproduct to perform the operation. If you have some other variable like (X − µ)2, you would use the values (ai − µ)2 and the same pi.

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Examples IV

Example

X 2 3 −1 1 X 2 4 9 1 1 (X − µ)2 (2 − 1/4)2 (3 − 1/4)2 (−1 − 1/4)2 (1 − 1/4)2 P(X) 1/2 1/8 1/4 1/8 Computing the expected values is below. µ = E(X) = (2) × (1/2) + (3) × (1/8) + (−1) × (1/4) + (1) × (1/8) = 1/4. Var(X) =(2 − 1/4)2 · (1/2) + (3 − 1/4)2 · (1/8) + (−1 − 1/4)2 · (1/4)+ +(1 − 1/4)2 · (1/8) = 47/16.

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Grouped Data

Example (Grouped Data)

Interval Frequency Midpoint xM 30-39 1 34.5 40-49 6 44.5 50-59 13 54.5 60-69 22 64.5 70-79 17 74.5 80-89 13 84.5 90-99 8 94.5 Find the standard deviation of these grouped data. In this case you must sum the x2

M multiplied by the frequencies, and

subtract 80 × x where x is for the full sample, (which is not in the table, you must get it from the full data).

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Chebyshev’s and Markov’s Inequality I

P(X ≥ a) ≤ E(X) a Markov. P (|X − µ| ≥ kσ) ≤ 1 k2 Chebyshev. In words, the probability that X is more than k standard deviations away from the mean is less than 1/k2.

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Chebyshev’s and Markov’s Inequality II

Example (from the practice prelim)

• 8. (14 points) Assume that the height in inches of American women

follows a normal distribution with mean Mu = 64′′ (5’4”) and standard deviation σ = 3′′. (a) (3 points) How many standard deviations above or below the mean is a height of 72” (6’0”)? (b) (4 points) What fraction of women are taller than 6 feet? (c) (4 points) In a room with 30 women, what is the probability that at least one of them is taller than 6 feet? (d) (3 points) What assumptions did you make when answering part (c)? Are there circumstances under which those assumptions would not be justified?

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Chebyshev’s and Markov’s Inequality III

Answer.

Say we don’t know what distribution it is. We can still use Markov’s and Chebyshev’s inequality. (a) 72−64

3

= 8

3, so closer to 3. Use 73 to get 3.

(b) Markov’s inequality says P(X ≥ 73) ≤ 64

• 73. To use Chebyshev’s

inequality we must write |X − 64| ≥ 3σ. Then P(|X − 64| ≥ 3σ) ≤ 1 9. In other words, k = 3. This includes not just X ≥ 73, but also X ≤ 55. Still we can say the probability is less than 1/9, because |X − 64| ≥ 9 is larger than X − 64 ≥ 9. (c) 1 − P( none are taller than 6 ) = 1 − (1 − P( one is not taller than 6 ))30.

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Continuous Random Variables I

In many applications it useful to assume that the random varuiable X may take any real value. The probability distribution for the case of finitely many values does not work. We assume that the sampe space is S = R all real number. The typical event (subset of S) is restricted to sets of the form A = {x | x ≤ a} and complements and intersections of such sets. For us they will be at most sets of the form A = {a ≤ x ≤ b}. The probability distribution is given as numbers P(X ≤ a); in other words a function which takes nonnegative values only, and we allow a = −∞ in which case the value is 0, and ∞, in which case the number is 1. For a continuous reandom variable, P(X = a) = 0 always, but the situation is not trivial.

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Continuous Random Variables II

Example (the uniform distribution of the interval (0, 1))

Let f (x) =      0 if a ≤ 0 1 if a < 0 < 1 0 if 1 ≤ a Define P(X ≤ a) = area between the x−axis and f (x), and before the vertical line x = a. See the picture in class, or in the text for the normal

• distribution. So

P(X ≤ a) =      0 if a ≤ 0 a if 0 < a < 1 1 if 1 ≤ a

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Continuous Random Variables III

Exercise

Do the same for f (x) =      0 if x ≤ 0 2x if 0 ≤ x ≤ 1 0 if 1 < x. You need the formula for the area of the triangle, Area = (base) × (height)/2.

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Normal Distribution I

Definition

Data are said to be normally distributed if the rate at which the frequencies fall off is proportional to the distance of the score from the mean, and to the frequencies themselves. This definition requires Calculus. We don’t assume or do Calculus in this

• course. We will however learn how to work with this distribution.

It is very useful in that many phenomena can be modeled by this. We will see how the binomial distribution is related to the normal distribution later in the chapter. Suppose you have a random variable X, and you would like to know about its mean µ. So you perform many n independent trials, and draw a

• histogram. The larger the n, the closer the outcome will look like the

curve f (x) =

1 √ 2πσe− (x−µ)2

2σ2 . The pictures in the text show what it looks

• like. The resulting probability is called N(µ, σ2), normal with mean µ and

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Normal Distribution II

variance σ2. There is a precise statement called the Central Limit Theorem which says that for large n, √n(Sn − µ) “looks” like a normal distribution N(0, σ2). it is used in practice to model large populations and “ errors”. There are many examples that can be approximated by normal

• distributions. Heights of people, and scores on tests are examples.

This is not a finite distribution. For a random variable that is normally distributed, we write N(µ, σ2), P(X ≤ a) = the area under the normal curve from − ∞ to a. This is tabulated for µ = 0 and σ = 1. The rest is computed by simple formulas involving arithmetic.

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Height Example I

Example (from the practice prelim)

• 8. (14 points) Assume that the height in inches of American women

follows a normal distribution with mean mu = 64′′ (5’4”) and standard deviation σ = 3′′. (a) (3 points) How many standard deviations above or below the mean is a height of 73” (6’1”)? (b) (4 points) What fraction of women are taller than 73 inches? (c) (4 points) In a room with 30 women, what is the probability that at least one of them is taller than 73”? (d) (3 points) What assumptions did you make when answering part (c)? Are there circumstances under which those assumptions would not be justified?

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Height Example II

Answer.

(a) same as before 3 standard deviations away. (b) P(X ≥ 73) = P(X − 64 ≥ 73 − 64 = 9 = 3σ) = P(X − µ σ ≥ 3) = =1 − P(X − µ σ ≤ 3) = 1 − 0.999 = 0.001. This is 1/1000. The random variable X has probability distribution N(64, 17). The probability P(X ≥ 73) comes from this normal distribution. To actually look it up in the tables, you rewrite it in terms of Z = X−64

3

which has probability distribution N(0, 1). This is the one in the tables. (c) P(at least 1/30 ≥ 73) =1 − P(30/30 ≤ 73) = 1 − P(X ≤ 73)30 = =1 − (0.998)30.

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z−value

The principle is X normal N(µ, σ2) ⇐ ⇒ Z = X−µ

σ

normal N(0, 1). So P(X ≤ a) = P(Z ≤ a − µ σ ). z = a−µ

σ

is called the z−value. This is what you look up in the tables.

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Example with Grades

Example

A professor (not this one!) of a course wants to give grades so that A top 8% F bottom 8% B next 20% below A D next 20% above the F C the rest The mean is µ = 67 and the standard deviation is σ = 17. Find the cutoffs.

Answer.

P(≤ A) = 0.92 z = 1.41 a = µ + zσ = 67 + 17 · 1.41 = 91 P(≤ B) = 0.72 z = 0.58 a = µ + zσ = 67 + 17 · 0.58 = 77 P(≤ C) = 0.28 z = −.59 a = µ + zσ = 67 + 17 · (−.59) = 57 P(≤ D) = 0.08 z = −1.39 a = µ + zσ = 67 + 17 · (−1.39) = 43 from the tables. In Excel or alike you can write norminv(0.92, 67, 17) ∼ = 91.

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