To Notes

IT 403 -- Oct 19, 2016

Review Exercises

1. How do you compute the expected value of a random variable?
   Ans: Compute the weighted average of the outcomes, weighted by the probability of each outcome:
         E(x) = x₁ p₁ + ... + x_n p_n
2. Intuitively, what does the expected value mean.
   Ans: The long run average of the outcomes of the random variable over a very large number of trys.
3. What is the expected value of a Bernoulli random variable?
         E(x) = x₁ p₁ + x₂ p₂ = 0(1 - p) + 1p = p
4. What is the theoretical standard deviation of a Bernoulli random variable?
   Ans: If x is a Bernoulli random variable,
         Var(x) = (x₁ - E(x))² P(x₁) + ... + (x_k - E(x))² P(x_k)
                  = (0 - p)² (1 - p) + (1 - p)² p = p(1 - p)(p + 1 - p) = p(1 - p)
   Then
              σ_x = √Var(x) = √p(1 - p)
   
5. State the Multiplication Rule for probabilities.
6. When flipping a fair coin, what is the probability of
   
   1. getting at exactly 5 heads in a row?
      Ans: (1/2)⁶ = 1/64. The sequence must be HHHHHT
   2. getting at least 5 heads in a row?
      Ans: (1/2)⁵ = 1/32. The sequence must be HHHHH; we don't care what the next outcome is.
   3. obtaining the sequence THTTH?
      Ans: (1/2)⁵ = 1/32
7. Some 18th century dice games:
   
   1. Roll a die 4 times. What is the probability of getting at least one ace (1)?
      Ans: 1 - (1 - 1/6)⁴ = 0.5177 = 51.8%
   2. Roll a pair of dice 24 times. What is the probability of obtaining snakeeyes (pair of aces at least once)?
      Ans: 1 - (1 - 1 / 36)²⁴ = 0.4914 = 49.1%
   Blaise Pascal (1623 - 1662) was the first to apply the laws of probability to gambling.
8. The expected value of a random variable x is 15. What is the expected sum of 100 outcomes of x?
   Ans: E(S) = nE(x) = 100(15) = 1,500.
9. The theoretical SD of a random variable is 3. What is the theoretical standard error of the sum of 100 independent outcomes of x?
   Ans: σ_S = √n σ_x = √100 3 = 30.
10. An American roulette wheel contains 38 pockets: 18 red pockets, 18 black pockets, and two green pockets labeled 0 and 00. European roulette wheels have only one green pocket labeled 0. Here is an online European roulette game to see what it is like.
    If you bet on red or black, here are the payoff tables for an American roulette wheel:
    

    Bet on Red
    Outcome	Payoff	Probability
    Red	+1	18/38 = 0.474
    Black	-1	18/38 = 0.474
    Green	-1	2/38 = 0.052

    
    

    Bet on Black
    Outcome	Payoff	Probability
    Red	-1	18/38 = 0.474
    Black	+1	18/38 = 0.474
    Green	-1	2/38 = 0.052

    
    What is the expected payoff of playing on red? playing on black.
    Ans: Playing on red is E(x) = 1(0.474) + (-1)(0.474) + (-1)(0.052) = -0.052. You lose about 5 cents each time you play on red, same for black.
11. The following boxes contain 6 tickets each. In which urns are letter independent from color. In other words, in which urns are the probabilities of the letters the same whether or not you know the color.

    
    Ans: In a and b, letter is independent of number. To show that letter and number are independent, you need to show that for each letter value L P(L) if you don't know the color = P(L) for a given color. For example, in Problem 8a, if you don't know the color, P(A) = P(B) = P(C) = 2/6 = 1/3. If you know the color is red, P(A) = P(B) = P(C) = 1/3. If you know the color is green, the probabilities are also 1/3.

    In Problem 8c, if you don't know the color, P(F) = 2/6 = 1/3. If you know that the color is green, P(F) = 2/3 = 2/3, so color and letter are not independent.

    To show that two events are not independent, it is enough to find one case where the probabilities are different; to show that two events are independent, you must show that the probabilities are the same for all choices of color and letter.

12. Are these events independent?
    Person A lives at least 20 more years after February 21, 2011.
    Person B lives at least 20 more years after February 21, 2011.
    Ans: It is hard to know. If the people know each other, their lives and life choices can affect each other so A and B are probably not independent. If the people do not know each other, A and B may be independent, but they may also be related in unexpected ways.
13. If the probability of getting struck by lightning in a year is 1/300,000, what is the probability of getting struck by lightning twice in a year?
    Ans: (1 /300,000)² = 1 / 90,000,000,000.

14. If the probability of getting a cold in a given week is 3%, what is the probability of getting at least one cold in a year?
    Ans: 1 - (1 - p)^n = 1 - (1 - 0.03)^52 = 0.7948 = 79%.
15. Is it possible for two events to be both independent and mutually exclusive at the same time?

    Ans: If A and B are independent, P(A and B) = P(A) P(B). But if A and B are mutually exclusive, P(A and B) = 0. Thus P(A) P(B) = 0 and either P(A) = 0 or P(B) = 0.

16. Estimate the probability that an amateur golfer that plays one round a week makes at least one hole in one on a certain par 3 hole in 50 years. The golfer hits the green 10% on that hole of the time, which has an area of about 4,000 square feet. The hole on a golf green has a radius of 4.25 inches.
    Ans: First estimate the probability of a hole in one in a single try. The area of the hole in square feet ipi*r²/144 = pi*4.25²/144 =0.3757 ft². If the area of the green is 4000 ft², the probability of a hole in one is 0.1 * 0.3757 / 4000 = 9.3925 × 10^-6. This the probability of getting at least one hole in one in 50 years is
          1 - (1 - 9.3925 × 10^-6)^{(50 * 52)} = 0.024 = 2.4%
17. How many ways are there to arrange n objects? Ans: n!
18. How many subsets are there of a set that contains n objects? Ans: 2ⁿ
19. How many ways are there to choose k objects out of n objects. Ans: nCk = n! / [k! (n-k)!]

Collect Data for Project 6

• Use the instructor's micrometer to measure the thicknesses of the two sheets of paper (Batch A and Batch B) that the instructor will give you.
   
  
   
• This micrometer measures thicknesses to the nearest thousandth of a millimeter (nearest micrometer = 10^-6 m).
   
• Before you make each measurement,
   
  
  make sure that the units are set to mm, not to inches (use the mm/in button to change units if necessary),
   
  close the distance between the spindle and the anvil until they are almost touching,
   
  tighten the spindle with the ratchet screw, so that the spindle and anvil touch the sheet of paper being measured,
   
  zero out the distance on the digital display (ON/OFF 0 Button).
  
  
• When you measure each paper thickness, use the ratchet screw to obtain an accurate reading.
   
• Write your name and measured thickness in mm on each sheet of paper.
   
• These paper thickness measurements will go into a dataset that you will analyze for the two-sample t-tests for Project 6.

Quiz 5

• Go over Quiz 5 of the D2L Quizzes.

Factorials and Counting Combinations

• The number of ways of drawing k objects from an urn that contains n distinct objects is
  
• The binomial coefficient defined in the previous bullet point is also written as nCk and read "n choose k."
• The binomial coefficients can also be obtained from Pascal's Triangle:
  

                      1
                   1     1
                1     2     1
              1    3     3     1
           1    4     6     4     1
        1     5    10    10    5     1
      1    6    15    20    15    6     1
    1   7    21    35    35    21    7     1
  1   8   28    56    70    56    28    8     1
  

• Each entry in Pascal's triangle can be obtained recursively by adding the two entries above it.
• nCk can be obtained from in Pascal's Triangle as entry k in row n.
• Practice Problems: How many ways are there of choosing
  
  1. What is the number of ways of choosing 2 out of 4 objects?
     Ans: 4C2 = 4! / (2!·2!) = (4·3·2·1) / (2·1·2·1) = (3·2) / (1·1) = 6
  2. What is the number of ways of choosing 3 out of 5 objects?
     Ans: 5C3 = 5! / (3!·2!) = (5·4) / (2·1) = 10
  3. What is the number of ways of choosing 6 out of 10 objects?
     Ans: 10C6 = 10! / (6!·4!) = (10·9·8·7) / (4·3·2·1) = 10·3·7 because 4 and 2 in the denominator cancel with 8 in the numerator, and 3 in the denominator cancels with 9 in the numerator, leaving 3 in the numerator. The answer is 210.
  4. What is the number of ways of choosing 13 out of 20 objects?
     Ans: 20C13 = 20! / (13!·7!) = (20·19·18·17·16·15·14) / (7·6·5·4·3·2·1) = 19·17·16·15. The 5 and 4 in the denominator cancel with 20 in the numerator; the 6 and 3 in the denominator with the 18 in the numerator; the 7 and 2 in the denominator cancel with 14 in the numerator. The factors 19, 17, 16, and 15 are left in the numerator. 19·17·16·15 = 77,520.

The Binomial Formula

• Suppose you have a Bernoulli distribution with probability of success = p. What is the probability of k successes out of n trials? Ans:
  
  
  
• This formula can also be written as
  P(successes in k trials) = nCk p^k (1 - p)^n-k
• This formula is valid because
  
  1. The trials are independent so the multiplication rule can be used.
  2. P(success) = p, P(failure) = 1 - p
  3. There are n choose k ways to choose k successes out of n trials.
• Practice Problems:
  
  1. When tossing a fair coin, what is the probability of obtaining
     
     1. 3 heads out of 5?<
        Ans: 5C3 (0.5)^3 (1-0.5)^(5-3) = 10(0.125)(0.25) = 0.03125 = 3%.
     2. 7 heads out of 9?
        Ans: 9C7 (0.5)^7 (1-0.5)^(9-7) = 36 (0.001953125) = 0.0703125 = 7%.
     3. 10 heads out of 14?
        Ans: 14C10 (0.5)^10 (1-0.5)^(14-10) = 1001×0.0000061352 = 0.0611 = 6%.
     4. 10 or 11 heads out of 14?
        Ans: P(10 out of 14) + P(11 out of 14) = 0.0611 + 0.0222 = 0.0833
     5. 10 or more heads out of 14 Ans: P(10 out of 14) + P(11 out of 14) + P(12 out of 14) + P(13 out of 14) + P(14 out of 14) = 0.0611 + 0.0222 + 0.00555 + 0.00085 + 0.0000610 = 0.089761.
  2. You roll a die 10 times. What is the probability of rolling exactly 5 ones (aces)?
     Ans: 10C5 (1/6)⁵ (1 - 1/6)^10-5 = 0.01302381 = 1.3%.
  3. If your true probability of shooting freethrows for practice is 70%, what is the probability that you shoot 16, 17, or 18 for 20?
     Ans: P(16 for 20) + P(17 for 20) + P(18 for 20) = 0.1304 + 0.0716 + 0.0278 = 0.2298 = 23%.
  4. If your true probability of shooting freethrows in practice is 70%, what is your expected number of freethrows made out of 20? what is your standard deviation?
     Ans: E(S) = np = 10(0.7) = 7; σ_S = √np(1-p) = √20(0.7)(1-0.7) = 2.049
  5. The probability that the Cubs can beat Team A in a single game is 60%. What is the probability that they win 7 or more games in a 10 game series?
     Ans: P(7 out of 10) + P(8 out of 10) + P(9 out of 10) + P(10 out of 10) =  0.2150 + 0.1209 + 0.0403 + 0.0060 = 0.3822 = 38%

Project 4

• Look at the Project 4 Description.
• We already saw how to simulating repeated trials from a Bernoulli random variable using SPSS using RV.BERNOULLI(0.5)
• Use RV.BINOM to compute generate sums of Bernoulli random variables for Part C of Project 4.

Finding a Confidence Interval for a Proportion

• Example 1:   Flip a coin 1,000 times and obtain 514 heads. Find a 95% confidence interval for the true value p of obtaining a head. This will give us an idea if the coin is fair.
  Ans: Flipping a coin once is a Bernoulli random variable with unknown probability of obtaining a head p. We also know that the sum S = 514, and the estimated probability of success p^{^} = S / n = 514 / 1,0000 = 0.514. We know that E(S) = nE(x) = np. Furthermore, we know that σ_S = √[np(1-p)]. Since p is unknown we estimate it with p^{^} and obtain σ_S  = √[np^{^}(1-p^{^})] = √[1000(0.514)(1 - 0.514)] = 15.81. Finally, we compute the z-score
        z = (S - E(S)) / σ_S = (514 - np) / 15.81.
  To obtain z, we have scaled S by subtracting its expectation and dividing by its standard error.  This means that E(z) = 0 and σ_z = 1. By the Central Limit Theorem, z is a standard normal random variable if n > 30, so -2 ≤ z ≤ 2 95% of the time. (More precisely, -1.96 ≤ z ≤ 1.96 95% of the time. We can then solve this inequality for for p:
  -2 ≤ z ≤ 2
  -2 ≤ (S - np) / σ_S ≤ 2
  -2 ≤ (514 - 1,000p) / 15.81 ≤ 2
  -31.62 ≤ 514 - 1,000p ≤ 31.62
  -31.62 - 514 ≤ - 1,000p ≤ 31.62 - 514
  -545.62 ≤ - 1,000p ≤ -475.62
  0.54562 ≥ p ≥ 0.47562
  This means that the 95% confidence interval is [0.47562, 0.54562] or [47.6%, 54.6%].

Tests of Hypotheses for Proportions

• A test of hypothesis can be used to determine if a coin or die is fair.
• For testing if a coin is fair, we test the null hypothesis H₀: p = 0.5 against the alternative hypothesis H₁: p ≠ 0.5.
• Here are the four steps for testing if the probability of success p for a Bernoulli random variable is equal to p₀:
  
  1. Write down the null and alternative hypothesis:
           H₀: p = p₀
           H₁: p ≠ p₀.
  2. Write down the test statistic, assuming the null hypothesis:
           z = (S - E(S)) / σ_S
  3. Write down a 95% confidence interval for z:
           I = [-1.96, 1.96]
  4. If z ∉ I, reject H₀; if z ∈ I, accept H₁.
• Example 2:   You flip a coin 100 times and obtain 43 heads. Is the coin fair?
  
  1. H₀: p = 0.5; H₁: p ≠ 0.5
  2. S = 43, E(S) = np, and σ_S = √[np(1-p)] = √[100(0.5)(1-0.5)] = 5.0, so
         z = (S - E(S)) / σ_S = (43 - 100(0.5)) / 5.0 = -1.4
  3. A 95% confidence interval for z is [-2, 2].
  4. -1.4 ∈ [-2, 2], so accept the null hypothesis. Not enough evidence to conclude that the coin is not fair.
• Example 3: To test if a four-faced die (shape of a tetrahedron) is fair, such a die is rolled 1600 times. 421 aces are obtained. Perform a test of hypothesis to test whether the die is fair.
  Here are the steps to perform the test of hypothesis:
  
  1. Write down the null hypothesis:    H₀: p = 1/4
  2. Compute the text statistic:    z = (S - E(S)) / SE_S
  3. Write down a 95% confidence interval:    I = (-1.96, 1.96)
  4. If z ∈ I, accept the null hypothesis that the die is fair. If z ∉ I reject the null hypothesis.
• Ans:
  
  1. H₀: p = 1/4.
     
  2. S = 421   n = 1,600   SE(S) = sqrt(1,600 * (1/4) * (1-1/4)) = 17.32
     z = (S - np) / SE(S) = (421 - 1,600 (1/4)) / 17.32 = 1.212
     
  3. The 95% confidence interval using the standard normal table is (-1.96, 1.96)
     
  4. 1.212 ∈ (-1.96, 1.96), so accept the null hypothesis that the die is fair; we don't have enough evidence to reject the null hypothesis.