To Notes

IT 223 -- Feb 25, 2026

Review Exercises

1. The probability that a part on an assembly line is defective is 5%. What is the probability that
   
   1. exactly 3 parts are defective out of 15. Use R to check your answers. Answer:
      P(k successes out of n) = nCk p^k (1-p)^n-k
            = 15C3 * 0.05³ * (1-0.05)^15-3
            = 15!/(3!*12!) * 0.05³ * 0.95¹²
            = (15*14*13) /(3*2*1) * 0.05³ * 0.95¹²
            = 5 * 7 * 13 * 0.000125 * 0.540360 = 0.03073298 = 3%.
      This probability can also be computed with the R dbinom function:
      

      > dbinom(3, 15, 0.05)
      [1] 0.03073298
      

   2. exactly 2 parts are defective out of 15. Answer:
      P(k successes out of n) = nCk p^k (1-p)^n-k
            = 15C2 * 0.05² * (1-0.05)^15-2
            = 15*14*13/(2!*13!) * 0.05² * 0.95¹³ = 13%.
      Using dbinom:
      

      > dbinom(2, 15, 0.05)
      [1] 0.1347523
      

   3. exactly 1 part is defective out of 15.Answer: P(k successes out of n) = nCk p^k (1-p)^n-k
            = 15C1 * 0.05¹ * (1-0.05)^15-1
            = 15!/(1!*14!) * 0.05¹ * 0.95¹⁴ = 0.3657562 = 37%.
      This probability is computed using R like this:
      

      > dbinom(1, 15, 0.05)
      [1] 0.3657562
      

   4. no parts are defective out of 15. Answer: P(k successes out of n) = nCk p^k (1-p)^n-k
            = 15C0 * 0.05⁰ * (1-0.05)^15-0
            = 15!/(0!*15!) * 0.05⁰ * 0.95¹⁵ = 0.4632912 = 46%.
      This probability is computed using R like this:
      

      > dbinom(0, 15, 0.05)
      [1] 0.4632912
      

   5. 3 or fewer parts are defective. Answer:
      P(3 or fewer defective parts:
      

      > pbinom(3, 15, 0.05) + pbinom(2, 15, 0.05) + 
      + pbinom(1, 15, 0.05) + pbinom(0, 15, 0.05)
      

      This gives us
          0.03073298 + 0.1347523 + 0.3657562 + 0.4632912 = 0.9945327 = 99%.
      To verify this answer with R:
      

      > # The R statement pbinom function computes the
      > # probability of 3 or fewer sucesses out of 15
      > # when the probability of success is 0.05.
      > pbinom(3, 15, 0.05)
      [1] 0.9945327
      

2. State the Law of Large Numbers (LLN). Who first proved it?
   Answer: Jakob Bernoulli in 1713.
3. State the Central Limit Theorem (CLT). Who first proved it?
   Answer: Abraham de Moivre proved the CLT for the binomial distribution in 1733.
   Alexandr Lyapunov proved the modern form of the CLT in 1901.
4. A random sample of 500 voters registered in Chicago is collected. 236 of them say they will vote for Candidate A for mayor. Find a 95% confidence interval for the true probability that a registered voter will vote for candidate A. Compute this confidence interval using S = the number of voters in the sample that said they will vote for Candiate A. X = S / n = 0.472. Compute the confidence interval with two methods: Method 1 uses the sum S and Method 2 uses the average X. You should obtain the same answer using either method.
   
   Answer: Method 1 is to compute the confidence interval for the true value of p using S:
   n = 500   S = 236   E(S) = np = 500*p
   p^{^} = S / n = 236 / 500 = 0.472
   σ_S = √n * p * (1-p) = √500 * 0.472 * (1-0.472) = 11.1628
   z_S = (S - E(S)) / σ_S = (244 - 500 * p) / 11.1628
   Now -2 ≤ Z_S ≤ 2 95% of the time. Solve for p:
   -2 ≤ (236 - 500 * p) / 11.1628 ≤ 2
   -2 * 11.1628 ≤ 236 - 500 * p ≤ 2 * 11.1628
   -22.3256 - 236 ≤ - 500 * p ≤ 22.3256 - 236
   -258.3256 ≤ -500 * p ≤ -213.6744
   -258.3256 / -500  ≥ p  ≥ -213.6744 / -500
   0.5166512 ≥ p ≥ 0.4273488.
   The 95% confidence interval for p is [0.4273488, 0.5166512] = [43%, 52%].
   Because values on either side of 50% are in the confidence interval, we don't have enough evidence to conclude if Candidate A will win the election.
   
   Method 2 is to compute the 95% confidence interval for the true value of p using the average X of the 500 outcomes instead of S.
   n = 500   X = 236 / 500 = 0.472   E(X) = p
   p^{^} = X = 236 / 500 = 0.472
   σ_xbar = √(p * (1-p)) / n = √0.472 * (1 - 0.472) / 500 = 0.0223256
   z_xbar = (x - E(x)) / σ_xbar = (0.236 - p) / 0.0223256.
   Now -2 ≤ z_xbar ≤ 2 95% of the time. Solve for p:
   -2 ≤ (0.472 - p) / 0.0223256 ≤ 2
   -2 * 0.0223256 ≤ 0.472 - p ≤ 2 * 0.0223256
   -0.0446512 - 0.472 ≤ -p ≤ 0.0446512 - 0.472
   -0.5166512 ≤ -p ≤ -0.4273488
   0.5166512 ≥ p ≥ 0.4273488
   0.4273488 ≤ p ≤ 0.5166512
   The 95% confidence interval for p is [0.4273488, 0.5166512] = [43%, 52%].
   
   This is the same result as we obtained with Method 1.
   
   Collecting a true random sample for election polling is not easy. When the Gallup polling company started predicting election results in 1936, other polling companies were collecting random samples from phone books and club membership lists. This skewed the sample because, in 1936, only wealthy people were likely to have phones or belong to clubs.
   
   The Gallup company used quota sampling from the population of census data, which means that persons are sampled from social and economic substrata of the population. Today, it is easier to select random names from phone directories because about 98% of the population has a phone.  Nevertheless, collecting a good random sample for predicting an election is not easy. Polling companies are always striving to improve their methods.

Tests of Hypothesis

• The fundamental question that a researcher conducting a test of hypothesis is trying to answer is: Is the result significant, or is it merely due to chance variation?
• Flip a coin 10,000 times and obtain 5,038 heads. Is the coin biased, or is it just chance variation?
• To reduce the bias, reduce the effect of lurking variables:
  If the lurking variable is known and can be controlled:
        -- Insure that the effect of the lurking variable is the same on all subjects.
• If the lurking variable is known, but cannot be controlled:
        -- Include the lurking variable as an independent variable in the model.
        -- Include the lurking variable as a panel variable.
        -- Only include observations with the same value of the lurking variable.
• If the lurking variable is unknown:
        -- Randomize the assignment of treatments to subjects to ensure that the effect of lurking variables is equally likely for all subjects.

         -- If possible, make the treatments double blind.

• The null hypotheses, denoted by H₀, states that the treatment or effect under investigation does not make a difference; the effect is merely due to chance.
• Some sample research questions phrased as null hypotheses:
  
  1. The number of heads obtained with a coin being investigated is not significantly different than a fair coin.
  2. There is no real difference in the autism rates between children that receive a vaccine and those that do not receive it.
  3. The electric and magnetic fields caused by high voltage power lines does not cause significant health risks to those living nearby.
  4. Eating irradiated food does not cause significant health risks.
  5. There is no real difference in reading scores between the students that use the new reading curriculum and those that do not.
  6. A specific die is fair, which means that the probability of any face coming up when rolled is 1/6.
  7. There is no real difference in network traffic speed between the old router and the new router.
  8. The new tax law is essentially revenue neutral.
• The steps for a test of hypothesis. The test is a α%-level z-test:
  
  1. State the null (H₀) and alternative hypotheses (H₁).
  2. Compute the test statistic T, assuming the null hypothesis is true.
  3. Write down a (1-α)% confidence interval for the test statistic I.
  4. If T ∈ I, accept the null hypothesis; if T ∉ I, reject the null hypothesis.
  5. (If possible) determine the p-value of the test statistic.
• Note: for Project 4, the z-tests are 5%-level tests.
• The p-value is the probability of obtaining a test statistic at least as extreme at the test statistic actually obtained, given that the null hypothesis is true.

The z-test

• Example 1:   To test whether a new tax bill is revenue neutral, test the new tax rules on a random sample 100 tax returns out of 100,000 tax returns on file. For each tax return, compute
        x_i = tax under new rules - tax under old rules
  Then compute x = -$219 and SD = $725. Let μ be the true change in tax revenue from the old rules to the new rules.
• The test of hypothesis:
  
  1. State the null and alternative hypotheses:
           H₀: μ = 0
           H₁: μ ≠ 0
  2. Compute the test statistic, assuming H₀ is true:
           z = (x - μ) / SE_ave = (-219 - 0) / (725 / √100) = -3.02
  3. Write down a 95% confidence interval for the test statistic, assuming H₀: [-1.96, 1.96].
  4. z = -3.02 ∉ (-1.96, 1.96), so reject H₀ and accept H₁; the difference is real and not merely due to chance.
  5. Compute the p-value, which is the probability that z is as extreme or more extreme the z which is actually obtained. In our case z = -3.02, and we want to find the probability in the tails. Now the area under the normal curve for the interval (∞, -3.02) is 0.0013. Thus the area in both tails is 2 × 0.0013 = 0.0026.
• The general form of the z-score for a z-test is
        z = (T - E(T)) / SE_T
  Two examples are
        z = (S - E(S)) / SE_S   and   z = (x - E(x)) / SE_x
• When performing a z-test for p, the test statistic is z = (S - np) / SE_S, where S is the number of successes from the Binomial Distribution (sum of n Bernouilli random numbers), and p is the true probability of success.
• When performing a z-test for μ, from the ideal measurement model, the test statistic is
        z = (x - μ) / SE_ave,
  where μ is the mu of the null hypothesis.
• Only use a z-test when the sample size n is greater than or equal to 30. This insures that, in the case of the ideal measurement model x_i = μ + e_i, the test statistic
        z = (x-μ) / SE_ave
  is approximately normally distributed according to the CLT, and
  SD⁺ in SE_ave = SD⁺ / √n is close to the true standard deviation σ of the population.
• In the case of a z-test for a probability p, n ≥ 30 insures that
  the test statistic S is approximately normally distributed (thanks to the Central Limit Theorem), and

  p^ = S/n in SE_S = √np(1-p) is close to the true value of p.

More about p-values

• In the old days (50 or more years ago) statisticians were content to know whether H₀ was accepted or rejected. Now they want to know the p-value, which gives more information.
• Recall that the p-value is the probability of obtaining a test statistic as extreme or more extreme that the value actually obtained, given that the null hypothesis is true.
• If p is close to zero, the evidence is overwhelming that the result was not due to chance.
• If p is slightly less than 0.05, H₀ was just barely rejected. The evidence is borderline as to whether the result is due to chance; more research is needed.
• If p is slightly more than 0.05, H₀ was just barely accepted. The evidence is borderline as to whether the result is due to chance; more research is required.
• Usually a researcher wants to reject H₀ to prove that the treatment that he or she is investigating is real, not just chance variation.
• If H₀ is accepted, it does not necessarily mean that we believe that H₀ is true, it means that there is not enough evidence to reject it.
• At the risk of complicating things, our formulation of the z-test is phrased as a two-tailed test, where
        H₀: μ = c      H₁ ≠ c For a 5% level test (95% confidence), this means that we reject H₀ when the test statistic is not in the confidence interval I = [-1.96, 1.96].
• It is also possible to phrase the z-test (and the t-tests we will discuss later) as one tailed tests, for which the null and alternative hypotheses are
        H₀: μ = c      H₁: μ > c
  or
        H₀: μ = c      H₁: μ < c
  In the case of H₁: μ > c, we reject H₀ when the test statistic is not in the interval I = (-∞, 1.645].
• Some researchers think that the one-tailed test is an improvement over the two-tailed test because the test of hypothesis is more precise.
• However, other researchers think that the one-tailed test is cheating, because it makes rejecting the null hypothesis easier.

Practice Problems

• Example 2. In 1999, it was reported that the mean serum cholesterol level for female undergraduates was 168 mg/dl. A recent study at Baylor university collected the following data for cholesterol levels for females:
        n = 25      x = 173.7     SD+ = 27
  Is there a real difference between the women in the Baylor study and the reported value in 1999? Perform the test at the 10%-level.
  
  1. H₀: 168     H₁: μ ≠ 168
  2. z = (x - μ) / SE_ave = (173.7 - 168) / (27 / √25) = 1.055556
  3. A 90% confidence interval for z is [-1.64,1.64].
  4. 1.056 ∈ [-1.64,1.64], so accept the null hypothesis.
  5. Find the area corresponding to the bin [-1.056,1.056]: 2 × 0.1455 = 0.2910.
• Example 3: if all high school seniors in California took the SAT test, the mean score would be equal to 450. To test this claim, select a random sample of 400 high school seniors and give them the test. Here are the data:
  n = 400     x = 461     SD+ = 100
  Is this result for the sample significantly different from 450 or is it just chance variation? Perform the test at the 99%-level.
  Ans: Here are the steps of the z-test:
  
  1. H₀: μ = 450      H₁: μ ≠ 450
  2. z = (x - μ) / SE_ave = (461 - 450) / (100 / √400) = 2.2.
  3. A 99% confidence interval for z is [-2.58,2.58].
  4. 2.2 ∉ [-2.58, 2.58], so reject the null hypothesis.
  5. The p-value is the probability of obtaining a z-value as extreme or more extreme than the one actually obtained. Find the area corresponding to the bin [-2.2,2.2]: 2 × 0.0139 = 0.0278.
  6. Example 4: Perform a test of hypotheses for the NIST 10 gram weight measurements. Remember that these measurements were read from the CSV file nist-10.txt.
     Here are the R statements:
     

     > setwd("c:/workspace")
     > getwd( )
     [1] "c:/workspace"
     > dir( )
     [1] "bears-2024-roster.txt" "bears-2026-roster.txt" "ht-wt.txt" 
     [4] "laundry-detergent.txt" "nist-10.txt"           "paper-thickness.txt" 
     > df <- read.csv("nist-10.txt")
     > df$Weight
     [1] 9.999591 9.999600 9.999594 9.999601 9.999598 9.999594 9.999599 9.999597
     [9] 9.999599 9.999597 9.999602 9.999597 9.999593 9.999598 9.999599 9.999601
     [17] 9.999600 9.999599 9.999595 9.999598 9.999592 9.999601 9.999601 9.999598
     [25] 9.999601 9.999603 9.999593 9.999599 9.999601 9.999599 9.999597 9.999600
     [33] 9.999590 9.999599 9.999593 9.999577 9.999594 9.999594 9.999598 9.999595
     [41] 9.999595 9.999591 9.999601 9.999598 9.999593 9.999594 9.999587 9.999591
     [49] 9.999596 9.999598 9.999596 9.999594 9.999593 9.999595 9.999589 9.999590
     [57] 9.999590 9.999590 9.999599 9.999598 9.999596 9.999595 9.999608 9.999593
     [65] 9.999594 9.999596 9.999597 9.999592 9.999596 9.999593 9.999588 9.999594
     [73] 9.999591 9.999600 9.999592 9.999596 9.999599 9.999596 9.999592 9.999594
     [81] 9.999592 9.999594 9.999599 9.999588 9.999607 9.999563 9.999582 9.999585
     [89] 9.999596 9.999599 9.999599 9.999593 9.999588 9.999625 9.999591 9.999594
     [97] 9.999602 9.999594 9.999597 9.999596
     > t.test(df$Weight, mu=10)
     
                One Sample t-test
     
     data: df$Weight
     t = -625.64, df = 99, p-value < 2.2e-16
     alternative hypothesis: true mean is not equal to 10
     95 percent confidence interval:
      9.999594 9.999597
     sample estimates:
     mean of x 
     9.999595 
     

     The p-value is 2.2 × 10^-15 so the evidence if overwhelming to reject H₀.