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IT 223 -- June 3, 2024

Review Exercises

1. If X is a random variable with expected value E(X), variance Var(X), and standard deviation σ_X, use the formulas that you know to obtain the following:
         E(S)   Var(S)   σ_S    z_S
         E(X)   Var(X)   σ_X    z_X
   Answer:
         E(S) = E(X₁)
         Var(S) = n * Var(X₁) if X₁, ... , X_n are independent.
         σ_S = √n * σ_X1 if X₁, ... , X_n are independent
         z = (S - E(S) ) / σ_S
         E(X) = E(X₁)
         Var(X) = E(X₁) / n, if X₁, ... , X_n are independent.
         σ_X = σ_X1 / √n
         z = (X - E(X)) / σ_X
2. Look at Problem 3a of Project 4.  Which R statement do you use to simulate 1,200 rolls of a fair die?
   Answer: You can simulate these 1,200 rolls using rbinom in two ways:
   

   > # Method 1: Repeat 1,200 times the 
   > # experiment of rolling a die once
   > rbinom(1200, 1, 1/6)
   

   The only problem with Method 1 is that we need to count how many ones are rolled with the die. Better: use the R the sum function to add up how many ones are obtained:
   

   > # Better method 1:
   > sum(rbinom(1200, 1, 1/6))
   

   The alternative is to conduct one experiment of flipping 1,200 coins once each:
   

   > # Method 2
   > rbinom(1, 1200, 1/6)
   

   This method call returns the number of successes (k) obtained in 1,200 (n) trys with probability p of success, where "success" is obtaining a one with a single die roll.
3. What are the five steps for the z-test? Answer:
   
   1. State the null (H₀) and alternative (H₁) hypotheses:
            H₀: μ = μ₀
            H₀: μ ≠ μ₀
   2. Compute the test statistic z:
            z = (x - μ) / (SD_x/√n)
   3. When can we use the z-test?
      Answer: we can use the t-test when n ≥ 30. Even if the observations are not normally distributed, by the Central Limit Theorem x is close to normally distributed, so the test statistic z also has close to a normal distribution.
   4. Write down confidence interval I for the test statistic. Usually we use the 95% confidence interval [-1.96, 1.96] or the 99% confidence interval [-2.58, 2.58].
   5. If z ∈ I, accept H₀; if z ∉ I, reject H₀.
   6. Compute the p-value, which is the probability of obtaining a test statistic value z as extreme or more extreme than the value of z actually, obtained, given that the null hypothesis is true.
4. Use the R function qnorm to verify that a 95% confidence interval for normally distributed data is [-1.96, 1.96].  Also verify that a 99% confidence interval confidence interval is [-2.58, 2.58].
   Answer: for a 95% confidence interval, the area in the two tails is 100% - 95% = 5% and the area of one tail is 5% / 2 = 2.5% = 0.025.  Using R:
   

   > qnorm(0.025)
   [1] -1.959964
   

   For a 99% confidence interval, the area of the two tails is 100% - 99% = 1% and area in one tail is 1% / 2 = 0.5% = 0.005. Using R:
   

   > qnorm(0.005)
   [1] -2.575829
   

5. 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 100 females:
         x = 173.7     SD+ = 27
   Is there a significant difference between the chloresterol levels of the women in the Baylor study and the reported value in 1999? Perform the test at the 5%-level;  at the 1% level.
6. Claim: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.
7. How is the t-test different than the z-test?
8. Who invented the t-test?
   Answer: William Gosset (1876 - 1937) invented the t-test to control the quality when brewing beer at the Guinness Company outside of Dublin Ireland.  Guinness did not allow employees to publish their research because it was proprietary. Gosset published his research on the t-test under the pen name "Student." Even today, the t-test is often called the Student's t-test.

The t-test

• Use the t-test for μ even when n < 30 and the data are fairly normally distributed.
• Construct a normal plot to check if the data are close to normal whenever a t-test is performed.
• Because n is small, SD might not be a good approximation of σ This increases the variability, which increases the size of the confidence interval. The t-table is used instead of the z-table to account for the extra variability.
• When performing t-tests, you loose one degree of freedom for each parameter that you estimate, so since you use x instead of μ when computing the sample standard deviation s_x, the degrees of freedom are n - 1. Keep this in mind when looking up the confidence interval in the t-table.
• Example 6: A technician makes five measurements of the concentration of carbon monoxide (CO):
        78    83    68    72    88
  The descriptive statistics are:
        n = 5    x = 77.8    SD⁺ = 8.07
  Is the average of these concentrations significantly different than 70 or is it just chance variation?
• There are two important differences between the t-test and the z-test:
  
  1. The t-table is used instead of the z-table.
  2. The p-value is hard to compute so we let R compute it.
• The form of the t-statistic is exactly the same as the z-statistic. The only difference is that since n < 30, we can no longer guarentee that t is normally distributed, so we use the t-table instead of the standard normal table.
• The tails of the t-density are fatter than the tails of a normal density.
• Here is the t-test for Example 2.
  
  1. State the null and alternative hypotheses:
           H₀: μ = 70
           H₁: μ ≠ 70
  2. Compute the test statistic:
           t = (x - μ) / (SE / √n) = (77.8 - 70) / (8.07 / √5) = 2.16
  3. Look up a 95% confidence interval using the t-table with n - 1 = 4 degrees of freedom. Use the upper-tail probability of 0.025 to obtain a two sided confidence interval of level 0.05 to obtain [-2.776, 2.776].
  4. Decide whether to accept or reject the null hypothesis:
           2.16 ∈ [-2.776, 2.776], so accept H₀.
• To perform this test using R, create a vector x and use the t.test function:
  

  > x <- c(78, 83, 68, 72, 88)
  > t.test(x, mu=70)
  
            One Sample t-test
  
  data: x
  t = 2.16, df = 4, p-value = 0.09689
  alternative hypothesis: true mean is not equal to 70
  95 percent confidence interval:
   67.774 87.826
  sample estimates:
  mean of x 
       77.8 
  

  Use the R qt function to obtain the 95% confidence interval for the t-statistic:
  

  > qt(0.025, 4)
  [1] -2.776445
  

Degrees of Freedom

• Degrees of Freedom (df) is a technical term that arises when using the t-test. We are using x to estimate μ in SD+. If we are computing the SD+, when we are computing the square of the deviations, once we know the first n - 1 deviations, we automatically know the nth deviation because the sum of the deviations is always zero. n - 1 is called the degrees of freedom because only n - 1 of the deviations are able to vary freely.
• The degrees of freedom for the t-test is related to the n - 1 that is used in the denominator of SD+.
• Taking df = n - 1 compensates for the additional variation introduced because the true mean μ is unknown and x is used to estimate it.

The Paired Sample t-test

• Goal: to test whether there is a significant difference between subjects from two different groups.
• Typically, one group is the treatment group and the other group is the control group.
• To use the paired sample t-test, each subject in one group is matched with a subject in the other group.
• Then compute the differences in the response variable and perform a one-sample t-test on the differences.
• Example 7:   To test whether a new type of shoe sole material (type B) is better than the old type (type A), manufacture 10 pair of shoes where one shoe is made of type A and the other of type B. Randomly assign the type of material to left or right. Here is the data:
  

  SoleMaterialA	SoleMaterialB
  13.2	14.0
  8.2	8.8
  10.9	11.2
  14.3	14.2
  10.7	11.8
  6.6	6.4
  9.5	9.8
  10.8	11.3
  8.8	9.3
  13.3	13.6

• Perform the paired-sample t-test to see if there is a real difference between the two sole materials, or if it is just chance variation.
• Use R to create a dataframe from this t-test2.txt:
  

  > setwd("c:/it223/sole-material")
  > df <- read.csv("t-test2.txt")
  > diff <- df$A - df$B
  > t.test(diff, mu=0)
  
           One Sample t-test
  
  data: diff
  t = -3.3489, df = 9, p-value = 0.008539
  alternative hypothesis: true mean is not equal to 0
  95 percent confidence interval:
   -0.6869539 -0.1330461
  sample estimates:
  mean of x 
       -0.41 
  

• Here are the five steps of the two-sample t-test:
  
  1. Write down the null and alternative hypothesis:
     H₀: SoleMaterial A = SoleMaterial B
     H₁: SoleMaterial A ≠ SoleMaterial B
• Obtain the test statistic from R: t = -3.349
• Using the t-table, obtain a 95% confidence interval with n - 1 = 10 - 1 = 9 degrees of freedom:
  I = [-2.26, 2.26]
• t ∉ I so reject H₀.
• Find the p-value from the R output: p = 0.009.
• The test statistic for the two-sample t-test obtained by computing the differences
        diff = SolematerialA - SoleMaterialB,
  then use R to perform a one-sample t-test on the variable diff.
• Here are the five steps of the one-sample t-test performed with the diff variable:
  
  1. Write down the null and alternative hypothesis:
           H₀: diff = 0
           H₁: diff ≠ 0
  2. Obtain the test statistic from R: t = -3.349
  3. Using the t-table, obtain a 95% confidence interval with n - 1 = 10 - 1 = 9 degrees of freedom:
           I = [-2.26, 2.26]
  4. t ∉ I so reject H₀.
  5. Find the p-value from the R output:
           p = 0.009.
• Notice that the test statistic t and the p-value are exactly the same whether a paired two-sample t-test is performed or whether a one-sample t-test on the differences is performed.

Perform the Pendulum Experiment

• Perform the pendulum experiment in groups of three.
• Take a pendulum consisting of a nut, thread, and paper that indicates the length of the pendulum in inches. Measure the time in seconds that it takes the pendulum to complete 15 complete periods.
• Use your phone stopwatch to measure the time for 15 periods. Alternatively, you can use this online stopwatch:
        www.online-stopwatch.com/
• Record your time to the nearest hundredth of a second.
• We will analyze the resulting regression model on Wednesday, June 5.

Inference for Simple Linear Regression

• Question: if y = ax + b is the regression equation, what is the difference between a, b and a^{^}, b^{^}?
• Example 8: The data file chem-reaction.txt contains two columns: (a) mass, the mass of the chemical used in the chemical reaction and (b) time, the time needed for the reaction to occur.
  
  1. Create the scatter plot of time vs. mass.
  2. Find the linear regression equation for predicting time from mass.
  3. Find the R-squared value for this equation. Interpret it.
  4. Create the boxplot of the residuals.
  5. Create the scatterplot of the residuals vs. the predicted values. Interpret it.
  6. Create the normal plot of the residuals. Interpret it.
  7. If y = ax + b is the true regression equation. Perform a t-test that tests the null hypothesis that the true value of the slope a is 0.
  8. If y = ax + b. Perform a t-test that tests the null hypothesis that the true value of the intercept b is 0.
  9. The the chemical reaction studied in this example, if the mass is 10.0, what is the predicted time for the chemical reaction?
• Example 2: the blood alchohol level for a random sample of college students is tested after they drink a few beers. The data file beer-bac.txt contains two columns: (a) the number of beers (beers) consumed and (b) their blood alchohol levels (bac) after they drink the beers. Analyze the regression model and graphs of the resulting data. Use the output and graphs produced from this R script: beer-bac.R
  
  1. Create the scatter plot of time vs. mass.
  2. Find the linear regression equation for predicting bac from beers.
  3. Find the R-squared value for this equation. Interpret it.
  4. Create the boxplot of the residuals.
  5. Create the scatterplot of the residuals vs. the predicted values. Interpret it.
  6. Create the normal plot of the residuals. Interpret it.
  7. If y = ax + b. Perform a t-test that tests the null hypothesis that the true value of the slope a is 0.
  8. If y = ax + b. Perform a t-test that tests the null hypothesis that the true value of the intercept b is 0.
  9. The the chemical reaction studied in this example, if the number of beers consumed is 4, what is the predicted time for the chemical reaction?

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 1:   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 2: 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.
• Answer:
  
  1. H₀: p = 1/4.
     
  2. S = 421   n = 1,600   SE(S) = √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.
• Example 3: Use R to simulate this situation: use the R function call rbinom(1, 500, 0.5) to simulate 500 outcomes of a Bernouilli random variable with p = 0.5. Test whether the random number generator is fair for the given probability.

One-tailed vs. Two-tailed Tests

1. 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].
2. 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].
3. Some researchers think that the one-tailed test is an improvement over the two-tailed test because the test of hypothesis is more precise.
4. However, other researchers think that the one-tailed test is cheating, because it makes rejecting the null hypothesis easier.