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IT 223 -- Mar 2, 2026

Review Exercises

1. Look at the formulas in the Final Exam Review Guide.
2. What is the difference between these three function calls?
   

   rbinom(20, 1, 0.25)  rbinom(1, 20, 0.25)  rbinom(20, 20, 0.25)
   

   Answer: the three parameters to rbinom represent (1) the number of binomial experiments to simulate, (2) the number of trials in one binomial simulation, and (3) the probability of success in one trial. Here are the results:
   

   # Simulate 20 experiments with one trial each:
   rbinom(20, 1, 0.25)
   [1] 0 0 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0
   # Simulate one experiment with 20 trials:
   # rbinom(1, 20, 0.25)
   [1] 8
   # Simulate 20 experiments with 20 trials each:
   # rbinom(20, 20, 0.25)
   [1] 2 8 2 7 9 2 3 6 5 5 4 5 4 6 3 4 5 5 5 4
   

3. What are the five steps for performing a z-test?
   
   1. State the null and alternative hypotheses.
   2. Compute the test statistic z.
   3. Find a 95% confidence interval I = (-2, 2) (or more precisely I = (-1.96, 1.96).
   4. If z ∈ I, accept the null hypothesis, if z ∉ i, reject the null hypothesis.
   5. Compute the p-value.
4. What is the p-value for a statistical test?
   Answer: the p-value is the probability of obtaining a test statistic as extreme or more extreme that the one actually obtained, given that the null hypothesis is true.
5. A lightbulb manufacturer claims its bulbs last 1,000 hours. A consumer group tests 40 bulbs and finds an average lifespan of 980 hours with SD+ = 50 hours. Test the claim with a z-test at a 5% significance level. Answer: here are the steps of the t-test:
   
   1. H₀: x = 1000    H₁: x ≠ 1,000
   2. z = (x / (SD+_x / √n) = (980 - 1000) / (50 / √40)  = -2.53
   3. Confidence interval is I = (-2, 2)
   4. z ∉ I, so reject the null hypothesis.
   5. p-value is 0.018, which is < 0.05, which is another way to see that the null hypothesis should be rejected.
6. The examTimes vector is defined by this R statement:
   

   examTimes <- c(18, 160, 234, 149, 145, 107, 
       197,  75, 201, 225, 211, 119, 157, 145, 
       107, 244, 163, 114, 145,  68, 111, 185, 
       202, 146, 203, 224, 213, 104, 178, 166)
   

   These are the times in minutes for students in a statistics class to complete the class final exam. Use the R function t.test to test the null hypotheses that the true average time for a typical student to take the exam is 175 minutes. Perform the test at these significance levels: 0.05, 0.01, 0.10.
   
   Note: R does not have a function that specifically performs z-tests. Use the t.test function instead. A z-test and a t-test use exactly the same test statistic:
   t = z = (x - μ) / (s_x/√n)
   The advantage of the t-test is that it works for any sample size, including that when n < 30. However, for smaller values of n, a wider confidence interval is used, computed with the t-distribution instead of the normal distribution. We will discuss t-tests later today.
   
   Answer:
   

   > t.test(exam_scores, mu=175, conf.level=0.95)
   
           One Sample t-test
   
   data:  exam_scores
   t = -1.7956, df = 29, p-value = 0.08298
   alternative hypothesis: true mean is not equal to 175
   95 percent confidence interval:
    136.9259 177.4741
   sample estimates:
   mean of x 
       157.2 
   

Test of Hypothesis for a Proportion

• 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 = [-2, 2]
  4. If z ∉ I, reject H₀; if z ∈ I, accept H₁.
  5. Compute the p-value.
• Example 1:   Flip a coin 100 times and obtain 43 heads. Is the coin fair? Answer:
  
  1. H₀: p = 0.5; H₁: p ≠ 0.5
  2.      n = 100, S = 43, E(S) = np,
          σ_S = √np(1-p) = √100(0.5)(1-0.5) = 5.0.
          Assuming the null hypotheses that p = 0.5,
          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 = (-2, 2)
  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 ∈ (-2, 2), 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 rbinom to simulate 500 outcomes of a Bernouilli random variable with p = 1/6. Test whether the random number generator is fair. Answer: here are the R statements and output:
  

  > sim <- rbinom(500, 1, 1/6)
  > t.test(sim, mu=1/6)
  
          One Sample t-test
  
  data:  sim
  t = 0.4321, df = 499, p-value = 0.6659
  alternative hypothesis: true mean is not equal to 0.1666667
  95 percent confidence interval:
   0.140656 0.207344
  sample estimates:
  mean of x 
      0.174  
  

  The p-value = 0.6659 > 0.05, so accept the null hypothesis.  There is not enough evidence to reject the conclusion that the simulation is fair.

The t-test

• Use the t-test for μ even when n < 30. If n < 30, x may not be normally distributed unless 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.
• Example 4: 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 from the t-table so we let R compute it instead.
• 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 4:
  
  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 probility of 0.025 to obtain a two sided confidence interval of 95% to obtain 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 dataset with variable x containing 78, 83, 68, 72, 88. The null hypothesis is
  H₀: x = μ = 70. Use the R t.test function:
  

  > conc <- c(78, 83, 68, 72, 88)
  > t.test(conc, mu=70)
  
          One Sample t-test
  
  data:  conc
  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 
  

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.
• Example 5. A manufacturer claims that their energy bars contain an average of 20 grams of protein. To verify this claim, a quality control team takes a random sample of 12 energy bars and measures the protein content. Is the manufacturer's claim accurate at a significance level of 0.05? Here are the protein content of each of the energy bars:
  

  20.69 27.46 22.15 19.85 21.29 24.75 
  20.75 22.91 25.34 20.33 21.54 21.08
  

  This problem and dataset was given by a Chrome AI Overview in response to the search string "example t-test problem with dataset."
  
  Answer:
  

  > protein <- scan( )
  1: 20.69 27.46 22.15 19.85 21.29 24.75 
  7: 20.75 22.91 25.34 20.33 21.54 21.08
  13: 
  Read 12 items
  
              One Sample t-test
  
  data: protein
  t = 3.4723, df = 11, p-value = 0.005219
  alternative hypothesis: true mean is not equal to 20
  95 percent confidence interval:
   20.85857 23.83143
  sample estimates:
  mean of x 
     22.345 
  

  Use the t-table to look up a 95% confidence interval for the t-statistic. Because n = 12, the degrees of freedom are
       df = n - 1 = 12 - 1 = 11.
  In the t-table, look at row df = 11 and column upper tail probability = 0.025. This gives 2.201 for the t-value.  Therefore, therefore 95% confidence interval is I = (-2.201, 2,201). Since t = 3.4723 ∉ I, reject the null hypothesis. Another way to come to this conclusion is to use the p-value. Since p = 0.005219 < 0.05, reject the null hypothesis H₀.