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Final Exam Practice Problems 2

Multiple Choice Questions

For each question, show your work or give a reason explaining your answer. 4 points for the reason, 1 point for the correct answer.

  1. What is the value of nC0, for all 0?
       a. 0    b. 1   c. n / 2    d. n

    Answer: b. Recall that nCk = n!/(k!(n-1)!, so nC0 = n!/0!(n-0)! = 1, because 0! = 1.
  2. Who first proved the Central Limit Theorem?
       a. DeMoivre    b. Fisher    c. Lyaponov    d. Pascal

    Answer: c. DeMoivre first formulated the CLT, but Lyaponov proved it in the early 1900s.
  3. What does the word homoscedastic mean in Greek?
       a. Same stretch    b. Same substance    c. Smooth surface    d. Smooth scatter

    Answer: a. It means same variance all the way across the residual plot.
  4. When are two events A and B both independent and mutually exclusive?
       a. Always    b. If P(A)=0 or P(B)=0    c. If P(A)=1 or P(B)=1    d. Never

    Answer: b. If A and B are independent, P(A and B) = P(A)P(B). But this is 0 because P(A and B) = 0 if A and B are mutually exclusive. This means that either P(A)=0 or P(B)=0.
  5. Using the following statistics, find a 90% confidence interval for μ using the t-table:
    n = 9     x = 21     SD+ = 3
       a. [-1.65, 1.95]   b. [-1.96, 1.96]    c. [19.14, 22.86]    d. [15.00, 23.00]
    Answer: The 90% confidence interval for the t-statistic is [-1.86,1.86], so
    -1.86 ≤ (21 - μ) / (3 / √9) ≤ 1.86

    22.86 ≥ μ 19.14

  6. An interviewer asks potential voters the question "Do you have a favorable impression of Candidate X?" If p is the probability that the interviewer gets an answer of "yes," find a 95% confidence interval for p, is this is the data obtained:
         n = 2,500    S = 1,698
       a. [67.9%,67.9%]    b. [66%,70%]    c. [60%,70%]    d. [56%,79%]
    Answer: b. First we need the estimated p: p^ = S / n = 1698 / 2500 = 0.6792; then we need SE(S) = sqrt(np^(1-p^)) = sqrt(2500(0.6792)(1-0.6792)) = 23.34. Then
    -1.96 ≤ (S - np)/SE(S) ≤ 1.96
    -1.96 ≤ (1698 - 2500p)/23.34 ≤ 1.96
    -45.75 ≤ 1698 - 2500p ≤ 45.75
    -1698 - 45.75 ≤ - 2500p ≤ -1698 + 45.75
    -1744 ≤ - 2500p ≤ -1652
    (-1744) / (-2500) ≥ p ≥ -1652 / (-2500)
    0.697 ≥ p ≥ 0.661
    The confidence interval is [0.661, 0.697] = [66%, 70%].
  7. How do the p-values for a one-tailed test and a two-tailed test compare?
    1. Two-tailed p-value is half as large as the one-tailed.
    2. Two-tailed p-value the same as the one-tailed.
    3. Two-tailed p-value is twice as large as the one-tailed.
    4. It depends on the degrees of freedom.
    Answer: c.

Problems

Show all of your work. You may use a calculator.

  1. Students are asked to rate the professor, in a large college class of 225 students, as 4 (excellent), 3 (good), 2 (fair), 1 (poor). Here is the probability distribution of the choices:
    Outcome Probability
    1 0.05
    2 0.15
    3 0.40
    4 0.40
    Use the normal approximation to estimate the probability that the professors average rating is greater than or equal to 3.2.

    Answer: Step 1: find the expected value.
    E(x) = 1×0.05 + 2×0.15 + 3×0.40 + 4×0.40 = 3.15

    Step 2: find the SE.
    σ = sqrt[(1-3.15)2(0.05) + (2-3.15)2(0.15) + (3-3.15)2(0.40) + (4-3.15)2(0.40)] = 0.853

    Step 3: find the formulas for the expected value and SE of the average of 225 ratings:
    E(ave) = E(x) = 3.15, SE(ave) = 0.853 / sqrt(225) = 0.0569

    Step 4: find the probability that the average is 3.2 or greater.
    z = (3.2 - 3.15) / 0.0569 = 0.88. Finding the area under the normal curve corresponding to the bin (-∞, 0.88] gives 0.1894 = 19%.

R Analysis

Perform the following analyses with R. Save your output file as a Word .docx file. Type any interpretation of the output into the output file itself. Questions marked with * require typed output.

  1. Random Number Simulation
    This problem simulates repeating 30 times the experiment of picking 10 Bernoulli random variables with p=0. This is a binomial random variable S with n=10 and p=0.5.
  2. Create and interpret the normal plot for x.
  3. One-sample t-test
    Cell phone users in the U.K spend an average of 8.2 hours per month listening to full-track music on cell phone. A random samples of cell phone users in the U.S. is selected. The number of hours that they spend listening to full-track music on their cell phone is
    5   6   0   4   11   9   2   3
    1. *State the null and alternative hypothesis.

      Answer: H0: μ = 8.2; H1: μ ≠ 8.2

    2. *Create and interpret a normal plot of the data.
    3. *Find the following information with R output: mean, SD, SE(ave), degrees of freedom, t-statistic.
      Answer: x = 5.0, SD+ = 3.63, SEave = 1.28, df = n - 1 = 7, t = -2.497, p = 0.041.
    4. *Do you accept or reject the null hypothesis? Why?

      Answer: Reject the null hypothesis because p < 0.05.