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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

    Ans: 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

    Ans: 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

    Ans: 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

    Ans: 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]

    Ans: 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%]

    Ans: 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.

    Ans: c.

Short Essay Questions

For full credit, use complete sentences and paragraphs.

  1. What constitutes a good regression model?
  2. Explain the difference between a z-test and a t-test.

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.

    Ans: 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%.

SPSS Analysis

Perform the following analyses with SPSS. 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 will simulate repeating 30 times the experiment of picking 10 Bernoulli random variables.
    1. Enter 30 values of a dummy variable called dummy. This is so that the dataset contains 30 rows for the next step.
      Ans: Enter the 30 values by hand, copy and paste from SPSS, or use the SPSS Range Syntax.
    2. Select Transform >> Compute Variable. Enter x as the variable name and Rv.Binomial(10, 0.25) as the expression to compute.
    3. *Create and interpret the normal plot for x.

      Ans: Use Analyze >> Descriptive Statistics >> Q-Q Plot.

  2. 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.

      Ans: H0: μ = 8.2; H1: μ ≠ 8.2

    2. *Create and interpret a normal plot of the data.

      Ans: Use Analyze >> Descriptive Statistics >> Q-Q Plot. If the normal plot is a straight line, the data are normally distributed.

    3. *Find the following information on the SPSS output: mean, SD, SE(ave), degrees of freedom, t-statistic.

      Ans: 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?

      Ans: Reject the null hypothesis because p < 0.05.