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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.
- 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.
- 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.
- 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.
- 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.
- 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
- 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%].
- How do the p-values for a one-tailed test and a two-tailed test compare?
- Two-tailed p-value is half as large as the one-tailed.
- Two-tailed p-value the same as the one-tailed.
- Two-tailed p-value is twice as large as the one-tailed.
- It depends on the degrees of freedom.
Answer: c.
Problems
Show all of your work. You may use a calculator.
- 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.
- 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.
- Create and interpret the normal plot for x.
- 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
- *State the null and alternative hypothesis.
Answer: H0: μ = 8.2;
H1: μ ≠ 8.2
- *Create and interpret a normal plot of the data.
- *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.
- *Do you accept or reject the null hypothesis? Why?
Answer: Reject the null hypothesis because p < 0.05.