Problem 1
Chapter 6 (Probability): Exercises 6.81
6.81 Your favorite team is in the final playoffs. You have assigned a probability of 60% that they will win the championship. Past records indicate that when teams win the championship, they win the first game of the series 70% of the time. When they lose the series, they win the first game 25% of the time. The first game is over; your team has lost. What is the probability that they will win the series?
Series
I Game / Win / Lose / Total
Win / 0.7 * 0.6 = 0.42 / 0.25 * 0.4 = 0.10 / 0.52
Lose / 0.6 - 0.42 = 0.18 / 0.4 - 0.1 = 0.30 / 0.48
Total / 0.6 / 0.4 / 1
P(Win the Series | Lose the I game) = 0.18/0.48 = 0.375
Problem 2
Chapter 10 (Introduction to Estimation): Exercises 10.11
10.11. a. A random sample of 25 was drawn from a normal distribution whose standard deviation is 5. The sample mean is 80. Determine the 95% confidence interval estimate of the population mean.
b. Repeat Part a with a sample size of 100.
c. Repeat Part a with a sample size of 400.
d. Describe what happens to the confidence interval estimate when the sample size increases.
n = 25
x-bar = 80
5
% = 95
Standard Error, SE = σ/n = 1.0000
z- score = 1.9600
Width of the confidence interval = z * SE = 1.9600
Lower Limit of the confidence interval = x-bar - width = 78.0400
Upper Limit of the confidence interval = x-bar + width = 81.9600
The confidence interval is [78.04, 81.96]
(b)
n = 100
x-bar = 80
5
% = 95
Standard Error, SE = σ/n = 0.5000
z- score = 1.9600
Width of the confidence interval = z * SE = 0.9800
Lower Limit of the confidence interval = x-bar - width = 79.0200
Upper Limit of the confidence interval = x-bar + width = 80.9800
The confidence interval is [79.02, 80.98]
(c)
n = 400
x-bar = 80
5
% = 95
Standard Error, SE = σ/n = 0.2500
z- score = 1.9600
Width of the confidence interval = z * SE = 0.4900
Lower Limit of the confidence interval = x-bar - width = 79.5100
Upper Limit of the confidence interval = x-bar + width = 80.4900
The confidence interval is [79.51, 80.49]
(d) It is clear from the above that as the sample size increases, the confidence interval narrows.
Problem 3
Chapter 11 (Introduction to hypothesis testing): Exercises 11.52
11.52 A statistics practitioner wants to test the following hypotheses with σ = 20 and n = 100:
a. Using α = .10 find the probability of a Type II error when μ = 102.
b. Repeat Part a with α = .02.
c. Describe the effect on β of decreasing α.
1 - α = 1 - 0.10 = 0.90
z- score for p = 0.90 is 1.282
x-bar = μ + z * (σ/√n)
= 100 + 1.282 * (20/√100)
= 102.564
Therefore, we will reject Ho for x-bar > 102.564
Now, we find the probability that x-bar < 102.564 given that μ = 102
z = (x-bar - μ)/(σ/√n)
= (102.564 - 102)/(20/√100)
= 0.282
P(x-bar < 102.564) = P(z < 0.282)
= 0.611
Thus, probability of a Type II error = 0.611
(b) Now, with α = 0.02
1 - α = 1 - 0.02 = 0.98
z- score for p = 0.98 is 2.054
x-bar = μ + z * (σ/√n)
= 100 + 2.054 * (20/√100)
= 104.108
Therefore, we will reject Ho for x-bar > 104.108
Now, we find the probability that x-bar < 104.108 given that μ = 102
z = (x-bar - μ)/(σ/√n)
= (104.108 - 102)/(20/√100)
= 1.054
P(x-bar < 104.108) = P(z < 1.054)
= 0.854
Thus, probability of a Type II error = 0.854
(c) We notice that as α decreased from 0.10 to 0.02, β increased from 0.611 to 0.854
Thus, decrease in α results in an increase in β. This is because the less Type I error we tolerate, the more Type II error we will be making.
Problem 4
Chapter 12 (Inference about a population): Exercises 12.74
12.74 Spam is of concern to anyone with an e-mail address. Several companies offer protection by eliminating Spam e-mails as soon as they hit an inbox. To examine one such product, a manager randomly sampled his daily e-mails for 50 days after installing spam software. A total of 374 e-mails was received, of which 3 were spam. Use the Wilson estimator to estimate with 90% confidence the proportion of spam e-mails that get through.
The standard error of p is SE = √{p(1 - p)/(n + 4)} = √{0.0132(1 - 0.0132)/(374 + 4)} = 0.0059
Z- score for 90% confidence = 1.645
Margin of error is E = z * SE = 1.645 * 0.0059 = 0.0097
The confidence interval for the population proportion is [p - E, p + E]
= [0.0132 - 0.0097, 0.0132 + 0.0097]
=[ 0.0035, 0.0229]
The 90% confidence interval is [0.0035, 0.0229]