(a)
To find: The people who are selected for the simple random sample.
(a)
Answer to Problem 12E
Solution: The selected people are those who are labeled as 19, 22, 39, and 50.
Explanation of Solution
Calculation:
The people are labeled as
The people selected are the people who are numbered with the same number of the selected random numbers.
To find: The proportion,
Answer to Problem 12E
Solution: The required proportion is obtained as 0.5.
Explanation of Solution
Calculation:
According to the question, the colored circle indicates the persons who are opposing the constitutional amendment and the white circle represents the people who are in favor of the amendment. The selected persons are 19, 22, 39, and 50. Among them, 19 and 50 are the colored circles and 22 and 39 are the white circles.
Therefore, the proportion
(b)
To find: The people who are selected for the simple random sample.
(b)
Answer to Problem 12E
Solution: The selected people are those people who are labeled with the same number of selected random number and the samples are shown in the below table.
Sample | Selected label |
Sample 1 | 45, 46, 77, 17 |
Sample 2 | 52, 71, 13, 88 |
Sample 3 | 95, 59, 29, 40 |
Sample 4 | 68, 41, 73, 50 |
Sample 5 | 82, 73, 95, 78 |
Sample 6 | 60, 94, 07, 20 |
Sample 7 | 36, 00, 91, 93 |
Sample 8 | 38, 44,84, 87 |
Sample 9 | 73, 67, 64, 71 |
Explanation of Solution
Calculation:
The people are labeled as
The people selected are the people who are numbered with the same number of the selected random numbers.
Similarly, the other eight samples of size 4 are selected and the samples are shown in the below table.
Sample | Selected label |
Sample 1 | 45, 46, 77, 17 |
Sample 2 | 52, 71, 13, 88 |
Sample 3 | 95, 59, 29, 40 |
Sample 4 | 68, 41, 73, 50 |
Sample 5 | 82, 73, 95, 78 |
Sample 6 | 60, 94, 07, 20 |
Sample 7 | 36, 00, 91, 93 |
Sample 8 | 38, 44, 84, 87 |
Here, the people selected in each set of sample are the people who are numbered with the same number of the selected random numbers.
To find: The proportion,
Answer to Problem 12E
Solution: The required proportions for the 10 samples are shown below.
Sample | Labeled value | Proportion |
Sample 1 | 45, 46, 77, 17 | 0.75 |
Sample 2 | 52, 71, 13, 88 | 1 |
Sample 3 | 95, 59, 29, 40 | 0.75 |
Sample 4 | 68, 41, 73, 50 | 0.5 |
Sample 5 | 82, 73, 95, 78 | 0.75 |
Sample 6 | 60, 94, 07, 20 | 0 |
Sample 7 | 36, 00, 91, 93 | 0.75 |
Sample 8 | 38, 44,84, 87 | 0.5 |
Sample 9 | 73, 67, 64, 71 | 0.5 |
Sample 10 | 19, 22, 39, 50 | 0.5 |
Explanation of Solution
Calculation:
The calculation of the sample proportion,
Sample | Labeled value | Colored circle | White circle | Proportion |
Sample 1 | 45, 46, 77, 17 | 77 | 45, 46, 17 | |
Sample 2 | 52, 71, 13, 88 | 88 | 52, 71, 13 | |
Sample 3 | 95, 59, 29, 40 | 29, 40, 95, 59 | ||
Sample 4 | 68, 41, 73, 50 | 73, 50 | 68, 41 | |
Sample 5 | 82, 73, 95, 78 | 73, | 82, 95, 78 | |
Sample 6 | 60, 94, 07, 20 | 60, 94, 07, 20 | ||
Sample 7 | 36, 00, 91, 93 | 91 | 00, 36, 93 | |
Sample 8 | 38, 44,84, 87 | 38, 87 | 44, 84 | |
Sample 9 | 73, 67, 64, 71 | 73, 64 | 67, 71 | |
Sample 10 | 19, 22, 39, 50 | 22, 39 | 19, 50 |
(c)
To graph: The histogram.
(c)
Answer to Problem 12E
Solution: The required histogram is shown below.
Explanation of Solution
Interpretation: From the hisogram it is clear that none of the sample has sample proportion of 0.25.
(d)
The number of sample proportion that estimates the population proportion as p = 0.5 .
(d)
Answer to Problem 12E
Solution: There are 4 samples that provide the proportion as 0.5.
Explanation of Solution
Whether the value 0.5 is in the center of the sample values.
Answer to Problem 12E
Solution: The original value 0.5 is in the center for the sample proportion values.
Explanation of Solution
To explain: The reason that the population proportion value will be in the center for the large samples.
Answer to Problem 12E
Solution: The value will be centered in 0.5 for large samples because the large samples have less variability than the small samples.
Explanation of Solution
Want to see more full solutions like this?
Chapter 3 Solutions
Statistics: Concepts and Controversies
- solve the question based on hw 1, 1.41arrow_forwardT1.4: Let ẞ(G) be the minimum size of a vertex cover, a(G) be the maximum size of an independent set and m(G) = |E(G)|. (i) Prove that if G is triangle free (no induced K3) then m(G) ≤ a(G)B(G). Hints - The neighborhood of a vertex in a triangle free graph must be independent; all edges have at least one end in a vertex cover. (ii) Show that all graphs of order n ≥ 3 and size m> [n2/4] contain a triangle. Hints - you may need to use either elementary calculus or the arithmetic-geometric mean inequality.arrow_forwardWe consider the one-period model studied in class as an example. Namely, we assumethat the current stock price is S0 = 10. At time T, the stock has either moved up toSt = 12 (with probability p = 0.6) or down towards St = 8 (with probability 1−p = 0.4).We consider a call option on this stock with maturity T and strike price K = 10. Theinterest rate on the money market is zero.As in class, we assume that you, as a customer, are willing to buy the call option on100 shares of stock for $120. The investor, who sold you the option, can adopt one of thefollowing strategies: Strategy 1: (seen in class) Buy 50 shares of stock and borrow $380. Strategy 2: Buy 55 shares of stock and borrow $430. Strategy 3: Buy 60 shares of stock and borrow $480. Strategy 4: Buy 40 shares of stock and borrow $280.(a) For each of strategies 2-4, describe the value of the investor’s portfolio at time 0,and at time T for each possible movement of the stock.(b) For each of strategies 2-4, does the investor have…arrow_forward
- Negate the following compound statement using De Morgans's laws.arrow_forwardNegate the following compound statement using De Morgans's laws.arrow_forwardQuestion 6: Negate the following compound statements, using De Morgan's laws. A) If Alberta was under water entirely then there should be no fossil of mammals.arrow_forward
- Negate the following compound statement using De Morgans's laws.arrow_forwardCharacterize (with proof) all connected graphs that contain no even cycles in terms oftheir blocks.arrow_forwardLet G be a connected graph that does not have P4 or C3 as an induced subgraph (i.e.,G is P4, C3 free). Prove that G is a complete bipartite grapharrow_forward
- Prove sufficiency of the condition for a graph to be bipartite that is, prove that if G hasno odd cycles then G is bipartite as follows:Assume that the statement is false and that G is an edge minimal counterexample. That is, Gsatisfies the conditions and is not bipartite but G − e is bipartite for any edge e. (Note thatthis is essentially induction, just using different terminology.) What does minimality say aboutconnectivity of G? Can G − e be disconnected? Explain why if there is an edge between twovertices in the same part of a bipartition of G − e then there is an odd cyclearrow_forwardLet G be a connected graph that does not have P4 or C4 as an induced subgraph (i.e.,G is P4, C4 free). Prove that G has a vertex adjacent to all othersarrow_forwardWe consider a one-period market with the following properties: the current stock priceis S0 = 4. At time T = 1 year, the stock has either moved up to S1 = 8 (with probability0.7) or down towards S1 = 2 (with probability 0.3). We consider a call option on thisstock with maturity T = 1 and strike price K = 5. The interest rate on the money marketis 25% yearly.(a) Find the replicating portfolio (φ, ψ) corresponding to this call option.(b) Find the risk-neutral (no-arbitrage) price of this call option.(c) We now consider a put option with maturity T = 1 and strike price K = 3 onthe same market. Find the risk-neutral price of this put option. Reminder: A putoption gives you the right to sell the stock for the strike price K.1(d) An investor with initial capital X0 = 0 wants to invest on this market. He buysα shares of the stock (or sells them if α is negative) and buys β call options (orsells them is β is negative). He invests the cash balance on the money market (orborrows if the amount is…arrow_forward
- MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons IncProbability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage LearningStatistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
- Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSONThe Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. FreemanIntroduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman