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Concept explainers
In recent years a number of research projects in extrasensory perception have examined the possibility that hypnosis may be helpful in bringing out ESP in persons who did not think they had any. The obvious way to test such a hypothesis is with a self-paired design: the ESP ability of a subject when he is awake is compared to his ability when hypnotized. In one study of this sort, fifteen college students were each asked to guess the identity of two hundred Zener cards (see Case Study 4.3.1). The same “sender”-that is, the person concentrating on the card-was used for each trial. For one hundred of the trials both the student and the sender were awake; for the other one hundred both were hypnotized. If chance were the only factor involved, the expected number of correct identifications in each set of one hundred trials would be twenty. The observed average numbers of correct guesses for subjects awake and subjects hypnotized were 18.9 and 21.7, respectively (24). Use the analysis of variance to determine whether that difference is statistically significant at the 0.05 level.
Number of Correct Responses (out of 100) in ESP Experiment | ||
Student | Sender and Student In Waking State | Sender and Student In Hypnotic state |
1 | 18 | 25 |
2 | 19 | 20 |
3 | 16 | 26 |
4 | 21 | 26 |
5 | 16 | 20 |
6 | 20 | 23 |
7 | 20 | 14 |
8 | 14 | 18 |
9 | 11 | 18 |
10 | 22 | 20 |
11 | 19 | 22 |
12 | 29 | 27 |
13 | 16 | 19 |
14 | 27 | 27 |
15 | 15 | 21 |
16 | 18.9 | 21.7 |
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Chapter 13 Solutions
An Introduction to Mathematical Statistics and Its Applications (6th Edition)
- Theorem 1.2 (1) Suppose that P(|X|≤b) = 1 for some b > 0, that EX = 0, and set Var X = 0². Then, for 0 0, P(X > x) ≤e-x+1²² P(|X|>x) ≤2e-1x+1²² (ii) Let X1, X2...., Xn be independent random variables with mean 0, suppose that P(X ≤b) = 1 for all k, and set oσ = Var X. Then, for x > 0. and 0x) ≤2 exp Σ k=1 (iii) If, in addition, X1, X2, X, are identically distributed, then P(S|x) ≤2 expl-tx+nt²o).arrow_forwardTheorem 5.1 (Jensen's inequality) state without proof the Jensen's Ineg. Let X be a random variable, g a convex function, and suppose that X and g(X) are integrable. Then g(EX) < Eg(X).arrow_forwardCan social media mistakes hurt your chances of finding a job? According to a survey of 1,000 hiring managers across many different industries, 76% claim that they use social media sites to research prospective candidates for any job. Calculate the probabilities of the following events. (Round your answers to three decimal places.) answer parts a-c. a) Out of 30 job listings, at least 19 will conduct social media screening. b) Out of 30 job listings, fewer than 17 will conduct social media screening. c) Out of 30 job listings, exactly between 19 and 22 (including 19 and 22) will conduct social media screening. show all steps for probabilities please. answer parts a-c.arrow_forward
- Question: we know that for rt. (x+ys s ا. 13. rs. and my so using this, show that it vye and EIXI, EIYO This : E (IX + Y) ≤2" (EIX (" + Ely!")arrow_forwardTheorem 2.4 (The Hölder inequality) Let p+q=1. If E|X|P < ∞ and E|Y| < ∞, then . |EXY ≤ E|XY|||X|| ||||qarrow_forwardTheorem 7.6 (Etemadi's inequality) Let X1, X2, X, be independent random variables. Then, for all x > 0, P(max |S|>3x) ≤3 max P(S| > x). Isk≤narrow_forward
- Theorem 7.2 Suppose that E X = 0 for all k, that Var X = 0} x) ≤ 2P(S>x 1≤k≤n S√2), -S√2). P(max Sk>x) ≤ 2P(|S|>x- 1arrow_forwardThree players (one divider and two choosers) are going to divide a cake fairly using the lone divider method. The divider cuts the cake into three slices (s1, s2, and s3).If the chooser's declarations are Chooser 1: {s3} and Chooser 2: {s3}, which of the following is a fair division of the cake?arrow_forwardTheorem 1.4 (Chebyshev's inequality) (i) Suppose that Var X x)≤- x > 0. 2 (ii) If X1, X2,..., X, are independent with mean 0 and finite variances, then Στη Var Xe P(|Sn| > x)≤ x > 0. (iii) If, in addition, X1, X2, Xn are identically distributed, then nVar Xi P(|Sn> x) ≤ x > 0. x²arrow_forwardTheorem 2.5 (The Lyapounov inequality) For 0arrow_forwardTheorem 1.6 (The Kolmogorov inequality) Let X1, X2, Xn be independent random variables with mean 0 and suppose that Var Xk 0, P(max Sk>x) ≤ Isk≤n Σ-Var X In particular, if X1, X2,..., X, are identically distributed, then P(max Sx) ≤ Isk≤n nVar X₁ x2arrow_forwardTheorem 3.1 (The Cauchy-Schwarz inequality) Suppose that X and Y have finite variances. Then |EXYarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_iosRecommended textbooks for you
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