Statistics 200, Homework 9 Due Thursday, November 19, 2020 by noon CA time NOTE: Just like the other homeworks, you may seek help from TAs, friends, books, etc., but please credit your sources! 1....

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Statistics 200, Homework 9 Due Thursday, November 19, 2020 by noon CA time NOTE: Just like the other homeworks, you may seek help from TAs, friends, books, etc., but please credit your sources! 1. Suppose X1, . . . , Xn are i.i.d. with density function fθ(x) = (1 + θ)x θI(0 ≤ x ≤ 1) . The densities are on the unit interval and the unknown parameter θ is non- negative. (i) For testing the null hypothesis H0 : θ = 1 against the simple alternative hypothesis H0 : θ = θ1, determine the most powerful level α test as explicitly as possible. (Here, θ1 is fixed and specified.) (ii) Describe how you would compute a critical value for the test in (i). Be as explicit as possible, even if your solution is an approximation. (iii) Does there exist a UMP level α test for testing H0 : θ = 1 against the two-sided alternative H1 : θ 6= 1? (iv) For the two-sided problem in (iv), find a generalized likelihood ratio test as explicitly as possible. 2. An Extra-Sensory Perception (ESP) game consists of a subject being shown five cards with different symbols on them (a triangle, a star, a square, a hexagon and a circle). The cards are turned over and shuffled, and the subject draws out one of them face down. She is then asked to guess which of the five symbols she has selected. Since the expected proportion of successes by random guessing is 1/5, an observed proportion that is significantly greater than this may be taken as evidence that the subject indeed has ESP. The experiment is repeated a total of 200 times. (i) Formulate this as a hypothesis testing problem and state H0 and H1. (ii) Suppose that a one-sided test rejects the null hypothesis whenever X ≥ 50, where X is the observed number of successes in n = 200 trials. Approxi- mate the probability of a Type 1 error. (iii) Suppose the true probability of success is 0.3. Approximate the proba- bility of a type II error. (iv) Rather than mix up the cards on each of the n trials, suppose that the experiment was done in the following way. On the first trial, the subject draws out a card as before, and it is then determined whether the subject was 1 correct by showing her the symbol on the card. But before reshuffling, the subject now guesses on the outcome when picking one of the four remaining cards as well (knowing it cannot be the previous card). Once this is done, all five cards are mixed up again and the process repeated 100 times, so that there are a total of 200 guesses. Let Y be the total number of correct guesses. Compute E(Y ), assuming the subject is guessing at random. (v) Compute V ar(Y ), assuming the subject is guessing at random. (vi) Approximate the Type 1 error if the test rejects the null hypothesis when Y ≥ 50. 3. Suppose X1, . . . , Xn are i.i.d. bernoulli trials with Pθ(Xi = 1} = θ = 1 − Pθ{Xi = 0}. Suppose n is large. Consider testing H0 : θ = 0.5 versus H1 : θ > 0.5. Let θ̂n = ∑ Xi/n be the MLE of θ. Consider the large sample test which rejects when 2 √ n(θ̂n−0.5) > z1−α, where z1−α is the 1−α quantile of the standard normal distribution. (i) What is the limiting probability of a Type 1 error? (ii) Show that the power tends to one against any fixed θ > 0.5 as n → ∞; that is, if θ is true, the test rejects H0 with probability tending to one. 4. ( Continuation of Problem 3) (i) An approximate p-value can be computed as the area under the normal curve to the right of 2 √ n(θ̂n − 0.5). That is, we can define the p-value p̂n by p̂n = 1− Φ[2 √ n(θ̂n − 0.5)] , where Φ is the standard normal c.d.f. If H0 is true, what is the limiting distribution of p̂n? (ii) If θ > 0.5, what is the limiting distribution of p̂n? 5. Suppose an investigator wants to compare a new drug designed to lower LDL cholesterol levels with a placebo. She conducts a well-designed double- blind study using 100 patients (50 receiving treatment and 50 control) in order to test the null hypothesis of no difference. She plans on using a difference in means test resulting in a p-value; call it p̂1. Independently, the same study is done at a hospital in another part of the country, with another 100 patients recruited (50 receiving treatment and 50 control), resulting in a p-value p̂2. Assume the drug is no better than the placebo. In fact, assume there are H hospitals, resulting in independent p-values, p̂1, . . . , p̂H . (i) If each test is carried out at level α = .05, what is the expected number of significant results? 2 (ii) If each test is carried out at level α = 0.05, what is the chance that at least one study will find a significant result? (iii) In order to evaluate results from all studies, you decide you will reject the null hypothesis that there is no difference between the drug and the placebo if there is at least one p-value that is less than c. What should c be so that, if there really is no difference between the drug and placebo, the chance that you reject the null hypothesis of no difference is 5 percent? 6. A thousand individuals sampled were classified according to sex and ac- cording to whether or not they were color-blind as follows: 442 were male and normal, 514 were female and normal, 38 were male and color-blind, and 6 were female and color-blind. According to the genetic model, these numbers should have relative frequencies given by: p/2, pq+ (p2/2), q/2 and q2/2, for some p and q = 1− p. (i) Assuming the model is correct, compute the maximum likelihood estima- tor of p. (ii) Are the data consistent with the model? (If you couldn’t do (i), suggest a reasonable estimator of p and continue as if it were the MLE.) (iii) Suppose an investigator now collects data based on three attributes: sex (male of female), whether or not the person is color-blind, and handedness (dominant right hand or not). Each of the three attributes has two pos- sibilities. If you wanted to test whether the three attributes are mutually independent, you can do a Chi-squared test. How many degrees of freedom should you use? 7. According to some folklore, if you examine n clover plants in a field and do not find any with four leaves, you may be reasonably confident that the fraction of four-leafed clovers is less than 3/n. Justify the folklore. You may assume n is reasonably large, but state your assumptions. 3
Answered Same DayNov 18, 2021

Answer To: Statistics 200, Homework 9 Due Thursday, November 19, 2020 by noon CA time NOTE: Just like the other...

Rajeswari answered on Nov 19 2021
149 Votes
72137 assignment
Qno.1
Given that
i) Here hypothesis is
H0: theta =1 vs theta ≠1
(Two tailed test at 5% significance level )
When theta =1, pdf is f(x) = 2x, 0Test statistic = , where sigma is population s
td dev of theta and theta s is the average of theta from the samples taken.
The test statistic would be very small if theta of sample is almost equal to 1 and std dev is very high.
Std dev high is not acceptable for consistency so we stick to theta of sample very near to 1.
In this case the chances of rejecting null hypothesis is remote hence we have to fix a power ful alpha.
It is possible not at 5% of course so let alpha be as low as possible such as 0.0001.
Thus alpha can be fixed as 0.0001.
ii) Critical value for 0.0001 from Z distribution is 3.71902
iii) When alpha is fixed as 0.0001, we get UMP = 1-0.0001 = 0.9999
iv) For obtaining MLE we consider the distributions of x1, x2…xn.
Since they are independent, the joint distn is product of these.
We take log of these and write the function as L
Note that each xi is less than 1, and hence log xi total would be negative.
Equate I derivative to 0
So we have
Q.no.2
    If each card is guessed then prob for correct guess p = 0.20 and q = 0.80
Each card is independent of other and hence X no of cards correctly guessed is binomial with mean =np
= 200*0.20 = 40
Variance = npq = 32
And std dev = 5.6564
i) H0: x bar = 40
Ha: x bar >40
Mean difference = 50-40 = 10
ii) Test statistic Z = 10/5.6564 = 1.7668
P value one tailed = 0.0386
Prob for type I error =0.038631.
iii) True probability = 0.3
P(x bar >40\p = 0.3 or np =60) <0.0001
Type II error = 0.0001
iv) If each time after one guess card is not replaced, then I trial p = 1/5, II trial ¼, III 1/3 , IV 1/2 and V trial 1. This repeats 40 times.
So here the variable would be x1+x2+x3+x4+x5 where each X1, x2 are binomial with p shown above.
Hence Mean
E(Y) = 40(0.2+0.25+0.333+0.5+1) = 91.32
v) Variance (Y) = 40 {0.2*0.8+0.25*0.75+0.333*0.667+0.5*0.5+0} = 32.78444
Std dev = 5,6257
vi) P(Y>50) = P(Z>(91.32-50/5.7258) = P(Z>7.21649) =0
P value one tailed =0
Type I error = 0
Qno.3
Bernoullie trials
For H0:...
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