Ho = 0,~Po(x) = { 3 otherwise Hz = 0,~p1(x) = 1 Osxs1 O otherwise
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- In a Right Tailed Hypothesis test, the test statistic was found to be Z=2.74The rejection region included values greater than the critical value Zc=2.02 The conclusion would be to... Reject the null hypothesis because the test statistic is NOT in the rejection region Fail to reject the null hypothesis because the test statistic is in the rejection region Fail to reject the null hypothesis because the test statistic is NOT in the rejection region Reject the null hypothesis because the test statistic is in the rejection region Accept the null hypothesis because the test statistic is NOT in the rejection regionSignal-to-Noise Ratio: If random variable X has mean μ ¹ 0 and standard deviation s > 0, the ratio r = |μ| /s is called the measurement signal-to-noise ratio (SNR) of X (sometimes SNR is defined as the square of this quantity.) The idea is that X can be expressed as X = μ+ (X − μ), with μ representing a deterministic constant-valued “signal”, and (X − μ) the random zero-mean “noise.” In applications, it is desirable to have an upper-bound, with a probabilistic guarantee, on the magnitude of the relative deviation of X from its mean μ, defined as D = |(X − μ) /μ| . That is, we are interested in finding a positive number a, such that P(D ≤ a) ≥ c where c is a pre-specified number representing our confidence in the upper-bound. If r = 10, provide an upper-bound a, with confidence level c = 0.95.Pfizer claims that their COVID-19 vaccine is 95percent effective. Suppose that in a clinical trial test, 960 out of 1000 subjects have received protection from the virus. Test the claim using two tailed test at 0.01 level.( Disregard the continuity correction in your solution). a.) State the null and alternative hypothesis b.) Calculate the critical region c.) What is the t-value/ z-value
- In a study with a non-directional hypothesis… A) There are two rejection regions that have the same area that adds up to the alpha level. B) There is a single rejection region with an area that corresponds to the alpha level. C) There are two rejection regions that have the same area that adds up to the p-value. D) There is a single rejection region with an area that corresponds to the p-value.A researcher is interested in testing whether annual house hold income in Philadelphia is normal. So she took a sample of 50 house holds and found that skewness (s)= 2.3190 and Kurtosis (k) = 6.7322. Use the Jarque- Bera Test to test , at alpha 0.05, whether income follows normal distribution. -Yes, population is normal because Chi-Square test is higher than critical value. -Yes, population is normal because Chi-Square test is less than critical value. -No, population is not normal because Chi-Square test is higher than critical value. -No, population is not normal because Chi-Square test is less than critical value.1. A factory collected data on the number of nonconforming parts to construct an np-chart. 15 samples of size 150 were collected. They determined that the average fraction was non-conforming as pbar=0.045. During production, a sample of 150 parts was taken, of which 18 were non-conforming. At the time this sample was taken: a. the upper control limit was 0.12 b. the lower control limit was 0 c. the sample was outside the control limits d. the sample was within the control limits. 2. A control chart was in control, but the operator adjusted the process anyway. What is this called? a. common cause variation b. special cause variation c. overcontrol d. undercontrol 3. A control chart was out of control, but the operator failed to adjust the process. What is this called? a. common cause variation b. undercontrol c. overcontrol d. special cause variation
- Texas Instruments produces computer chips in production runs of 1 million at a time. It has found that the fraction of defective modules can be very different in different production runs. These differences are caused by small variations in the set-up of each production run. Managers have observed that defective rates are roughly triangular, with a lower bound of 0%, and an upper bound of 50%. Defects more likely to be near 10% than any other single value in their range. Now suppose that we have taken a sample of 10 modules, and 2 of them are defective. i. What is the conditional probability of a defective rate less than 25% in this production run?ii. For what number M would you say that the defective rate is equally likely to be above or below M?The random variable X has a Bernoulli distribution with parameter p. A random sampleX1, X2, . . . , Xn of size n is taken of X. Show that the sample proportionX1 + X2 + · · · + Xnnis a minimum variance unbiased estimator of p.Given that X1, X2, . . . , Xn forms a random sample of size n from a geometric population withparameter p, show thatY =n∑j=1Let A and B represent two variants (alleles) of the DNA at a certain locus on the genome. Let p represent the proportion of alleles in a population that are of type A, and let q represent the proportion of alleles that are of type B. The Hardy–Weinberg equilibrium principle states that the proportion PAB of organisms that are of type AB is equal to pq. In a population survey of a particular species, the proportion of alleles of type A is estimated to be 0.360 ± 0.048 and the proportion of alleles of type B is independently estimated to be 0.250 ± 0.043. a) Estimate the proportion of organisms that are of type AB, and find the uncertainty in the estimate. b) Find the relative uncertainty in the estimated proportion. c) Which would provide a greater reduction in the uncertainty in the proportion: reducing the uncertainty in the type A proportion to 0.02 or reducing the uncertainty in the type B proportion to 0.02?
- (a) State H 0 and H a identify the claim. (b) Determine the critical value, X^2 0, and the rejection region. (c) Calculate the test statistic.a man is investigating the populaion of bear in two areas. Area 1 and Area 2. He expect the number of bear to be X and Y in area 1 and area 2 to be Poisson- distributeted. He expect the number og bear to be λ1 = 3 in area 1 and λ2 = 5 in area 2. Find P(X = 2) and P(X ≥3) and find an approximate value expression for P(X = Y)Area to the Right of the Critical Value Degrees of Freedom 0.995 0.99 0.975 0.95 0.90 0.10 0.05 0.025 0.01 0.005 Degrees of Freedom 1 - - 0.001 0.004 0.016 2.706 3.841 5.024 6.635 7.879 1 2 0.010 0.020 0.051 0.103 0.211 4.605 5.991 7.378 9.210 10.597 2 3 0.072 0.115 0.216 0.352 0.584 6.251 7.815 9.348 11.345 12.838 3 4 0.207 0.297 0.484 0.711 1.064 7.779 9.488 11.143 13.277 14.860 4 5 0.412 0.554 0.831 1.145 1.610 9.236 11.071 12.833 15.086 16.750 5 6 0.676 0.872 1.237 1.635 2.204 10.645 12.592 14.449 16.812 18.548 6 7 0.989 1.239 1.690 2.167 2.833 12.017 14.067 16.013 18.475 20.278 7 8 1.344 1.646 2.180 2.733 3.490 13.362 15.507 17.535 20.090 21.955 8 9 1.735 2.088 2.700 3.325 4.168 14.684 16.919 19.023 21.666 23.589 9 10 2.156 2.558 3.247 3.940 4.865 15.987 18.307 20.483 23.209 25.188 10 11 2.603 3.053 3.816 4.575 5.578 17.275 19.675 21.920 24.725 26.757 11 12 3.074 3.571 4.404 5.226 6.304 18.549 21.026 23.337 26.217 28.299 12 13 3.565 4.107 5.009 5.892 7.042 19.812 22.362 24.736…