Let xy X2 ..., Xn be a random sample from a Poisson distribution with parameter A. Find ML estimator of A. Also find its variance.
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distribution with parameter 2. Find ML estimator of 2.
Also find its variance.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4fa94e4-9a27-46f1-a875-f9f96d0d0cdb%2Fc18cd0e3-0da7-4a9c-8e1b-49316516d18d%2F3ysiw47_processed.jpeg&w=3840&q=75)
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- Let wages denote hourly wages, educ years of education, and exper years of experience, and suppose log(wages) = f1 + 62educ + B3exper +e where E[eleduc, exper] = 0. Let b1, b2, and bz be our estimates for B1, B2, and ß3 and r23 denote the sample correlation between {educ,}", and {exper, }". Which of the following statements is true? II O a. The variance of b3 does not depend on r23. O b. The variance of b, is increasing in E" (educ, – educ,)² (all else equal). O c. The variance of b2 is at least as large as o²/ E", (educ, - educ,)2 for o? Var(eleduc, exper). O d. The variance of b2 depends on the sign of r23. O e. The variance of b2 always increases as r23 increases (all else equal).Suppose that contamination particle size (in microm- eters) can be modeled as f(x)=2x³ for 1 < x. Determine the mean of X. What can you conclude about the variance of X?The frequency distribution table below is the net weight of a sample of candied fruit products in the food industry cans "ABC": - How many cans have a net weight of less than 20.2 grams. - If we want to compare about the distribution of net weight with the food company "XYZ" for the same type of product. And it is known that the average and variance is 20 grams and 0.04 grams. Which food company has an even distribution of net weight.
- Records from previous years for a casualty insurance company show that its clients average a combined total of 1.9 auto accidents per day, with a variance of 0.31. The actuaries of the company claim that the current variance, oʻ, of the number of accidents per day is not equal to 0.31. A random sample of 17 recent days had a mean of 2 accidents per day with a variance of 0.62. If we assume that the number of accidents per day is approximately normally distributed, is there sufficient evidence to conclude, at the 0.05 level of significance, that the actuaries are correct? Perform a two-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places and round your answers as specified below. (If necessary, consult a list of formulas.) (a) State the null hypothesis H, and the alternative hypothesis H,. p H, :0 H, :0 (b) Determine the type of test statistic to use. (Choose one) ▼ D=0 OSO O20 (c) Find the value of the test statistic. (Round…a sample of n = 10 score has ss= 108 what is the variance for this sampleLet X1, X2, X3, ..., X, be a random sample from a distribution with known variance Var(X,) = o², and unknown mean EX, = 0. Find a (1 – a) confidence interval for 0. Assume that n is large.
- Suppose Y = 6x-8, where x is a random variable with mean 2 and a variance of 4/3. Determine the variance of the random variable yLet S1^2 be the sampling variance for a random sample of twelve values (amount of mercury in the blood) and let S2^2 be the sampling variance for a random sample of ten values (amount of lead in the blood); samples from the same population were used. The population variance for mercury measurements is assumed to be twice the corresponding population variance for lead measurements. We will further assume that S1^2 is independent of S2^2. 1. Find a number b such that P [(S1^2/S2^2)<=b]=0.95 enter such a b to three decimal places b= 2. Consider the number b calculated above, find a number a such that P[a<=(S1^2/S2^2)<=b]=0.90 enter said a to three decimal places. a=One study claims that the variance in the resting heart rates of smokers is different than the variance in the resting heart rates of nonsmokers. A medical student decides to test this claim. The sample variance of resting heart rates, measured in beats per minute, for a random sample of 9 smokers is 480.7. The sample variance for a random sample of 9 nonsmokers is 131.6. Assume that both population distributions are approximately normal and test the study’s claim using a 0.10 level of significance. Does the evidence support the study’s claim? Let smokers be Population 1 and let nonsmokers be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. H0: σ21=σ22: Ha: σ21⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯σ22 Step 2 of 3: Compute the value of the test statistic. Round your answer to four decimal places. Step 3 of 3: Draw a conclusion and interpret the decision.
- Table 2 gives the output for 8 cropping seasons of Sweet Spring Country Farm that used each of 4 fertilizers. Assume that the outputs with each fertilizer are normally distributed with equal variance. (d) Test the hypothesis that the population means are the same at the 5% level of significance.2. The average zone of inhibition (in mm) for mouthwash L as tested by the medical technology students has been known to be 9mm. A random sample of 10 mouthwash L was tested and the test yielded an average zone of inhibition of 7.5mm with a variance of 25 mm. Is there enough reason to believe that the anti-bacterial property of the mouthwash has decreased? Test the hypothesis that the average zone of inhibition of the mouthwash is no less than 9mm using 0.05 level of significance. A. State the hypotheses. B. Determine the test statistic to use. C. Determine the level of significance, critical value, and the decision rule. D. Compute the value of the test statistic. E. Make a decision. F. Draw a conclusion.Let Y be the life in hours of a battery. Assume Y is normally distributed with mean 200 and variance o?. If a purchaser of such batteries requires that at least 90% of the batteries have lives exceeding 150 hours, what is the largest value o can be and still have the purchaser satisfied?
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