1) Suppose tha Y is a random variable having an gamma distribution with a = 1, find the value of k such that an interval from 0 to ky is a (1 - a)100% confidence interval for the parameter ß.
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- Suppose X is a random variable with standard beta distribution and parameters (alpha) = 8 and (beta) = 2 What is (PX>=0.6)?A company is doing a hypothesis test on the variation of quality from two suppliers. Both distributions are normal, and the populations are independent. use a = 0.01 2 H0:01 HA: 01 > 2 2 02² 10₂² The standard deviation of the first sample is 2.2077 5.1485 is the standard deviation of the second sample. The test statistic is =You are performing a right-tailed matched-pairs test with 9 pairs of data.If α=.005α=.005, find the critical value, to two decimal places.
- (Mathematical Statistics) Let X1,X2,...,X5 be a random sample of the Gamma distribution with parameters a = and ß = 0, e > 0. Determine the 95% confidence interval for the parameter e based on the statistic E(i=1 - 5) Xi. 2A random sample of size n = 12 from a normal pop-ulation has the mean x = 27.8 and the variance s2 = 3.24. If we base our decision on the statistic of Theorem 13, canwe say that the given information supports the claim thatthe mean of the population is μ = 28.5?The weight (in ounces) of a frozen food package produced by a given manufacturer is a normal random variable with mean u and variance o? (both unknown). We weigh n = 10 of these packages, selected at random and independently from those produced by this manufacturer and find 10 10 E*; = 159, E 4 = 2531. i) Estimate the true average weight of all packages produced by this manufacturer using a 95% confidence interval. ii) Find a 95% upper limit confidence interval for u. iii) Determine a 95% confidence interval for the true standard devi- ation, o.
- You have taken a random sample of sizen & 95of a normal population that has a population mean ofμ = 140and a population standard deviation ofo = 23. Your sample, which is Sample 1 in the following table, has a mean ofx = 141.3. (In the table, Sample 1 is indicated by "M1", Sample 2 by "M2", and so on.) (to) Based on Sample 1, plot the confidence intervals of80%and95%for the population mean. Use1,282as the critical value for the confidence interval of 80%and use1960 as the critical value for the confidence interval of95%. (If necessary, you can refer to a list of formulas .) • Write the upper limit and the lower limit on the graphs to indicate each confidence interval. Write the answers with one decimal place. • For the points (♦and ◆), write the population mean, μ = 140. 128.0 128.0 80% confidence interval 139.0 X Ś 150.0 150.0 128.0 128.0 95% confidence interval 139.0 X Ś 150.0 150.0Suppose a marketing company randomly surveyed 404 households and found that in 214 of them, the woman made the majority of the purchasing decisions. Construct a 90% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions.p'=α2=zα2=Margin of Error: E=We are 90% confident that the proportion of households in the population where women make the majority of purchasing decisions is between___ and ___.A random sample of n = 19 winter days in Denver gave a sample mean pollution index x1 = 43. Previous studies show that o1 = 10. For Englewood (a suburb of Denver), a random sample of n2 = 18 winter days gave a sample mean pollution index of x2 = 34. Previous studies show that o2 = 13. Assume the pollution index is normally distributed in both Englewood and Denver. Do these data indicate that the mean population pollution index of Englewood is different (either way) from that of Denver in the winter? Use a 1% level of significance. (a) What is the level of significance? State the null and alternate hypotheses. O Ho: H1 H2 O Ho: H1 = l2; H1: H1 < µ2 (b) What sampling distribution will you use? What assumptions are you making? O The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations. O The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations. O The Student's…
- 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?A proton emitter produces proton beams with changing kinetic energy that is uniformly distributed between three and eight joules. Suppose that it is possible to adjust the upper limit of the kinetic energy (currently set to eight joules). • What is the mean kinetic energy? What is the variance of the kinetic energy? What is the probability that a proton beam has a kinetic energy of exactly 3.5 joules? a. E(X) = 5.5, V(X) = 2.0833, probability is zero (0) because X has a continuous distribution O b. E(X) = 5, V(X) = 1.33, probability is zero (0) because X has a uniform distribution O c. NONE O d. E(X) = 5.5, V(X) = 2.0833, probability is zero (0) because X has a uniform distribution