The service times for customers coming through a checkout counter in a retail store are independent random variables with mean 1.5 minutes and variance 1.0. Find th the value Yo, such that Pr(1.5 – Yo <Ý < 1.5+ yo) = 0.95, where Y is the average service time of 100 randomly selected customers. 0.196
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- Let X be a random sample from Exp(O). a. Find the CRLB for variance of any estimator b. Find CSS. c. Find the UMVUE using (a) as estimator and (b) as CSS d. Is the estimator in (b) efficient? Show/ Justify?A company is doing a hypothesis test on the variation in quality from two suppliers. Both distributions are normal, and the populations are independent. Use a = 0.05. A sample of 23 products were selected from Supplier 1 and a standard deviation of quality was found to be 6.659. A sample of 25 products were selected from Supplier 2 and a standard deviation of quality was found to be 2.8432. Test to see if the variance in quality for Supplier 1 is larger than Supplier 2. What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: Select an answer Select an answer ✓ H₁: Select an answer Select an answer Based on the hypotheses, compute the following: Round answers to at least 4 decimal places. The test statistic is = The p-value is = The decision is to Select an answer The correct summary would be: [Select an answer quality for Supplier 1 is larger than Supplier 2. that the variance inHighway engineers in Ohio are painting white stripes on a highway. The stripes are supposed to be approximately 10 feet long. However, because of the machine, the operator, and the motion of the vehicle carrying the equipment, considerable variation occurs among the stripe lengths. Engineers claim that the variance of stripes should be less than 16 inches squared. At α = .05, use the sample lengths given here from 12 measured stripes (in feet and inches) to test the variance claim. Assume stripe length is normally distributed. (in feet) (in inches) 9.85 118.2 9.7 116.4 9.9 118.8 9.5 114 9.15 109.8 10.1 121.2 10 120 9.8 117.6 9.9 118.8 10.3 123.6 10.1 121.2 10.2 122.4 The appropriate test for this question is: a Z-test for the true variance of stripe length. a t-test for the true mean of stripe length. a Chi-square test…
- A company is doing a hypothesis test on the variation in quality from two suppliers. Both distributions are normal, and the populations are independent. Use a = 0.01. A sample of 26 products were selected from Supplier 1 and a standard deviation of quality was found to be 6.0042. A sample of 25 products were selected from Supplier 2 and a standard deviation of quality was found to be 1.802. Test to see if the variance in quality for Supplier 1 is larger than Supplier 2. What are the correct hypotheses? Note this may view better in full screen mode. Select the correct symbols in the order they appear in the problem. Ho: Select an answer ✓ ? Select an answer H₁: Select an answer ? ✓ Select an answer Based on the hypotheses, compute the following: Round answers to at least 4 decimal places. The test statistic is = The p-value is The decision is to Select an answer The correct summary would be: Select an answer quality for Supplier 1 is larger than Supplier 2. ✓that the variance inLet x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately ?2 = 47.1. However, a random sample of 18 colleges and universities in Kansas showed that x has a sample variance s2 = 86.1. Use a 5% level of significance to test the claim that the variance for colleges and universities in Kansas is greater than 47.1. Find a 95% confidence interval for the population variance. (a) What is the level of significance?State the null and alternate hypotheses. Ho: ?2 = 47.1; H1: ?2 < 47.1Ho: ?2 = 47.1; H1: ?2 > 47.1 Ho: ?2 = 47.1; H1: ?2 ≠ 47.1Ho: ?2 < 47.1; H1: ?2 = 47.1 (b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)What are the degrees of freedom?What assumptions are you making about the original distribution? We assume a uniform population distribution.We…The t test for the mean difference between 2 related populations assumes that the population sizes are equal. sample variances are equal. population of differences is approximately normal or sample sizes are large enough. All of the above.
- The average growth of a certain kind of tree is 1.63 inches in three years. A biologist claims that a new kind will have greater three-year growth. A random sample of 25 of the new variety has an average three-year growth of 3.56 inches and a variance of 11.77 inches. The test statistic is: (write only the number in the blank round it to 4 decimal points if needed) Answer: Next pageA correlation of +.20 means that [blank] of the variance in the outcome variable is accounted for by the predictor variable.can you fill in the blanks
- Let x represent the average annual salary of college and university professors (in thousands of dollars) in the United States. For all colleges and universities in the United States, the population variance of x is approximately o² = 47.1. However, a random sample of 17 colleges and universities in Kansas showed that x has a sample variance s² = 82.7. Use a 5% level of significance 1 test the claim that the variance for colleges and universities in Kansas is greater than 47.1. (a) What is the level of significance? State the null and alternate hypotheses. Ho: o²=47.1; H: o> 47.1 Ho: o²=47.1; H: o2 0.100 0.050 a, we fail to reject the null hypothesis. Since the P-value > x, we reject the null hypothesis. Since the P-values ∞, we reject the null hypothesis. Since the P-values a, we fail to reject the null hypothesis. (e) Interpret your conclusion in the context of the application. At the 5% level of significance, there is insufficient evidence to conclude the variance of annual salaries…2. Let X - tp. Verify the mean and variance formulas. (Hint: Let X = p/2UV-1/2.)A physical therapist wanted to know whether the mean step pulse of men was less than the mean step pulse of women. She randomly selected 54 men and 70 women to participate in the study. Each subject was required to step up and down a 6-inch platform. The pulse of each subject was then recorded. The following results were obtained. Two sample T for Men vs Women N Mean StDev SE Mean Men Women 98% CI for mu Men - mu Women (- 12.20, - 1.00) T-Test mu Men = mu Women (vs H2 O C. Ho: H1 = H2; Ha: H1 #H2 (b) Identify the P-value and state the researcher's conclusion if the level of significance was a = 0.01. What is the P-value? P-value =