3. In a test of Ho: µ = 85 against Ha= %3D the p-value for the test.
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- If the test of H0: = 19 against Ha: ≠ 19 based on an SRS of 15 observations from a Normal populationproduces the statistic of t = –1.75. The P-value isIn a test of H0: µ=150 against HA: µ<150, a sample of size 250 produces Z = -0.65 for the value of the test statistic. Thus the p-value is approximately equal toThe average weight of a Coastal male Grizzly Bear is approximately normal with E(x); =795 pounds and SD * (x) = 80 pounds. 8. How likely is it to randomly select 64 Coastal male Grizzly Bears with a sample average weight of 810 pounds or more? Which density curve is the best model for this problem?
- Q1 A) List down the measures of central tendency and measures of dispersion 2) The operations manager of a plant that manufactures tires wants to compare the actual inner diameters of two grades of tires, each of B) which is expected to be 575 millimeters. A sample of five tires of each grade was selected, and the results representing the inner diameters of the tires, ranked from smallest to largest, are as follows. Grade X grade Y 568 570 575 578 584 573 574 575 577 578 requirement. a) for each of the tow grades of tries, compute the mwan, median, and standred deviation. b) which grade of tire providing better quality? explain. c) what would be the effect on your answer in (a) and (b) if the last value for grade Y were 588 insert 578 explain. C) The file contins the overall miles per gallon (MPG) OF 2010 family sedan: 24 21 22 23 24 34 34 34 20 20 22 22 44 32 20 20 22 20 39 20 Source:…Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800.Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Let ?1 denote true average tread life for a premium brand of P205/65R15 radial tire, and let ?2 denote the true average tread life for an economy brand of the same size. Test H0: ?1 − ?2 = 5000 versus Ha: ?1 − ?2 > 5000 at level 0.01, using the following data: m = 45, x = 42,100, s1 = 2400, n = 45, y = 36,200, and s2 = 1800. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Fail to reject H0. The data does not suggest that the difference in average tread life exceeds 5000.Reject H0. The data suggests that the difference in average tread life exceeds 5000. Fail to reject H0. The data suggests that the difference in average tread life exceeds 5000.Reject H0. The data does not suggest that the difference in average tread life exceeds 5000.
- A survey of 90 recently delivered women on the rolls of a county welfare department revealed that 27 had a history of intrapartum or postpartum infection. What is the critical value of z if we need to conclude that the population proportion with a history of intrapartum or postpartum infection is less than 0.25.A study assessed the lung destructive index among smokers and non- smokers provided the following data Smokers : overline x 1 =14.5 n 1 =12 S P =2.50 Non- Smokers: overline x 2 =9.5 n 2 =10 a = .05 From the available data, can we conclude that smokers have greater lung damage than non-smokers?In a test of H0:μ = 100 against H1:μ>100, a sample of size 80 producers z= 0.8 for the value of the test statistic. The P-value of the test is equal to?
- A researcher interested in exercise endurance in older adults with coronary heart disease measured the endurance of 450 older adults during an exercise routine on a scale of 0-10. He obtained a mean (SD) rating of 6.2 (3.4). Compute the 85% C. I: _____________.In a test of H0:p=0.4 against Ha:p≠0.4, a sample of size 100 produces z=1.28 for the value of the test statistic. Thus the p-value of the test is approximately equal to?