Question

Your answer is partially correct. Try again. Supercavitation is a propulsion technology for undersea vehicles that can greatly increase their speed. It occurs above approximately 50 water to dissociate into water vapor forming a gas bubble behind the vehicle. When the gas bubble completely encloses the vehicle, supercavitation is said to occur. Eight (n = 8) tests were conducted on a scale model of an undersea vehicle in a towing basin with the average observed speed = 102.2 meters per second. Assume that speed is normally distributed with o = 4 meters per second. Use a = 0.05. (a) Test Ho:u= 100 versus H : < 100. cannot reject null hypothesis (b) Compute the power of this test if the true mean speed is low as 95 meters per second. Round your answer to four decimal places (e.g. 98.7654). T0.9706 (c) What sample size would be required if we want to detect a true mean speed as low as 94 meters per second if we wanted the power of the test to be at least 0.85. 3.2041 Statistical Tables and Charts Click if you would like to Show Work for this question: Open Show Work

Answer 1

Solution: Given that m/s Average observed Speed π = 102.2 Assuming that . Normally distributed ( ) = 4 m/s use, P = 0.05 Number of tests conducted (n)= 8 @ To test Ho : M = 100 Versus Hi ell < 100 Here Null hypothests. Ho : M = 100 (Mo) Alternative hypothesis, Hi : 4 < 100 (Mo) we know test Statistic is given by that, formalla to compute the ã-llo za - 102.2 -.100 212 1.4142 vs. z = 1.5556 2.2 = 1:56] .2 1.54

we know, from the one-sided the standard normal table, the critical value for @ text at & = 0.05 level of Significance is Zd=-1.645 Cleft tailed test) So, test statistic value is not 1.56 > -1:645 Hence in rejection region cit 5% level of hypothes fail to reject the null There fore Significance 6 To compute the power of the test 1. When mean Speed = (M) = 95 so, S = M- Mo : 95-100 8 = -5 from the formula, - 1.645 = 1-0 (1689) = 1- ( = NORMSDSIT

1 -0.9706 ß = 0.0294 power of the test i given by 1- B - - 0.0294 -0.9706 power of the test = 0.9706 © Given that power, l-B = 0.85 1. B = 0.15 let s= M-lo Mean Speed 4 =95 , = 95-100 8=-5 By formula (ExtZo) on nz n =(70.05 + Z0.15 Cust (95-10072 (1:645 +1-04)°C16) (2.685) x 16 25 25 na 7.2092 x 16 - 2 - 4.61 25 115. 3476 25 n = 4:61 25 1. Hence the sample size is 5.