During the 1999 and 2000 baseball seasons, there was much
speculation that the unusually large number of home runs hit was
due at least in part to a livelier ball. One way to test the
“liveliness” of a baseball is to launch the ball at a vertical
surface with a known velocity VL and measure the ratio of the
outgoing velocity VO of the ball to VL. The ratio R=VO/VL is called
the coefficient of restitution. Following are measurements of the
coefficient of restitution for 40 randomly selected baseballs. The
balls were thrown from a pitching machine at an oak surface. 0.6248
0.6237 0.6118 0.6159 0.6298 0.6192 0.6520 0.6368 0.6220 0.6151
0.6121 0.6548 0.6226 0.6280 0.6096 0.6300 0.6107 0.6392 0.6230
0.6131 0.6223 0.6297 0.6435 0.5978 0.6351 0.6275 0.6261 0.6262
0.6262 0.6314 0.6128 0.6403 0.6521 0.6049 0.6170 0.6134 0.6310
0.6065 0.6214 0.6141
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- During The 1999 And 2000 Baseball Seasons There Was Much Speculation That The Unusually Large Number Of Home Runs Hit W 1 (165.94 KiB) Viewed 34 times
During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs hit was due at least in part to a livelier ball. One way to test the "liveliness of a baseball is to launch the ball at a vertical surface with a known velocity V and measure the ratio of the outgoing velocity Vo of the ball to VL. The ratio R = V/V is called the coefficient of restitution. Following are measurements of the coefficient of restitution for 40 randomly selected baseballs. The balls were thrown from a pitching machine at an oak surface. 0.6248 0.6237 0.6118 0.6159 0.6298 0.6192 0.6520 0.6368 0.6220 0.6151 0.6121 0.6548 0.6226 0.6280 0.6096 0.6300 0.6107 0.6392 0.6230 0.6131 0.6223 0.6297 0.6435 0.5978 0.6351 0.6275 0.6261 0.6262 0.6262 0.6314 0.6128 0.6403 0.6521 0.6049 0.6170 0.6134 0.6310 0.6065 0.6214 0.6141 (a) Test the hypothesis Ho: u = 0.635 versus H :p > 0.635. Use a = 0.05. Find to Round your answer to two decimal places (e.g. 98.76). to = Is it possible to reject Ho hypothesis at the 0.05 level of significance? O Yes. O No. Find the P-value. O 0.9995 < P-value O 0.999 <P – value < 0.9995 O 0.9975 <P - value < 0.999 O 0.995 <P - value < 0.9975 (b) Check the normality assumption, using the normal probability plot. Probability Plot of Baseball Coeff of Restitution Normal 95- 90 Percent 80 70 60 50 - 40 30 20 10- . 5 1 0.59 0.60 0.65 0.66 0.61 0.62 0.63 0.64 Baseball Coeff of Restitution Data appear to be normally distributed. Data appear to be not normally distributed. (c) Compute the power of the test if the true mean coefficient of restitution is as high as 0.64. O Power = 0.60 O Power = 0.75 O Power = 0.40 O Power = 0.25 (d) What sample size would be required to detect a true mean coefficient of restitution as high as 0.638 if you wanted the power of the test to be at least 0.75? Round your answer to the nearest ten if needed. on = 20 On = 50 On = 70 On = 100 (e) Which confidence interval should be used to consider Hhypothesis? 0 95% one-sided confidence interval. О 99% two-sided confidence interval. 0 95% two-sided confidence interval. О 99% one-sided confidence interval. Find lower confidence bound for this interval. Round your answers to two decimal places (e.g. 98.76). i <u