During the 1999 and 2000 baseball seasons, there was much speculation that the u

Question

During the 1999 and 2000 baseball seasons, there was much speculation that the unusually large number of home runs that were 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 / L 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 Suppose that any baseball that has a coefficient of restitution that exceeds 0.640 is considered too lively. Based on the available data, what proportion of the baseballs in the sampled population are too lively? Find a 95% lower conndence bound on this proportion. Assume population is approximately normally distributed Round your answers to 3 decimal places. 10.125 P2 0.040

Explanation / Answer

Question 1

Here p^ = 0.125

Lower 95% bound = p^ - 1.96 * sqrt [p^ * (1-p^)/N] = 0.125 - 1.645 * sqrt [0.125 * 0.875/30]

= 0.125 - 0.099

= 0.026

p > = 0.026

Quesion 2

Here sample mean x = 13.655

population standard deviation = 1.5

99% two sided confidence interval = 13.655 +- Z99% /sqrt (n)

= 13.655 +- 2.576* (1.5/sqrt(11)

= 13.655 +- 1.165

= (12.49, 14.82)

(b) 95% lower confidence bound = x - Z95% (/sqrt (n) = 13.655 - 1.645 * (1.5/sqrt(11) = 13.655 - 0.744 = 12.91

(c) Here the sample size used is n

Here error < 1.4

Here error = critical test statistic * standard error of the mean

1.4 > Z95% * /sqrt (n)

1.4 > 1.96 * 1.5/ sqrt(n)

sqrt(n) > 1.96 * 1.5/1.4

sqrt (n) > 2.1

n > 4.41

so n shall be 5 or greater than 5.

(d) Total width of confidence intervak = 1.5

2 * critical test statistic * standard error of mean = 1.5

2 * 1.96 * 1.5/sqrt(n) = 1.5

sqrt(n) = 3.92

n = 15.3664

so n would be 16 here.

QUestion 3

HEre standard deviation s = 0.09 seconds

n = 17

dF = 17 -1 = 16

99% confidence interval for standard deviation will have

lower two sided 99% confidence bound = sqrt [(n-1)s2/X20.005] = sqrt [(17-1) * 0.092/34.2672] = 0.061

Upper two sded 99% confidence bound = sqrt [(n-1)s2 /X20.995] = sqrt [ (17 - 1) * 0.09 * 0.09/5.1422] = 0.159

so, 99% confidence interval

0.061 < < 0.159

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