A confidence interval estimate is desired for the gain in a circuit on a semicon

Question

A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation = 30.

(a) How large must n be if the length of the 95% CI is to be not greater than 60?

(b) How large must n be if the length of the 99% CI is to be not greater than 60?

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An article in Medicine and Science in Sports and Exercise “Electrostimulation Training Effects on the Physical Performance of Ice Hockey Players,” (2005, Vol. 37, pp. 455–460) considered the use of electromyostimulation (EMS) as a method to train healthy skeletal muscle. EMS sessions consisted of 30 contractions (four-second duration, 85 Hz) and were carried out three times per week for three weeks on 17 ice hockey players. The ten-meter skating performance test showed a standard deviation of 0.09 seconds. Construct a 99% confidence interval of the standard deviation of the skating performance test. Assume population is approximately normally distributed.

Round your answers to 3 decimal places.

Explanation / Answer

CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ^2 right < ^2 < (n-1) s^2 / ^2 left
where,
s = standard deviation
^2 right = (1 - confidence level)/2
^2 left = 1 - ^2 right
n = sample size
since alpha =0.05
^2 right = (1 - confidence level)/2 = (1 - 0.95)/2 = 0.05/2 = 0.025
^2 left = 1 - ^2 right = 1 - 0.025 = 0.975
the two critical values ^2 left, ^2 right at 16 df are 28.8454 , 6.908
s.d( s^2 )=0.09
sample size(n)=17
confidence interval for ^2= [ 16 * 0.0081/28.8454 < ^2 < 16 * 0.0081/6.908 ]
= [ 0.1296/28.8454 < ^2 < 0.1296/6.9077 ]
[ 0.0045 < ^2 < 0.0188 ]
and confidence interval for = sqrt(lower) < < sqrt(upper)
= [ sqrt (0.0045) < < sqrt(0.0188), ]
= [ 0.067 < < 0.137 ]

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