Question

Adinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older S ects. Aner treatment with the drug. 28 Subjects had a mean wake time of 6.7 mm and a standard devon of 43.4 min Assume that the 28 sample values appear to be from a normally distributed population and construct a 95% confidence interval estate of the standard deviation of the wake me for a population with the drug treatments Does the result indicate whether the treatment is effective? Find the confidence interval estimate min< < min (Round to two decimal places as needed)

Answer 1

Solution: Given that n = 28, s = 43.4, s^2 = 1883.56 95% Confidence interval
df = n-1 = 27, X^R = 43.1945 X^L = 34.3129

=> 95% Confidence interval for the standard deviation :
= sqrt((n-1)*s^2/X^2R) < σ < sqrt((n-1)*s^2/X^2L)
= sqrt(27*1883.56/43.1945) < σ < sqrt(27*18883.56/34.3129)
= 34.3129 < σ < 59.0733
= (34.31 < σ < 59.07) rounded
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Solution: Given that information :
Bank A : 6.5,6.6,6.7,6.8,7.1,7.2,7.6,7.8,7.8,7.8
n = 10, s = 0.528, s^2 = 0.2788, 99% Confidence interval
df = 9, X^2R = 23.5893 X^2L = 1.7349

99% Confidence interval for the population standard deviation :
= sqrt(9*0.2788/23.5893) < σ < sqrt(9*0.2788/1.7349)
= 0.3261 < σ < 1.2026
= (0.33 < σ < 1.20) rounded

Bank B : 4.1,5.4,5.8,6.3,6.7,7.8,7.8,8.5,9.4,10.0
s = 1.8546, s^2 = 3.4395
=> sqrt(9*3.4395/23.5893) < σ < sqrt(9*3.4395/1.7349)
= (1.15 < σ < 4.22)

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Solution: Given that n = 21, s = 0.22, 95% Confidence interval

df = n-1 = 20

X^2L = 9.591

X^2R = 34.170