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A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with theâ drug, 20 subjects had a mean wake time of 90.7 min and a standard deviation of 43.7 min. Assume that the 20 sample values appear to be from a normally distributed population and construct a 98â% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Find the confidence interval estimate.

Sagot :

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Answer:

67.9 min < μ < 113.5 min

Step-by-step explanation:

Calculation to determine the confidence interval estimate

First step is to find alpha (α/2)

α =1-0.98

α=0.02/2

α= 0.01

Second step is to determine ± zα/2

z-score of 1.0000 - 0.0100 = .9900 using normal distribution table

z-score = ± 2.33

Now Let determine the confidence interval estimate using this formula

X ± zα/2(σ/√n)

Where,

Confidence Interval (CI) = .98

X = 90.7

n = 20

σ = 43.7

Let plug in the formula

Confidence interval estimate = 90.7 ± 2.33 (43.7/√20)

Confidence interval estimate= 90.7 ± 22.76786775

Confidence interval estimate=90.7 - 22.76786775 < μ < 90.7 + 22.76786775

Confidence interval estimate= 67.9 min < μ < 113.5 min

Therefore the confidence interval estimate will be 67.9 min < μ < 113.5 min