Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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 :

abiolataiwo2015

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

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.