Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment, 16 subjects had a mean wake time of 100.0 min. After​ treatment, the 16 subjects had a mean wake time of 75.5 min and a standard deviation of 24.2 min. Assume that the 16 sample values appear to be from a normally distributed population and construct a 95​% confidence interval estimate of the mean wake time for a population with drug treatments. What does the result suggest about the mean wake time of 100.0 min before the​ treatment? Does the drug appear to be​ effective?
Construct the 95​% confidence interval estimate of the mean wake time for a population with the treatment.

Answer 1

The 95​% confidence interval estimate of the mean wake time is equal to (62.61, 88.39).

Given the following data:

• Mean wake time = 75.5 min.

• Standard deviation = 24.2 min.

• Number of sample, n = 16.

How to construct a 95​% confidence interval?

Mathematically, a confidence interval of 95% is given by;

α = 1 - 0.95

α = 0.05.

α/2 = 0.05/2 = 0.025.

Also, the degrees of freedom (df) is given by:

Degrees of freedom (df) = n - 1

Degrees of freedom (df) = 16 - 1

Degrees of freedom (df) = 15.

From the Student's t-distribution table, a critical value at t₀.₀₂₅, ₁₅ = 2.131.

Mathematically, the confidence interval for mean is given by:

Mean ± (t-critical × (standard deviation/√(sample size)))

75.5 ± (2.131 × (24.2/√(16))

75.5 ± (2.131 × 6.05)

For the upper end, we have:

75.5 + 12.89 = 88.39

For the lower end, we have:

75.5 - 12.89 = 62.61.

In conclusion, we can deduce that there isn't a significant difference between the mean wake time before and after treatment and as such, this drug isn't effective at the significance level of 0.05.

Read more on confidence interval here: brainly.com/question/25779324

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