Question

A clinical trial was conducted to test the effectiveness of a drug for treating insomnia in older subjects. Before​ treatment,

20

subjects had a mean wake time of

101.0

min. After​ treatment, the

20

subjects had a mean wake time of

92.4

min and a standard deviation of

21.1

min. Assume that the

20

sample values appear to be from a normally distributed population and construct a

99%

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

101.0

min before the​ treatment? Does the drug appear to be​ effective?

Construct the

99%

confidence interval estimate of the mean wake time for a population with the treatment.

_ min < u < min

The confidence interval

▼

the mean wake time of

101.0

min before the​ treatment, so the means before and after the treatment

▼

This result suggests that the drug treatment

▼

a significant effect.

First blank: DOES NOT INCLUDE or INCLUDE

Second blank: ARE DIFFERENT or COULD BE THE SAME

Third Blank: HAS or DOES NOT HAVE

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 92.4

sample standard deviation = s = 21.1

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t_0.005,19 = 2.861

Margin of error = E = t_/2,df * (s /n)

= 2.861 * (21.1 / 20)

Margin of error = E = 13.5

The 99% confidence interval estimate of the population mean is,

- E < < + E

92.4 - 13.5 < < 92.4 + 13.5

(78.9 < < 105.9)

The confidence interval include the mean wake time of 101.0 min.before the treatment so the means before and after the treatment are same this result suggests that the drug treatment does not have a significant effect