Question

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

1616

subjects had a mean wake time of

102.0102.0

min. After​ treatment, the

1616

subjects had a mean wake time of

97.597.5

min and a standard deviation of

20.320.3

min. Assume that the

1616

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

9090​%

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

102.0102.0

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

Answer 1

Solution :

degrees of freedom = n - 1 = 16 - 1 = 15

t/2,df = t0.05,15 = 1.753

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

= 1.753 * ( 20.3/ 16)

Margin of error = E = 8.9

The 90% confidence interval estimate of the population mean is,

- E < < + E

97.5 - 8.9 < < 97.5 + 8.9

88.6 < < 106.4

The confidence interval include the mean wake time of 102.0 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