Question

A clinical trial was conducted to test whether a drug for treating insomnia is effective. Researchers administered the drug and measured the time it took 23 randomly selected subjects to fall asleep. The subjects had a mean wake time of 96.6 min and a sample standard deviation of 24.5 min. Assume that the population of wake times is normally distributed.

(a) Construct a 95% confidence interval. You may use your calculator. Round to 1 decimals.



(b) Are the assumptions met? Explain.




(c) Interpret the interval.





(d) Suppose unmedicated insomniacs typically have an average wake time of 110 minutes. Run a hypothesis test, at the 0.05 level of significance, to determine whether this drug treatment was able to decrease the amount of time it takes subjects to fall asleep.

Hypotheses: ___________________________ ___________________________

p-value (round to 4 decimals) = _______________

State your conclusion in terms of the problem

Answer 1

Solution :

Given that,

$\bar x$ = 96.6

s = 24.5

n = 23

Degrees of freedom = df = n - 1 = 23 - 1 = 22

At 95% confidence level the t is ,

$\alpha$  = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t$\alpha$ /2,df = t0.025,22 =2.074  

Margin of error = E = t$\alpha$/2,df * (s /$\sqrt$n)

= 2.074 * (24.5 / $\sqrt$ 23)

=10.6

Margin of error = 10.6

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

96.6 - 10.6 < $\mu$ < 96.6 + 10.6

86.0 < $\mu$ < 107.2

(86.0, 107.2 )

d ) This is the left tailed test .

The null and alternative hypothesis is ,

H0 :  $\mu$  = 110

Ha : $\mu$ < 110

Test statistic = t

= (
T
- $\mu$ ) / S / $\sqrt$ n

= (96.6 -110) / 24.5 / $\sqrt$ 23

= −2.623

Test statistic = t =  −2.623

P-value =0.0078

$\alpha$ = 0.05  

P-value < $\alpha$

0.0078 < 0.05

Reject the null hypothesis .

There is sufficient evidence to claim that the population mean μ is less than 110, at the 0.05 significance level.