Question

A clinical trial was conducted to test the effectiveness of a drug used for treating insomnia in older subjects. After treatment with the​ drug, 21 subjects had a mean wake time of 97.5 min and a standard deviation of 44.1 min. Assume that the 21 sample values appear to be from a normally distributed population and construct a 95​% confidence interval estimate of the standard deviation of the wake times for a population with the drug treatments. Does the result indicate whether the treatment is​ effective?

Answer 1

Answer:

$97.5-2.086\frac{44.1}{\sqrt{21}}=77.426$    

$97.5+2.086\frac{44.1}{\sqrt{21}}=117.574$    

So on this case the 95% confidence interval would be given by (77.426;117.574)    

Assuming this previous info:Before treatment with zopiclone, 21 subjects  had a mean wake time of 102.8 min

Since the lower bound for the confidence interval is lower than 102.8 we don't have enough evidence to conclude that we have a significant effect.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X =97.5$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=44.2 represent the sample standard deviation

n=21 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=21-1=20$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,20)".And we see that $t_{\alpha/2}=2.086$

Now we have everything in order to replace into formula (1):

$97.5-2.086\frac{44.1}{\sqrt{21}}=77.426$    

$97.5+2.086\frac{44.1}{\sqrt{21}}=117.574$    

So on this case the 95% confidence interval would be given by (77.426;117.574)    

Assuming this previous info:Before treatment with zopiclone, 21 subjects  had a mean wake time of 102.8 min

Since the lower bound for the confidence interval is lower than 102.8 we don't have enough evidence to conclude that we have a significant effect.