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

`97.5-2.086(44.1)/(√(21))=77.426`

`97.5+2.086(44.1)/(√(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.

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)(s)/(√(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(44.1)/(√(21))=77.426`

`97.5+2.086(44.1)/(√(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.