Question

A clinic wishes to look at two different treatments for headaches, meditation and pain relievers (specifically acetaminophen).

A simple random sample of 32 headache sufferers were taught meditation techniques to treat their pain.

This group reported an average of 4.83 headache-free days with a standard deviation of 1.65 days.


A different random sample of 35 headache sufferers was given acetaminophen to treat their pain and reported an average of 5.02 headache-free days with a standard deviation of 2.18 days.




Find the 98% confidence interval for the mean difference in headache-free days between the two populations of headache sufferers.

Answer 1

The 98% confidence interval would be given by
`-1.313 \leq \mu_1 -\mu_2 \leq 0.933`

Explanation:

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_1 =4.83` represent the sample mean 1

`\bar X_2 =5.02` represent the sample mean 2

n1=32 represent the sample 1 size

n2=35 represent the sample 2 size

`s_1 =1.65` population standard deviation for sample 1

`s_2 =2.18` population standard deviation for sample 2

`\mu_1 -\mu_2` parameter of interest.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

`(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_1)/(n_1)+(s^2_2)/(n_2)}` (1)

The point of estimate for
`\mu_1 -\mu_2` is just given by:

`\bar X_1 -\bar X_2 =4.83-5.02=-0.19`

Now we need to find the degrees of freedom given by:

`df=n_1 +n_2 -2= 32+35-2=65`

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,65)".And we see that
`t_(\alpha/2)=2.39`

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

`-0.19-2.39\sqrt{(1.65^2)/(32)+(2.18^2)/(35)}=-1.313`

`-0.19+2.39\sqrt{(1.65^2)/(32)+(2.18^2)/(35)}=0.933`

So on this case the 98% confidence interval would be given by
`-1.313 \leq \mu_1 -\mu_2 \leq 0.933`