Question

A random sample of size n1equals25​, taken from a normal population with a standard deviation sigma1equals6​, has a mean x overbar 1equals87. A second random sample of size n2equals33​, taken from a different normal population with a standard deviation sigma2equals4​, has a mean x overbar 2equals36. Find a 96​% confidence interval for mu1minusmu2.

Answer 1

The 96% confidence interval would be given by
`48.156 \leq \mu_1 -\mu_2 \leq 53.844`

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_1 =87` represent the sample mean 1

`\bar X_2 =36` represent the sample mean 2

n1=25 represent the sample 1 size

n2=33 represent the sample 2 size

`\sigma_1 =6` sample standard deviation for sample 1

`\sigma_2 =4` sample 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 z_(\alpha/2)\sqrt{(\sigma^2_1)/(n_1)+(\sigma^2_2)/(n_2)}` (1)

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

`\bar X_1 -\bar X_2 =87-36=51`

Since the Confidence is 0.96 or 96%, the value of
`\alpha=0.04` and
`\alpha/2 =0.02`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.02,0,1)".And we see that
`z_(\alpha/2)=2.05`

The standard error is given by the following formula:

`SE=\sqrt{(\sigma^2_1)/(n_1)+(\sigma^2_2)/(n_2)}`

And replacing we have:

`SE=\sqrt{(6^2)/(25)+(4^2)/(33)}=1.387`

Confidence interval

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

`51-2.05\sqrt{(6^2)/(25)+(4^2)/(33)}=48.156`

`51+2.05\sqrt{(6^2)/(25)+(4^2)/(33)}=53.844`

So on this case the 96% confidence interval would be given by
`48.156 \leq \mu_1 -\mu_2 \leq 53.844`