Question

A random sample of size n1= 25, taken from a normal population with a standard deviation σ1= 5, has a mean X1= 80. A second random sample of size n2=36, taken from a different normal population with a standard deviation σ2=3, has a mean X1= 75. Find a 94% confidence interval for μ1-μ2.

Answer 1

The 94% confidence interval would be given by
`2.898 \leq \mu_1 -\mu_2 \leq 7.102`

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 =80` represent the sample mean 1

`\bar X_2 =75` represent the sample mean 2

n1=25 represent the sample 1 size

n2=36 represent the sample 2 size

`\sigma_1 =5` sample standard deviation for sample 1

`\sigma_2 =3` sample standard deviation for sample 2

`\mu_1 -\mu_2` parameter of interest.

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 =80-75=5`

Since the Confidence is 0.94 or 94%, the value of
`\alpha=0.06` and
`\alpha/2 =0.03`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.03,0,1)".And we see that
`z_(\alpha/2)=1.88`

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{(5^2)/(25)+(3^2)/(36)}=1.118`

Confidence interval

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

`5-1.88\sqrt{(5^2)/(25)+(3^2)/(36)}=2.898`

`5+1.8\sqrt{(6^2)/(36)+(7^)/(49)}=7.102`

So on this case the 94% confidence interval would be given by
`2.898 \leq \mu_1 -\mu_2 \leq 7.102`