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

Answer:

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

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".  

$\bar X_1 =80$ 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:  

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2}\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$ (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:

$SE=\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}$

And replacing we have:

$SE=\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=1.118$

Confidence interval

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

$5-1.88\sqrt{\frac{5^2}{25}+\frac{3^2}{36}}=2.898$  

$5+1.8\sqrt{\frac{6^2}{36}+\frac{7^}{49}}=7.102$  

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