Question

A high-tech company wants to estimate the mean number of years of college education its employees have completed. A sample of 15 employees had a mean of 4 years with a standard deviation of .7 years. Find a 95% confidence interval for the true mean.

Answer 1

Answer:

$4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61$

$4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39$

The 95% confidence interval is given by (3.61;4.39)

Step-by-step explanation:

Notation and definitions

n=15 represent the sample size

$\bar X=4$ represent the sample mean  

$s=0.7$ represent the sample standard deviation

m represent the margin of error

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".  

Calculate the critical value tc

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. The degrees of freedom are given by:  

$df=n-1=15-1=14$  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.025,14)" for $t_{\alpha/2}=-2.14$  

"=T.INV(1-0.025,14)" for $t_{1-\alpha/2}=2.14$  

The critical value $tc=\pm 2.14$

Calculate the margin of error (m)

The margin of error for the sample mean is given by this formula:

$m=t_c \frac{s}{\sqrt{n}}$

$m=2.14 \frac{0.7}{\sqrt{15}}=0.387$

Calculate the confidence interval  

The interval for the mean is given by this formula:

$\bar X \pm t_{c} \frac{s}{\sqrt{n}}$

And calculating the limits we got:

$4 - 2.14 \frac{0.7}{\sqrt{15}}=3.61$

$4 + 2.14 \frac{0.7}{\sqrt{15}}=4.39$

The 95% confidence interval is given by (3.61;4.39)