Question

The sick days of employees every two years in a company are normally distributed with a population standard deviation of 7 days and an unknown population mean. If a random sample of 20 employees is taken and results in a sample mean of 21 days, find a 95% confidence interval for the population mean. Round your answer to TWO decimal places. z0.10 z0.05 z0.025 z0.01 z0.005 1.282 1.645 1.960 2.326 2.576 You may use a calculator or the common z values above.

Answer 1

Answer:

The 95% confidence interval for the population mean is between 17.93 days and 24.07 days.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$.

So it is z with a pvalue of $1-0.025 = 0.975$, so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96\frac{7}{\sqrt{20}} = 3.07$

The lower end of the interval is the sample mean subtracted by M. So it is 21 - 3.07 = 17.93 days

The upper end of the interval is the sample mean added to M. So it is 21 + 3.07 = 24.07 days

The 95% confidence interval for the population mean is between 17.93 days and 24.07 days.

Answer 2

Answer:

$21-1.96\frac{7}{\sqrt{20}}=17.93$    

$21+1.96\frac{7}{\sqrt{20}}=24.07$    

So on this case the 95% confidence interval would be given by (17.93;24.07)    

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".

$\bar X=21$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma=7$ represent the population standard deviation

n=20 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

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

$21-1.96\frac{7}{\sqrt{20}}=17.93$    

$21+1.96\frac{7}{\sqrt{20}}=24.07$    

So on this case the 95% confidence interval would be given by (17.93;24.07)