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.

Respuesta :

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 [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.95}{2} = 0.025[/tex]

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

So it is z with a pvalue of [tex]1-0.025 = 0.975[/tex], so [tex]z = 1.96[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 1.96\frac{7}{\sqrt{20}} = 3.07[/tex]

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:

[tex]21-1.96\frac{7}{\sqrt{20}}=17.93[/tex]    

[tex]21+1.96\frac{7}{\sqrt{20}}=24.07[/tex]    

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".

[tex]\bar X=21[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=7[/tex] 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:

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

Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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 [tex]z_{\alpha/2}=1.96[/tex]

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

[tex]21-1.96\frac{7}{\sqrt{20}}=17.93[/tex]    

[tex]21+1.96\frac{7}{\sqrt{20}}=24.07[/tex]    

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