Question

The average time entities spend in the system in five simulation runs are: 25.2, 19.7, 23.6, 18.6, and 21.4 minutes, respectively. Five more simulations are run and the following average times in the system are obtained 22.1, 26.0, 20.2, 16.4, and 17.9 minutes. a). Build a 95% confidence interval for the mean time in the system using the first five averages collected.

Answer 1

Answer:

a) $21.7-2.776\frac{2.718}{\sqrt{5}}=18.33$    

$21.7+2.776\frac{2.718}{\sqrt{5}}=25.07$  

So on this case the 95% confidence interval would be given by (18.33;25.07)

b) $20.52-2.776\frac{3.757}{\sqrt{5}}=18.84$    

$20.52+2.776\frac{3.757}{\sqrt{5}}=22.20$    

So on this case the 95% confidence interval would be given by (18.84;22.20)

c) $21.11-2.262\frac{3.154}{\sqrt{10}}=18.85$    

$21.11+2.262\frac{3.154}{\sqrt{10}}=23.37$    

So on this case the 95% confidence interval would be given by (18.85;23.37)

And as we can see the confidence intervals are very similar for the 3 cases.

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".  

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

$\mu$ population mean (variable of interest)  

s represent the sample standard deviation  

n represent the sample size  

Part a)  Build a 95% confidence interval for the mean time in the system using the first five averages collected.

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

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

Data: 25.2, 19.7, 23.6, 18.6, and 21.4

In order to calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)    

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)    

The mean calculated for this case is $\bar X=21.7$  

The sample deviation calculated $s=2.718$  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n-1=5-1=4$  

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: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$  

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

$21.7-2.776\frac{2.718}{\sqrt{5}}=18.33$    

$21.7+2.776\frac{2.718}{\sqrt{5}}=25.07$  

So on this case the 95% confidence interval would be given by (18.33;25.07)

Part b: Build a 95% confidence interval for the mean time in the system using the second set of five averages collected.

Data: 22.1, 26.0, 20.2, 16.4, and 17.9

The mean calculated for this case is $\bar X=20.52$  

The sample deviation calculated $s=3.757$  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n-1=5-1=4$  

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: "=-T.INV(0.025,4)".And we see that $t_{\alpha/2}=2.776$  

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

$20.52-2.776\frac{3.757}{\sqrt{5}}=18.84$    

$20.52+2.776\frac{3.757}{\sqrt{5}}=22.20$    

So on this case the 95% confidence interval would be given by (18.84;22.20)

Part c: Build a 95% confidence interval for the mean time in the system using all ten averages collected.

Data: 25.2, 19.7, 23.6, 18.6, 21.4, 22.1, 26.0, 20.2, 16.4, and 17.9

The mean calculated for this case is $\bar X=21.11$  

The sample deviation calculated $s=3.154$  

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:  

$df=n-1=10-1=9$  

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: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.262$  

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

$21.11-2.262\frac{3.154}{\sqrt{10}}=18.85$    

$21.11+2.262\frac{3.154}{\sqrt{10}}=23.37$    

So on this case the 95% confidence interval would be given by (18.85;23.37)

And as we can see the confidence intervals are very similar for the 3 cases.