Question

You research commute times to work and find that the population standard deviation is 9.3 minutes. Repeat Exercise, using the standard normal distribution with the appropriate calculations for a standard deviation that is known. Compare the results. In a random sample of eight people, the mean commute time to work was 35.5 minutes and the standard deviation was 7.2 minutes.

Answer 1

Answer:

Case $s =7.2$

$35.5-2.36\frac{7.2}{\sqrt{8}}=29.49$    

$35.5+2.36\frac{7.2}{\sqrt{8}}=41.51$    

So on this case the 95% confidence interval would be given by (29.49;41.51)    

Case $\sigma =9.3$

$35.5-1.96\frac{9.3}{\sqrt{8}}=29.06$    

$35.5+1.96\frac{9.3}{\sqrt{8}}=41.94$    

So on this case the 95% confidence interval would be given by (29.06;41.94)

And we conclude that the intervals are very similar.  

Step-by-step explanation:

If we assume that for this question we need to find a confidence interval for the population mean. We have the following procedure:

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".

$\bar X= 35.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.2 represent the sample standard deviation

n=8 represent the sample size  

Solution to the problem

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

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

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

$df=n-1=8-1=7$

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(0.025,7)".And we see that $t_{\alpha/2}=2.36$

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

$35.5-2.36\frac{7.2}{\sqrt{8}}=29.49$    

$35.5+2.36\frac{7.2}{\sqrt{8}}=41.51$    

So on this case the 95% confidence interval would be given by (29.49;41.51)  

If we assume that the real population standard deviation is $\sigma =9.3$ the confidence interval is given by:

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

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 $z_{\alpha/2}=1.96$

$35.5-1.96\frac{9.3}{\sqrt{8}}=29.06$    

$35.5+1.96\frac{9.3}{\sqrt{8}}=41.94$    

So on this case the 95% confidence interval would be given by (29.06;41.94)

And we conclude that the intervals are very similar.