Question

In 2003, the Accreditation Council for Graduate Medical Education (ACGME) implemented new rules limiting work hours for all residents. A key component of these rules is that residents should work no more than 80 hours per week. The following is the number of weekly hours worked in 2016 by a sample of residents at the Tidelands Medical Center. 78 85 82 79 77 82 81 82 81 81 82 78
a. What is the point estimate of the population mean?
b. Develop a 99% confidence interval for the population mean.
c. Is the company’s claim that the "typical customer" spends $80 per month reasonable? Yes/No

Answer 1

Answer:

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

The mean calculated for this case is $\bar X=80.67$

b) The sample deviation calculated $s=2.270$

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

$df=n-1=12-1=11$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,11)".And we see that $t_{\alpha/2}=3.106$

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

$80.67-3.106\frac{2.270}{\sqrt{12}}=78.635$    

$80.67+3.106\frac{2.270}{\sqrt{12}}=82.705$    

So on this case the 99% confidence interval would be given by (78.635;82.705)    

c) Yes since the confidence interval calculated include the value of 80 for a possible value for the true mean of interest

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

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 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 $\bar X=80.67$

Part b

The sample deviation calculated $s=2.270$

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

$df=n-1=12-1=11$

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,11)".And we see that $t_{\alpha/2}=3.106$

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

$80.67-3.106\frac{2.270}{\sqrt{12}}=78.635$    

$80.67+3.106\frac{2.270}{\sqrt{12}}=82.705$    

So on this case the 99% confidence interval would be given by (78.635;82.705)    

Part c

Yes since the confidence interval calculated include the value of 80 for a possible value for the true mean of interest