Question

The mean waiting time at the drive-through of a fast-food restaurant from the time the food is ordered to when it is received is 85 seconds. A manager devises a new system that he believes will decrease the wait time. He implements the new system and measures the wait time for 10 randomly sampled orders. They are provided below:
109 67 58 76 65 80 96 86 71 72
Assume the population is normally distributed.
(a) Calculate the mean and standard deviation of the wait times for the 10 orders.
(b) Construct a 99% confidence interval for the mean waiting time of the new system.

Answer 1

Answer:

a) And if we replace we got: $\bar X= 78$

$s = 15.391$

b) $78-3.25\frac{15.391}{\sqrt{10}}=62.182$    

$78-3.25\frac{15.391}{\sqrt{10}}=93.818$    

So on this case the 99% confidence interval would be given by (62.182;93.818)    

Step-by-step explanation:

Dataset given: 109 67 58 76 65 80 96 86 71 72

Part a

For this case we can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X= 78$

And the deviation is given by:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got $s = 15.391$

Part b

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:

$df=n-1=10-1=9$

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(0.005,9)".And we see that $t_{\alpha/2}=3.25$

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

$78-3.25\frac{15.391}{\sqrt{10}}=62.182$    

$78-3.25\frac{15.391}{\sqrt{10}}=93.818$    

So on this case the 99% confidence interval would be given by (62.182;93.818)