Question

According to the U.S. National Transportation Safety Board, the number of airline accidents by year from 1983 to 2006 were 23, 16, 21, 24, 34, 30, 28, 24, 26, 18, 23, 23, 36, 37, 49, 50, 51, 56, 46, 41, 54, 30, 40, and 31.


For the sample data, compute the mean and its standard error (from the standard deviation), and the median.

Answer 1

Answer:

$\bar X= 33.79167$

$s= 12.06497$

$SE= \frac{s}{\sqrt{n}}= \frac{12.06497}{\sqrt{24}}=2.463$

$Median= \frac{30+31}{2}= 30.5$

Step-by-step explanation:

For this case we have the following data:

23, 16, 21, 24, 34, 30, 28, 24, 26, 18, 23, 23, 36, 37, 49, 50, 51, 56, 46, 41, 54, 30, 40, and 31

We can calculate the sample mean with the following formula:

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

And replacing we got:

$\bar X= 33.79167$

Then we can calculate the standard deviation with this formula:

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

And replacing we got:

$s= 12.06497$

And the sample size for this case is n =24. We can calculate the standard error with this formula:

$SE= \frac{s}{\sqrt{n}}= \frac{12.06497}{\sqrt{24}}=2.463$

And the median for this case since the sample size is 24 first we need to sort the data on increasing order and we got:

16 18 21 23 23 23 24 24 26 28 30 30 31 34 36 37 40  41 46 49 50 51 54 56

And for this case the median would be the average from the position 12 and 13 and we got:

$Median= \frac{30+31}{2}= 30.5$