Question

Problem 3: Migrating Snowy Owls (16 points) Snowy Owls are an Artic species that are sometimes observed at Northeastern Marine Center on Nahant in the winter. Bird banding data has measured their migration distances (in km) and found the following data for a sample of 20 owls: ## 317 323 379 398 442 456 517 539 563 573 584 609 646 731 735 770 801 833 944 956 Calculate the mean, standard deviation, standard error and 95% confidence intervals on migra- tion distance. Show your work by setting up a table where each row is an observation, the first column is the data and the second colum is the squared deviation from the mean. 4 points a)

B) Draw an approximate boxplot for the data, where the whiskers of the boxplot extend the range of the data. Your y-axis should span from 0 to 1200.

Answer 1

a)

Here, mean is computed using the formula:

$\dpi{80} Mean = \frac{317+323+......+944+956}{20}={\color{Red} 605.8}$

And Squared Deviation from Mean, (eg. for 1st observation)

Squared Deviation = (317 - 605.8)² = 83405.4

Summing the squared deviations and dividing the sum by ( No. of observation - 1) = (20-1) = 19, we obtain the variance.

$\dpi{80} Variance = \frac{(317-605.8)^{2}+(323-605.8)^{2}+......(956-605.8)^{2}}{19}$

  $\dpi{80} =\frac{710699.2}{19}=37405.2$

And Standard deviation is nothing but the square root of the variance.

Standard deviation = $\dpi{80} \sqrt{37405.2}={\color{Red} 193.4}$

Standard error is computed using the formula:

Standard error = Standard deviation / $\dpi{80} \sqrt{n}$

  $\dpi{80} =\frac{193.4}{\sqrt{19}}= {\color{Red} 44.37}$

b)

where the upper whisker corresponds to the value 956 and the lower whisker to the value 317.

Bar plot:

d) Comparing boxplot and barplot, the box plot reveals the median, quartiles and the range of the values in the whole dataset, whereas in bar plot we have plotted the mean and its range of error of 1 SE.

The data appears to be slightly positively skewed.The 95% confidence interval for mean is given by:

$\dpi{80} 605.8\mp 2.58*44.37=(491.3,720.3)$

The mean is supposed to lie within this interval for about 99% of the time.