Question

2. A computer software package calculated some summaries of sam- ple data. The results are below. Please fill in the blanks and estimate the mean of the population from which the sample was drawn.

Variable N Mean SE of Mean Std Dev Variance Sum of Squares

x ? ? 2.05 10.25 ? 3761.70

3. EXTRA CREDIT (5 points) Suppose that two independent ran- dom samples of size n1 and n2 from two normal distributions are available. Ex- plain how you would estimate the standard error of the difference in sample mean with the bootstrap method (you can write a pseudo code).

4. EXTRA CREDIT (5 points). Assuming your helicopter data repre- sents the prior distribution, compute a Bayesian estimator for the mean by taking another 5 observations as in step 2 and updating your estimate using the new mean (from your new observations). Use the variance of the 10 previous and 5 subsequent observations as the true variance of the population.

Answer 1

Q 2)

Variable	N	Mean	SE of mean	St. dev	Variance	Sum of Squares
X	?	?	2.05	10.25	?	3761.70

The SE of mean = 10.25

$n =\left ( \frac{10.25}{2.05} \right )^2 = 25$

$Variance (\sigma ^{2}) = (10.25)^2 = 105.0625$

$\sigma ^{2} = \frac{1}{n-1}\left [ \sum X^2 -\frac{(\sum X)^2}{n} \right ]$

$105.0625 = \frac{1}{24}\left [ 3761.70 -\frac{(\sum X)^2}{25} \right ]$

$\Rightarrow 105.0625*24 = \left [ 3761.70 -\frac{(\sum X)^2}{25} \right ]$

$\Rightarrow 2521.5-3761.70 =-\frac{(\sum X)^2}{25}$

$\Rightarrow -1240.2 =-\frac{(\sum X)^2}{25}$

$\therefore (\sum X)^2 =1240.2*25 =31005$

$\sum X=\sqrt{31005} =176.0824$

Mean

$\overline{X} =\frac{1}{n}\sum X=\frac{176.0824}{25}=7.0433$

Answer: n = 25

Mean = 7.0433

Variance =105.0625