Question

There are 9 observatories tracking a near-Earth asteroid, and each independently estimates how close it will come to Earth next year, in millions of miles, with the results. 481, .439,448,446,618, 411, 250, 604, 414. Previous experience suggests these are unbiased estimates of the true distance d, with a standard deviation of 15. a) We estimate the true mean of these distance estimates by d = i. b) Assuming this is a large enough sample, write down a 94% confidence interval for the true distance d.

Answer 1

Answer:

94% Confidence interval:  (0.3771 ,0.5365)

Step-by-step explanation:

We are given the following data set:

0.481, 0.439,0.448,0.446,0.618, 0.411, 0.250, 0.604, 0.414

a) True mean of these distance estimates by d

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{4.111}{9} = 0.4568$

b)

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$  

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.  

Sum of squares of differences = 0.0954

$S.D = \sqrt{\frac{0.0954}{8}} = 0.1092$

94% Confidence interval:  

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$  

Putting the values, we get,  

$t_{critical}\text{ at degree of freedom 8 and}~\alpha_{0.06} = \pm 2.1891$  

$0.4568 \pm 2.1891(\frac{0.1092}{\sqrt{9}} ) = 0.4568 \pm 0.0797 = (0.3771 ,0.5365)$