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

To calculate the 94% confidence interval for the true mean distance of an asteroid, one needs to compute the sample mean, standard error, and then apply the appropriate Z-score to determine the range.

Step-by-step explanation:

The student's question is related to calculating a confidence interval for the true mean distance of a near-Earth asteroid. With the given unbiased estimates of the distances and a standard deviation of 15, we will calculate the 94% confidence interval for the true distance d.

Firstly, calculate the sample mean X by adding up all the given distance estimates and dividing by the number of observatories, which is 9.

Next, find the standard error (SE) of the mean by dividing the standard deviation by the square root of the number of observatories (n).

To determine the 94% confidence interval, we'll use the Z-score associated with it (approximately 1.88) and the standard error.

The confidence interval formula is X ± (Z * SE). We calculate the upper and lower bounds of the interval accordingly.

By following these steps, the student can establish the 94% confidence interval, giving them a range within which the true mean distance of the asteroid is likely to fall.

Answer 2

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(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{(0.0954)/(8)} = 0.1092`

94% Confidence interval:

`\bar{x} \pm t_(critical)\displaystyle(s)/(√(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((0.1092)/(√(9)) ) = 0.4568 \pm 0.0797 = (0.3771 ,0.5365)`