exercise:558b27e5ee: Difference between revisions
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(Created page with "<div class="d-none"><math> \newcommand{\NA}{{\rm NA}} \newcommand{\mat}[1]{{\bf#1}} \newcommand{\exref}[1]{\ref{##1}} \newcommand{\secstoprocess}{\all} \newcommand{\NA}{{\rm NA}} \newcommand{\mathds}{\mathbb}</math></div> An astronomer makes <math>n</math> measurements of the distance between Jupiter and a particular one of its moons. Experience with the instruments used leads her to believe that for the proper units the measurements will be normally distributed...") |
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An astronomer makes <math>n</math> measurements of the distance between Jupiter and a particular one of its moons. Experience with the instruments used leads | |||
Jupiter and a particular one of its moons. Experience with the instruments used leads | |||
her to believe that for the proper units the measurements will be normally | her to believe that for the proper units the measurements will be normally | ||
distributed with mean <math>d</math>, the true distance, and variance 16. She performs a | distributed with mean <math>d</math>, the true distance, and variance 16. She performs a | ||
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</math> | </math> | ||
be the average of these measurements. | be the average of these measurements. | ||
<ul><li> Show that | <ul style="list-style-type:lower-alpha"><li> Show that | ||
<math display="block"> | <math display="block"> | ||
Latest revision as of 00:32, 15 June 2024
An astronomer makes [math]n[/math] measurements of the distance between Jupiter and a particular one of its moons. Experience with the instruments used leads her to believe that for the proper units the measurements will be normally distributed with mean [math]d[/math], the true distance, and variance 16. She performs a series of [math]n[/math] measurements. Let
[[math]]
A_n = \frac {X_1 + X_2 +\cdots+ X_n}n
[[/math]]
be the average of these measurements.
- Show that
[[math]] P\left(A_n - \frac 8{\sqrt n} \leq d \leq A_n + \frac 8{\sqrt n}\right) \approx .95. [[/math]]
- When nine measurements were taken, the average of the distances turned out to be 23.2 units. Putting the observed values in (a) gives the 95 percent confidence interval for the unknown distance [math]d[/math]. Compute this interval.
- Why not say in (b) more simply that the probability is .95 that the value of [math]d[/math] lies in the computed confidence interval?
- What changes would you make in the above procedure if you wanted to compute a 99~percent confidence interval?