Exam 02 - M316 ELEMENTARY STATISTICS Second Midterm Unique...

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M316 ELEMENTARY STATISTICS Unique number 58610 Second Midterm April 8th 2008 NAME .... SKETCHY S. OLUTION ................ - Are authorized: calculators, ”Tables and Formulas”. - No other document (book, hand written notes, etc ) is authorized. - Feel free to ask for scratch paper if necessary. Good Luck. PART I. Each question is worth 5 points. In an assessment process several measurements are made on each photographic plate to determine the distance between two points. For one plate we obtained the following four measurements:1.221, 1.220, 1.205, 1.214. Assume that the population of measurements has a normal distribution. The mean of the population of all measurements is the unknown distance between the two points that we are measuring. 1) Why is it reasonable or not to consider the population of measurements to have a normal distribution? Why might the standard deviation be known? We are dealing with measurements in physics. These are well known to follow a normal distribution. The size of the sample is not relevant. True, X n sample mean is normal when n is large enough, but here we are interested in the population of measurements themselves and the distribution of the random variable X (measurement). See the document “When is a variable normal?” Only a handful of students understood the problem. Being able to ‘feel’ when a variable is (or is not) normal is one of the most important things for the statistician apprentice. It’s way more important than remembering a formula or a specific computation. A few more were able to give a reasonable answer for the standard deviation: if the measurements are made with some tool, the standard deviation is a known constant for the tool. Other possibility: the experience of a large number of measurements showed that the standard deviation of the measurements is 0.01 inches. To avoid a systematic loss of 5 points for most of the students, I gave 1 or 2 points 1
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for inappropriate but somewhat meaningful answers and gave 6 points for the 3 following computational questions usually well understood. 2) Determine a 95% confidence interval for the distance between the two points, if the standard deviation for the population of measurements is known to be 0.01 inches. x = 1 . 215 Table C, 95% , z * = 1 . 960 , m = z * σ/ n = 1 . 960 × 0 . 01 / 4 = 0 . 0098 . 1 . 215 ± 0 . 0098 or [1 . 205; 1 . 225]. 3) If we specify the length of a 95% confidence interval to be 0.015 inches, how many measurements should be made on each plate? (Be careful! what is the relation between the length of the interval and the margin of error?) length l = 0 . 0015 , m = 1 / 2 l = 0 . 0075 = z * σ/ n n = ( z * σ/m ) 2 = ( 1 . 96 × 0 . 01 / 0 . 0075 ) 2 n = 7 n is the smallest integer larger than ( 1 . 96 × 0 . 01 / 0 . 0075 ) 2 4) By using the above four measurements, determine a 95% confidence interval for the distance between the two points, if the standard deviation is not known.
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