UTSA CS 4953 Experimentation in Computer Science Spring 2003

CS 4953 Experimentation in Computer Science Assignment 3 Comments: Spring 2003

This was graded on a basis of 3 points each for a total of 78 points. Fewer points were deducted if you made the same error on tow questions. There are different ways of calculating percentiles for sample data. The method given in the notes is one of the standard ways, but there are others. The formula given there assumes the sample points are indexed starting at 1 so if you indexed the points at 0, you must adjust. The formula given for the median given in the same table for the case of an odd nuber of points is incorrect. The subscript should be (n+1)/2. Part 1: 33 data points

By the formula from the notes, the 10th percentile should be at position floor(1+.1(32)) = 4 and the 90th percentile should be at position floor(1+.9(32)) = 29. These are the values 14 and 36.
The range should be specified by giving the min and max, not just the difference between the two.
For the SIQR, the 25th percentile should be at position floor(1+.25(32)) = 9 and the 75th percentile should be at position floor(1+.75(32)) = 25. These values are 21 and 34, giving SIQR = 6.5.
Do not use the normal distribution to calculate the SIQR of sample data.
The 90% confidence interval for the mean is (24.19,29.63).
The 99% confidence interval for the mean is (22.65,31.17).
To find the probability that the mean is less than 25, this corresponds to a z value of -1.2145. Don't forget the sqrt(n) in the denominator when converting to a standard normal distribution. P(z > 1.2145) = .113

Part 2: 10 data points

The 10th percentile is at position 1 and the 90th percentile is at position 9. These values are 14 and 42.
The 25th percentile is at position 3 and the 75th percentile is at position 7 giving values 23 and 33 for a SIQR of 5.
For the 90% confidence interval, use a t-distribution with 9 degrees of freedom, and a t-value of 1.833. Note that this comes from the .95 column of the table I gave you. Don't forget the sqrt(n) in the denominator. The confidence interval is (24,19,29.63).
For the 99% confidence interval, use a t-distribution with 9 degrees of freedom, and a t-value of 3.25. Note that this comes from the .995 column of the table I gave you. Don't forget the sqrt(n) in the denominator. The confidence interval is (22.65,31.17).
To find the probability that the mean is less than 25, you use a t-distribution. The corresponding t value is (25-29)/((10.41)/sqrt(10)) = -1.24. You need to find P(t < -1.215). You would find this from the table I gave you by taking 1 - P(t < 1.215). The best you can do from the table (using 9 degrees of freedom) is P(t < 1.383) = .9, giving an answer of 1 - .9 = .1. A better result can be obtained from a t-distribution calculator such as this one. This gives and answer of .1276.