20 Most Recent Institute of Mathematics and Statistics Nonlinear Regression Analysis - DataFitting Questions & Answers

The median of this distribution is found as follows. The number of data elements is even, so there is no (actual) middle element. We will need a virtual one. The two elements in the middle are the 5th and 6th elements. Between the two of them we insert a (virtual) element which will be called the median. Its value is the average value of the the 5th and 6th elements. 1,2,3,4,5,5.5,6,7,8,9,10

First you need the Std Error of the mean value, a measure of the dispersion of that mean value. SE = sample std deviation / sqrt (sample size) = 100 / sqrt (64) = 100 / 8 = 12.5 Then we use a figure for the number of std errors either side of the mean value, which make up a 95 % confidence interval. This is ± 1.96 std errors, from tables of the Normal Distribution. So the confidence interval is 350 ± 1.96 * 12.5 or 374.5 to 325.5 .

First you need the Std Error of the mean value, a measure of the dispersion of that mean value. SE = sample std deviation / sqrt (sample size) = 8 / ? 64 = 1 Then we use a figure for the number of std errors either side of the mean value, which make up a 99 % confidence interval. This is ± 2.58 std errors, from tables of the Normal Distribution. So the confidence interval is 125 ± 2.58 * 1 or 127.58 to 122.42 .

Ray is a bit off track, there are indeed a few mathematicians around Fixya. Here you are dealing with the standard error of a proportion. The proportion of students of interest in the sample is 540/970 = 0.5567. This is usually denoted by p. Now you want the standard error of this proportion, s. It is SQRT ( p (1-p) / N ) where N is sample size. Note the error is smaller for big samples. Here s = 0.0160. Now we turn to the Normal Distribution. We want the Z-statistic which corresponds to a confidence level of 92%, or 0.92. Z is the number of Standard Deviations on the Normal Curve which takes in 92% of the area under the curve. You can look this up in any table of the Normal Distribution, or with an online calculator http://onlinestatbook.com/2/calculators/normal.html This gives Z = 1.751. So the true result is somewhere in p ± (1.751 * 0.0160) or 0.5567 ± 0.0279. The margin of error is 0.0279. Now I don't know here if your tutor wants you to use a Continuity Correction, which is sometimes applied because we are approximating the Normal Distribution here. If so this is 0.5 / N = 0.5 / 970 = 0.0005. This small amount is added to the Margin of Error, to make 0.0284. A good reference for you to use is http://onlinestatbook.com/2/estimation/proportion_ci.html http://onlinestatbook.com/2/index.html

No. At the 0.05 level you cannot conclude that this sample is significantly different.

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