Discussion Questions Week 4 - Essay Example

ASSIGNMENT CENTRAL LIMITED THEOREM I would easily demonstrate the central limit theorem to my mates by using the example of coin tosses.Say I toss a coin so that every time it lands on its head I score a 1 and every time its tails, I score a 0. Now it I conducted batches of 100 experiments, and added up the total scores, the scores can theoretically vary from 0-100. Now if the experiment is repeated many, many times, I would end up with a situation as shown in the diagram below, where the horizontal axis shows the score and the vertical axis the number of times each score occurs : By the central limit theorem, by performing the experiment many times, I would end up with a normal distribution, and at very high numbers of experiments, the mean, the mode and the median scores would all tend to be 50 – or a perfect Bells curve.

The jaggedness and asymmetry of the curve would also disappear on numerous repetitions of the experiment.2. Confidence intervals apply to samples ( in the above example, the 100 toss batches) drawn from a large population ( a very large number of batches thrown) . It means the range of scores ( say 45 – 55 in the above example), when the population mean of 50 is likely to be present in a high proportion of cases. In normal statistical work an arbitrary figure, 95 % is taken as the proportion to define the confidence intervals, which is referred to as the confidence level.

The most controllable method to increase the precision (narrowing) of the confidence interval is to increase the sample size. For instance, in the current example, if the coin is thrown a 1000 times every batch, the confidence intervals could be significantly narrowed (say, 48 – 52).3. In the sample poll, as 572 likely voters said they would vote for the current mayor, my initial hunch is that it is impossible to predict from these results that the mayor would win by a majority. Even if we assume that the sample size of 1100 is normally distributed ( although in polls like this it is rarely so) and that the sample mean of 572 is a true estimation of a population which votes for the mayor by a majority, we can than then test the null hypothesis that this sample H1 is different from a larger population sample H0 where the mayor only has a 50 % chance of winning ( sample mean 550).

Let us assume (1) that the standard deviation of the samples is 11, as then, the sample mean of H1 is just within 2 standard deviations away from the population mean of 550 ( i.e. 550 + 11x2). Then the hypothesis is H0 : µ = 550 and H1 : µ > 550 as an unlikely majority for the mayor.As discussed sample mean (X) = 572 and population mean (x) = 550 and standard deviation s = 11. Do we have evidence that the sample mean is higher than the population average of 550 ? We have z = ( 572-550)/(11 x √1100) = 0.

06, where z is a z-statistic.Since z is so low, the probability that H0 is true is extremely high so we accept H0 and reject H1.Therefore it is extremely likely (95 % confidence) that the 572 likely votes in this poll does not imply that the mayor would win with a majority.Reference1 Hypothesis Testing For a Population Mean (2008) : Extracted from the internet on February 27, 2008 from : http://www.ltcconline.net/greenl/courses/201/hyptest/hypmean.htm