I. The Sampling Distribution of ProportionsII. The Sampling Distribution of MeansSample of 50Sample of 200III. Confidence Interval for the Population Mean with a Known Population VarianceIV. Confidence Interval for a Population Mean with an Unknown Population VarianceIt is not realistic to presume we do not know  but know , but if that is the case, then we can use the formula above. If we do not know , then we use the standard deviation of the sample, s, as an estimate of , whereV. Exercise: Due in One WeekOct. 8, 2003 LAB #3 ECON 240A-1 L. PhillipsSampling Distributions I. The Sampling Distribution of ProportionsAs an example, consider random samples of size 50 potential voters who are asked whether they will vote against Governor Davis( yes for recall), yes coded one, or for Governor Davis i.e. no on recall (or not yes) coded zero. For each sample, the sample proportion, pˆ, is calculated where pˆ= k/n, i.e. equals the fraction saying yes out of fifty. As we discussed in Lecture Four, these are repeated Bernoulli trials, and k should have the binomial distribution with mean np and variance np(1 – p). So, pˆhas mean p and variance [p (1 – p)]/n. If p is approximately 0.5, then np easily satisfies np5, as does n(1 – p), and we could use the normal distribution to approximate the binomial. SopˆN[p, {p (1 – p)}/n].Using a value of p equals 0.5, for example, we can plot this theoretical normal density function for the sample proportion pˆ, N(0.5, 0.005), i.e normal with mean 0.5, and variance 0.005 (equivalent to a standard deviation of 0.07071. Use Excel and in the first column, type in values for the sample proportion, ranging from zero to one with increments of 0.01. Then, in the adjacent column, use normdist to calculate the density function. Plot this density against the sample proportion. We can also use the random number generator feature to simulate the distribution of the sample proportion. Go to the “tools” menu, “data analysis”, and scroll down to “random number generation”. Hit OK and in the dialog box, use “10” for the number of variables entry. This is the number of samples. Use “50” for the number of random numbers entry. This is the sample size. For the distribution choose Bernoulli. In each column (i.e. for each sample), this will generate 50 numbers of zero or one, just like a random sample of 50 voters responding yes or no. For the parameter choose “ p = 0.5”. Note, you could repeat this simulation presuming p = 0.6 or 0.4, for the population parameter. Choose a random seed entry of “10”, so we get the same random numbers, i.e.we have replication from experimenter to experimenter and are all on the same page. Check “new worksheet ply” for the results.To calculate the sample proportion for each column, select cell A51, go to the menu bar and type “=”, insert function, and choose mean with an input range of A1-A50. Extend across all ten columns to obtain all ten sample proportions. To obtain a histogram,Oct. 8, 2003 LAB #3 ECON 240A-2 L. PhillipsSampling Distributions in cell A52, type in 0.05 for the upper limit of the first bin and go across in increments of 0.05. You can plot a histogram of these sample proportions. Go to the “tools” menu, “dataanalysis”, and “histogram”, and hit OK. For the input data range, type A51-J51, and for the bin range, type A52-T52. Check “new worksheet ply” and “chart output” and hit OK.In cell M53 type in “mean” for the average of the sample proportion distribution, and in cell N53 type in “std dev”. Select cell M54, go to the menu bar, type “=”, insert “function”, select “average”, and for the input range use A51-J51.Hit OK. The mean of this simulated sample of sample proportions is 0.444, compared to the theoretical value of 0.5. Follow a similar procedure, selecting cell N54 to get stdev, with a calculated valueof 0.066533, compared to the theoretical value of 0.07071. Repeat this process but for 50 simulated samples of 50 voters, i.e. 50 columns of zeros and ones. For the 50 simulated sample proportions, I calculate a mean of 0.4876, closer to 0.5, and a standard deviation of 0.07493.Repeat this process but for 200 simulated samples of 50 voters. For the 200 simulated sample proportions, I calculate a mean of 0.4861, and a standard deviation of 0.0677.------------------------------------------------------------------------------------Table: Average Sample Proportion for 50 Voters Versus Replications 10 Replications 50 Replications 200 Replications PopulationAv. Sample Proportion0.444 0.4876 0.4861 0.5Standard Deviation0.0655 0.07493 0.06767 0.07071---------------------------------------------------------------------------------The histogram of sample proportions becomes bell shaped by 50 replications, and there is not a great deal of difference as we go from 50 replications to 200 replications. The average sample proportion approaches the population proportion of 0.5 by 50 replications as well. So the simulation was working pretty well, i.e. according to theory, by 50 replications. Note that we never varied the number of voters, 50, in a sample. Increasing this would decrease the expected standard deviation for the sample proportion,since it varies with 1/n .Oct. 8, 2003 LAB #3 ECON 240A-3 L. PhillipsSampling Distributions In the ten replications, the minimum sample proportion was 0.32, with a standard deviation of 50/]68.0)(32.0[( equal to 0.066, so the 95 % confidence interval for the population proportion is 0.32 +/- 1.96*0.066, or P[0.1945.0 p] = 0.95, and does not include the true population proportion of 0.5. So there are occasions when a 95% confidence interval on the population proportion will not include it and, on average this occurs about 5% of the time. The other nine sample proportions have 95% confidence intervals that include the population proportion.II. The Sampling Distribution of MeansWe will take a sample of 50 observations from the uniform distribution, calculate the sample mean, and replicate this 50 times to show that the distribution of the 50 sample means is not flat but bell-shaped. In this case, x is uniform with mean 0.5 and variance 1/12, as we saw in Lecture Five. The sample mean, x, has mean 0.5 and variance 1/12 divided by 50, 0.00167, with a corresponding standard deviation of 0.0408.E( x) = {501)]([ ixE}/50 = 0.5,VAR(x) = VAR {{501)]([ ixE}/50 }= (1/n2) 501)]([ ixVAR=