Statistical Calculation and Analyze the Attributes of the Data

Questions: 1. For a series of random samples of 60, are the mean values of these random samples normally distributed? Explain. 2. Calculate the standard error of the mean and explain the meaning of this value. 3. Determine the 95% confidence interval and explain its meaning in the context of the overall problem. 4. What is the probability that a sample of 60 hospitals selected at random in the Melbourne area will have a mean greater than 7000.00 admissions? 5. If the admissions times were more variable, what effect would this have on the confidence interval? Answers: 1. The given series of 60 observations are indeed drawn from a Normal Population. The conclusion about the distribution they follow has been made on the basis of the p-value that has been provided with the other descriptive statistics. For the test of Normality of a given sample of observations, the test is stated as H0: The sample is drawn from Normal Population, against H1: The sample is not drawn from Normal Population. If the p-value of the test comes out to be less than 0.05, then at 5% level of significance we conclude that the null hypothesis is false and the sample has not been drawn from Normal Population. However, in this case the p-value of the test is 0.168 which is greater than 0.05 and thus we accept H0 and conclude that the sample has been drawn from Normal Population. Thus, as the sample is drawn from Normal distribution, it is well known that their mean also follow Normal Distribution. Hence the mean values of these random samples are normally distributed. 2. Let s denote the standard deviation of the sample that has been obtained and n denote the sample size. Now if the sample is drawn from Normal Population, then an estimate of the standard error of the mean of the population is given by . Here s =6995.56 n =60 Hence standard error of the mean = =903.1229. where s is the sample standard deviation. Standard error gives us the measure of variability in the estimation of a parameter. In this case the parameter is the population mean. Hence if we repeatedly draw a sample of size 60 from the population and each time estimate the population mean by the sample mean, then the variability in the sample means obtained is approximately equal to the standard error (if the process is repeated a large no. of times). Standard error gives us the efficiency of an estimator. If an estimator is efficient in estimating the parameter, then there should be small variation and hence small standard error. 3. The 95% confidence interval of the population mean is given by [m - *tn-1;0.025, m + *tn-1;0.025] m: sample mean =6959.00 s: sample standard deviation =6995.56 n: sample size =60 tn-1,0.025: the upper 2.5% point of tn-1 distribution =2.000995 Hence we have LCL =6959.00 * 2.000995 =5151.856 UCL =6959.00 + )* * 2.000995 =8766.144 Interpretation: There is 95% chance that the actual population mean would fall in the range [5151.856, 8766.144]. 4. The mean of the sample of 60 observations follow a Normal distribution with mean =00 and standard deviation =903.1229. Let us denote the mean by X. We are thus required to find out Probability (X7000) = Probability ((X-6959)/903.1229 (7000-6959)/903.1229) = Probability (Z 0.04539803) , where Z is the standard Normal Variable (Z follows N(0,1)) = 0.481895. 5. If the admission times were more variable that is if the standard deviation was higher, then we would have LCL = m - *tn-1;0.025, which decrease as s increases, and UCL = m + *tn-1;0.025, which increase as s increases Hence, if s was increased then the confidence interval would have been wider.