Lecture 3 Research Methods n Lecture 10 Additional Main Reading Business Statistics by Sonia Taylor 2001, Palgrave, New York n Statistics for Economics, Accounting and Business by Mike Barrow, 3rd Edition, Prentice Hall. n Topics: The Normal Distribution Confidence Interval 2 The Normal Distribution n n n n n Mean and Variance n “Bell shaped” Symmetrical Mean, median and mode are equal Asymptotic X is normally distributed Mean Median Mode f(X) n m X n n Mean X = 10 Variance X = 4 X~N (10, 4) Mean Y = 50 Standard Deviation Y = 3 Y~N (50, 9) n n 3 How to Find Probabilities? V(3X) = 32.V(X) = 9 V(X) V(3+X) = V(3) + V(X) = V(X) V(X+Y) = V(X) + V(Y) 4 n pre-calculated But there are different size of normal distribution because means and variances differ \ use the standardise normal distribution n Standardise using: Z = n f(X) d E(4X)= 4 E(X) E(4+X) = 4 + E(X) E(X+Y) = E(X) + E(Y) § Areas are difficult to estimate, \ need to be P (c £ X £ d ) c n hX The Standardised Normal Distribution Probability is the area under the curve! P ( X £ c) n Mean of X = Expected value of X: E(X) = X = Variance of X = V(X) = s X X n n P( X ³ d ) 5 n X -hX sX X~N (0, 1) Proof!! Hence only one table is needed 6 Dr N. Gooroochurn, Nov 03 Lecture 3 The Standardised Normal Distribution The Standardised Normal Distribution Table A Table B P (X ³ Z) P(0 £ X £ Z) 7 8 X is the weight of students and follow a normal distribution with mean 56 kg and variance 81. Find P(X £ 60). If mean is 5 and the standard deviation is 10. Find P(X ³ 6.2). Normal Distribution (i) Standardise the data and find the value of Z. s = 10 (i) Z= X -h X sX 60 - 56 = = 0.44 9 s = 6.2 - 5 = 0.12 10 Standardised Normal Distribution Normal Distribution Standardised Normal Distribution X 60 X -m Z= (ii) From the Z table find the probability value. m = 56 Example 2a s = 10 sZ =1 sZ =1 (ii) From the table A when z= 0.44, p = 0.3300 \ P(X £ 60) = 1-0.33 = 0.67 From table B when z = 0.44, p = 0.1700 hZ = 0 \ P(X £ 60) = 0.50+0.17 = 0.67 0.44 9 Z Find P (2.9 £ X £ 7.1) Z= s 2.9 - 5 = = -.21 10 Z = 6.2 X P(X ³ 6.2) = 0.4522 mZ = 0 0.12 Z 10 Sampling Distributions: Some properties Example 2b X -m m =5 X -m s = 7.1 - 5 = .21 10 Unbiasedness Unbiased (h = h x) Biased (h = h x) Standardised Normal Distribution Normal Distribution s = 10 .0832 sZ =1 .0832 2.9 m =5 7.1 X -0.21 P (2.9 £ X £ 7.1) = 0.1664 mZ = 0 0.21 Z 11 h hx X 12 Dr N. Gooroochurn, Nov 03 Lecture 3 Variability Effect of Large Sample Larger sample size Low Variance High Variance Smaller sample size m m X 13 Central Limit Theorem How Large is Large Enough? … the sampling distribution becomes almost normal regardless of shape of population As sample size gets large enough… n For most distributions, n>30 n For fairly symmetric distributions, n>15 n For normal distribution, the sampling distribution of the mean is always normally distributed X 15 16 Estimation Process Population Random Sample Mean, h, is unknown Mean X = 50 Point Estimates I am 95% confident that h is between 40 & 60. Point Estimate is a single value that is obtained from sample data and is used as the best guess of the corresponding population parameter Population Mean: Sample Interval Estimate Point Estimate 14 17 Sample hX = X Variance: s 2 = s 2 Standard Deviation: s = s 18 Dr N. Gooroochurn, Nov 03 Lecture 3 Interval Estimates n n Confidence Interval Estimates It is an interval centered on the point estimate within which we expect the point estimate to lie. Provides range of values n n n n n Take into consideration variation in sample statistics from sample to sample Based on observation from 1 sample Give information about closeness to unknown population parameters Stated in terms of level of confidence n Never 100% sure Referred to as ‘Confidence Interval’ Mean s Known 19 Standard Error of X (SE X ) SE: Shows the variation of h across different samples n s: shows the variation of the observations (X’s) s n s known: SE X = n s2 s = s unknown: SE X = n n n = n n n n n n Confidence interval estimate for h: 21 22 Interpretation of Confidence Interval X - z.SE X £ h £ X + z.SE X n Population SD (s) in known Population is normally distributed If population is not normal, sample is large We know the level of confidence X - z.SE X £ h £ X + z.SE X Elements of Confidence Interval n 20 Assumptions n s n s Unknown Differences: 1. Calculating the standard error of the mean 2. Whether to use normal distribution or student tdistribution within one sample SE depends on whether population standard deviation is known or not 2 Proportion Confidence Interval for h (s known) n n Confidence Intervals n n Level of confidence: z n Confidence in which the interval will contain the unknown population parameter. n High confidence, larger z and wider the range Precision (range): SE n Closeness to the unknown parameter n Large SE, wider range Sample size: n n Smaller sample, less precise, wider range n n Say 95% Confidence Interval It does not mean that there is a 95% chance that the true mean will lie between the range Probability is for random variables and not for parameters such as h If we take many samples (same size) from a population and calculate CI for each, h will lie within 95% of the calculated intervals 2.5%=0.025 95% 2.5%=0.025 (0.95) X - z.SE X 23 X X + z.SEX What value of z will give probability of 0.025? Look in table A and the answer is 1.96 24 Dr N. Gooroochurn, Nov 03 Lecture 3 Degrees of Freedom (df ) Confidence Interval for h (s unknown) n n n Assumptions n Population standard deviation is unknown Use Student’s t Distribution Confidence Interval Estimate n n Number of observations that are free to vary after sample mean has been calculated Example n X - t (a , n -1).SE X £ h £ X + t (a , n -1).SE X 2 2 X 1 = 1 (or any number) X 2 = 2 (or any number) X 3 = 3 (cannot vary) Degrees of Freedom Probability 95%: a= 0.05, a/2 = 0.025 n Mean of 3 numbers is 2 But as sample gets large, t-distribution is the same as normal distribution and t = z degrees of freedom = n -1 = 3 -1 =2 25 26 V is degrees of freedom Student’s t Distribution a = a/2 Standard Normal Bell-Shaped Symmetric ‘Fatter’ Tails t (df = 13) t (df = 5) 0 Z t 27 Learning Outcomes Example A random sample of n = 25 has X = 50 and S = 8. Set up a 95% confidence interval estimate for m S S £ m £ X + ta /2,n-1 n n 8 8 2.064 50 - 2.0639 £ m £ 50 + 2.0639 2.064 25 25 46.69 £ m £ 53.30 X - ta /2,n-1 29 n You should be able to understand n The normal distribution n Point estimate n Confidence interval n the Student’s t distribution 30 Dr N. Gooroochurn, Nov 03