<p>10.1. Population Mean: Large Sample Case ( n  30 ) Motivating example: In the survey conducted by CJW, Inc., a mail-order firm, the satisfaction scores (1~100) of 100 customers (n=100) are obtained. Suppose   20 is known. Also,  20 x  82,  x    2 . n 100 Derivation of 95% confidence interval: Since 2 . X  N(, X ) Thus, X    Z (~ N(0,1)) .  X</p><p>Then,  X       X  0.95  PZ  1.96  P  1.96  P  1.96        X   X     X   P1.96   1.96  P 1.96    X  1.96    X X    X </p><p> PX 1.96 X    X 1.96 X   There is an approximate 95% chance that the population mean  will fall between and , i.e., X 1.96 X X 1.96 X</p><p>,  falls in the interval Pu X 1.96 X , X 1.96 X   0.95</p><p>X 1.96 X , X 1.96 X  with a chance close to 0.95.</p><p>Note: in the above equation,  X    0.95  P  1.96  P X    1.96     X  X </p><p>1 There is an approximate 95% chance that the sample mean will provide a sampling error of 1.96 X .</p><p>Example (continue)</p><p>In the above example, since  X  2 , thus P(X 1.96 * 2    X 1.96* 2)  PX  3.92    X  3.92  0.95 There is an approximate 95% chance that the population mean  will fall between X  3.92 and X  3.92 . </p><p>95% confidence interval: Suppose the sample size is large.  As  is known,      x 1.96 X  x 1.96  x 1.96 , x 1.96  n  n n  is a 95% confidence interval estimate of the population mean  .  As  is unknown, s  s s  x 1.96sX  x 1.96  x 1.96 , x 1.96  n  n n  is a 95% confidence interval estimate of the population mean  ,</p><p>2 s  where s is the sample variance and X is the estimate of X . Example (continue)</p><p>In the above example, since x  82 ,  X  2 , and n  100 , thus  20 x 1.96  82 1.96  82  3.92  78.08, 85.92 n 100 is a 95% confidence interval estimate of the population mean  . General confidence interval:</p><p>Definition of z : 2 Let Z be the standard normal random variable. Then,    P Z  z   .  2  2 2 As   0.5 ,   P Z  z   1 .  2 </p><p> z z 2 2 Example:   PZ  1.64  0.9  1 0.1  P Z  z0.1   1 0.1  z0.1  z0.05  1.64  2  2   PZ  1.96  0.95  1 0.05  P Z  z0.05   1 0.05  z0.05  z0.025  1.96  2  2   PZ  2.576  0.99  1 0.01  P Z  z0.01   1 0.01  z0.01  z0.005  2.576  2  2 In summary,  1  z 2 2</p><p>0.1 0.9 0.05 z0.05  1.64</p><p>0.05 0.95 0.025 z0.025  1.96</p><p>0.01 0.99 0.005 z0.005  2.576 Derivation of 1 100% confidence interval: As the sample size is large, </p><p>3        X       X  1  P Z  z   P  z  P  z  2    2    2   X   X       X     P  z   z  P z  X    X  z  X   2 2   2 2    X     P X  z  X    X  z  X   2 2   There is an approximate 1 100% chance that the population mean  will fall between X  z  X and X  z  X , i.e., 2 2     ,  falls in the interval Pu  X  z  X , X  z  X  1   2 2 </p><p>  X  z  X , X  z  X with a chance close to 1 .  2 2  Note: as   0.05 , the above derivations are exactly the same as the ones for 95% confidence interval estimate. </p><p>Motivating Example (continue) As   0.1, </p><p>P(X  z0.1 * X    X  z0.1 * X )  PX  z0.05 *2    X  z0.05 *2 2 2  PX  3.28    X  3.28 1 0.1  0.9</p><p>There is an approximate 90% chance that the population mean  will fall between X  3.28 and X  3.28 .</p><p>1 100% confidence interval: Suppose the sample size is large.  As  is known,      x  z  X  x  z  x  z , x  z  2 2 n  2 n 2 n  is a 1 100% confidence interval estimate of the population mean  .</p><p>4  As  is unknown, s  s s  x  z s X  x  z  x  z , x  z  2 2 n  2 n 2 n  is a 1 100% confidence interval estimate of the population mean  .</p><p>Motivating Example (continue) As   0.1,  20 20 x  z  82  z0.05  82 1.64  78.72, 85.28 2 n 100 100 is a 90% confidence interval estimate of the population mean  . Note: in the CJW, Inc. example, the 95% confidence interval is wider than the 90% confidence interval. Intuitively, if we want to make sure that we will make less mistakes, we should speak vaguely (wider confidence interval). For instance, if we want to get a 100% confidence interval (for sure), the interval  ,  would make us not make any mistake. </p><p> s Note: the length of the confidence interval is 2z or 2z . 2 n 2 n Therefore, a larger sample size n will provide a narrow interval and a greater precision. Example A: A random sample of 81 workers at a company showed that they work an average of 100 hours per month with a standard deviation of 27 hours. Compute a 95% confidence interval for the mean hours per month all workers at the company work. [solution:] As   0.05 ,  27 27 x  z 100  z0.025 100 1.96  94.12, 105.88 2 n 81 9 is a 95% confidence interval estimate of the population mean  .</p><p>Online Exercise: Exercise 10 .1 .1</p><p>5 Exercise 10 .1 .2</p><p>6</p>