1. Friday Oct 8 • Confidence interval for  • How to obtain Z/2 • Interpretation of confidence interval • What n  30, can replace  by s • Read 5.2 • Next time: read 5.3

3. Similarly P(-1.96 < Z < 1.96) = 0.95 • Thus P(Ў-1.96Ў <  < Ў+1.96Ў) = 0.95 • For a 99% confidence interval: • P(-2.58 < Z < 2.58) =0.99. Z0.005=2.58 • Thus P(Ў -2.58Ў <  < Ў+2.58Ў) = 0.99

4. Formula for 100(1-)% confidence interval for  • when  is given is • [Ў Z/2* Ў] where Ў = /n • Z/2 is a value of Z having a tail area of /2 to its right Histogram of C1, with Normal Curve f(y) 3 a/2 a/2 2 m +Za/2 s y m-Za/2 s 1 0

5. A few frequently used Z value • 90% C. I.  = .1 Z/2 = Z.05 = 1.645 • 95% C. I.  =.05 Z/2 = Z.025 = 1.96 • 99% C. I.  =.01 Z/2 = Z.005 = 2.575 • 97% C. I.  =.03 Z/2 = Z.015 = 2.17

6. Example • E.g. 5.10 Recent data from a national survey of 1350 women indicated that the average woman goes to a hair salon once every 5 weeks and spends on the average $26.49. With a standard deviation (s) of $12.00. Use these data to construct a 99% C.I. for  (average amount spent by women in America on hair salon). • n = 1350, Ў = 26.40,  = s = 12,  = 0.01 99% C.I. : Ў  Z0.005.121350 = 26.40  2.58 . 0.3266 (25.56, 27.24) 99% C.I. for  • Verbal interpretation of the 99% C.I. : We are 99% confident that the average amount a woman in America spends in a hair salon is between $25.56 and $27.24.

7. e.g. 5.15 900 high-school graduates sampled. Objected of interest: distance between the high school attended and present address. Ў = 430 miles, s = 262 miles. Find • a. 95% CI for “average distance between a person’s present address and high school”  = 1 - 0.95 = 0.05 Ў ± Z/2*n, Ў =430, Z/2 = Z0.025 =1.96. n = 900 430 ± 1.96*262900 (412.88, 447.12) width = 34.24 • b. 99% Confidence interval 430 ± 2.58* 262900 (407.468; 452.532) width = 45.064