Question

For the following data: 1 2 3 4 5 6 16 20 18 17 18 21 a. Calculate the sample mean and sample variance b. Calculate the probability that the population mean is between 9 and 10 if the population standard deviation is known to be 2.3. What is the 95% confidence interval for the population mean if the population standard deviation is known to be 2.3? d. Calculate the 95% confidence interval using the sample standard deviation. c.

Answer 1

4 Answer Given data ५ 2 3 4 5 6 16 20 18 17.182) a) Sample mean x & Exi n (+2+3+4+5+6+6+20 +18+17+18+2) 162 N 131 12 ă = 10.92. Sample variance is š = 369 -2* -T = (-10.92)+(2-10.925+(3-10 425 *(4-10 9232 +(5-10-1252 +(6-10.925 +(6-10.92}+(20-10.923 +10-10.925 + (9-10.929 +(18-10.92; 7 (21-10.947 12-1

= (-9.92) + (-8.92) + (-7-92)* + (-6-9232 +(-5.92) +(-4.923+ (5-08) + (9.083 +67.885* + (6-B8)* +G-88] 4 (10.08) = 98.40) +679.56)+(62-72) +(47.88) + (35.04) +(24.20) +05.65) +82-44)+(50.12) +(36.46)+(50112)+(101.60) 11 694.69 1) $2.4 63.153. The es where a=2.3. probability that the populations mean is between 9 and 10. PlacM210) =) A (10 <-42-9) - P(10.92-10 27-M 210.92-9) =) PL <x-M 2.3/12 Yon 2.3 V5 zp 10.92-0 10.42-9 0.3 2-383.46 < 4.92 :) 23x3.40 el 0.93 7.958 LIM con 21.92 7.950

P ( 1.386 <Z <2.892) p(222-892) -P < 1.386) 0.998) -0.917) 30.08 1: C) The 95% Confidence internal for the populations mean is ă 11.9642.3 12 X † 1.96 2-3 3.46 & I 1.96 * 0.664 4 + 1-302 10.92 + 1.302 (10.92 92 -|-302, 10.92 + 1.302) (9.618, 12272) the d) The 95% Confidence interval Sample Standard deuiation is using given by, a I tallents So, the - Here, the degree of freedom as 12-1=1 The Critical value . tis 2.2 di 95% Confidence interval is 5 10.32 † 2-201 * 7.998 UTZ 10.42 +2.201 * 7.958 3.46

* (0 :42 + 3 201 * 2-3 = [b : 42- * 5- 06 (109) - 506, 10. 92+ 5- 06 ) (5-4 , 15.9g) - - . . - 0 4 12 4 2 - x 1 ) 2 (.)