Question

Use the given confidence level and sample data to find (a) the margin of error and (b) the confidence interval for the population mean u. Assume that the population has a normal distribution. Weight lost on a diet: 90% confidence; n 20, x= 7.0 kg, s 9.2 kg. Click the icon to view a table of critical t-values. kg (Round to one decimal place as needed.) (a) E (b) What is the confidence interval for the population mean u? kg<u kg (Round to one decimal place as needed.)

Answer 1

Solution I given values N=20 *=7.0 S: 9.2 CI= 0.90 =90% (top= tonos df-19= 1.729) las Margin of corel = tem a = 1.129 * 900 23.5569 E - 20708B 3.5569 3.6kg 6 90% of confidence intervalis XIV - 7.0 + 1,729 * 92 20 3.4431 to 10,55697 I giverdata . 3.4kg < M < 100.6 kg a) Salary viewers Y. Y 039 93 1395 8649 225 17.64 5.8 225 33.64 1.4 42 900 1.96 10.9 109 118.81 9.9 99 98.01 123.3 187.69 4.8 9.6 23.04 Ex=180, EY= 65.7, Ex.y = 1911.1 EX=10186 Sy? 105.19 х, у х. Х 15 4,2 46:2 121 87 10 100 100 13.7

1 n.EXY - EXEY [nex?- (sx)?]. [ney?-(87)? 91 80191.1) - 180(651) 18 C10180) - 1807. [86705.79) – 65.73 in 0.4288 20.429. b) The cortical values are df-n-2 8-2 = 16 2=0.05. for two-tailed test is r = corpora 0,707 or < 0.707 c) yes, because the absolute value of the correlation cofficient is cleve than the critical valere XY 343 ay 3.5 98 47.6 49 6,8 7 81.9 6.3 13 12 10.2 122.9 3 8.5 25.5 4.4 57.2 10 5.4 4 X:X 1 x 9604 169 9 169 13 100 2.7 Ex = 158, EY-442, Ex.4-701, Ex?=10248 a sy. EX_ SX-EXY 44.2 (10246) - 158(701) 6.00, 8 10248)-1582 neay 21 (S X 12 bah sxy. sy.Ey 80701-158 (44.2) 2-0.024 ncs x)2-(EX) 2 8C10248 )-(195) y a tb.x = 6.001 -0.024 (2) by predicted value is 226 The best Y = 6.001 -b (0.024)(6) = 5.85125.97