Question

8.1.7 Questi 1 4 The sample data below have been collected based on a simple random sample from a normally distributed population. Complete parts a and b. 0 2 9 3 9 4 3 1 a. Compute a 98% confidence interval estimate for the population mean The 98% confidence interval for the population mean is from to (Round to two decimal places as needed. Use ascending order.)

Answer 1

Sol: Given data : The Sample Random Variables 4 0 2 9 3 1 3 94 No. of Variables in) The mean & standard deviation for the fiven data is 1+ 4 + + 2 +9+3 +4 + 4 + 3+1 x ID ID ä 36 - ID 3| 3-6 S- E(2-a)? N-1 (1-36)+(4-36)+(-38)+(2-38)+(4-3-6)4(3-34)464-35) +(4-3-6)+(3-3-6)+(1-3-6) 2 10 -1 (-2.6)* +(0:4)+(-3-6)+(16)+(5.4)+(-0.6)+(5.4)*7(0.4)* (-0.6)*+(-2-6) 2 १ 6:76 +0.16 +12.96 +2.56 +29.16 +0.36+29.16 +0.16 +0.36+6.76 9 88.4 9.83 5- 3.1352 9 CS Scanned with CamScanner

the 98% of confidence interval for the population mean is, So, 1 - C. I *+++ ( 5 ) 98.1 1- 48 100 1- 0.98 ( level of Significance) 0.02 0.02 0.02 0.01 2 from T Critical value Calculator. DEN-I-10-1=9 Dr. 9, 20.02 T= Tx 2 +0.02 2.821 2 Then, 7t til) = 3.6 2.821 ( 5.1352 Vio 3-6 + 2.821 (0.991431292) 3.6 + 2.79 - (3.6 – 2.79) ; (3.6 +2.79 (0.81). (6-39) Ans: : The 98% CI for the population mean is from 10.81 to 6.39 Please Rate my work ... Pls Support me with the thumb
thank you for your considerations............stay safe dear student