Question

Construct 90%, 95%, and 99% confidence intervals to estimate μ from the following data. State the point estimate. Assume the data come from a normally distributed population.

12.1	11.6	11.9	12.3	12.5	11.4	12.0
11.7	11.8	12.1

Appendix A Statistical Tables

(Round the intermediate values to 4 decimal places. Round your answers to 2 decimal places.)

90% confidence interval:

≤ μ ≤

95% confidence interval:

≤ μ ≤

99% confidence interval:

≤ μ ≤

The point estimate is

.

Answer 1

solution & civen that ? point Estimate is . The mean oso - 120.5 2 12.05 10 storderd dewahon csson - ID:1845 = 0.43 D 90 % confidence interval 220-10 Liq0005 dof acn-LJ 29 txiq? 183 - from fortable = ( dan. Sirn <mcf+4212 Sha) =(1205-1083 & Ous cu< 12.05 +1.83 x 0143 VIO =C12.05 – 1.83 * 0.14 cm 21205 +1.831014 - C12.05-0.25 < W< 12-05 +0.25) (11.80 Luc 12.030)

23 95% confidence interval 2 20:05 X2= 0.025 difa (0-1)=9 +212= 2.26 = from table a coctai2. Sirm aulcst * 205/80 - (12-05-226 Oous cuec12.05 + 2.26% - 0.43 , 2012.05 - 2-26 * 0.14 Lu212.05+2.20 xalul =(12.05 - 003076 < WRC 12.05+0.3076 > Clậu 1 cu < (2.36 ) 3) qqe/o confidence interval X=0-01 Lin20.005 d;f=(n +3=9 +42,04 23-25 from t-table = Getra Slim <e <s+tă,25/5n = (12.05 -3.25x Pues cucc 12-05 +3.25.00 = (12.05-3-25* Onlucuk 12.05+3.25x0olu) 10

Ponoalomy 19.11) = Toinhoo +50.7) >m > binno - 50 •31) =