Question

You must estimate the mean temperature (in degrees Fahrenheit) with the following sample temperatures:

60.4

80.3

63.9

49.5

77.5

66.3

Find the 90% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

90% C.I. =

Answer should be obtained without any preliminary rounding.

Answer 1

Answer: Given data X = 60.4,80:3, 63:9,49.5, 11:5, 66:3 N=9 We have to find 90% confidence interwal estimate for population mean, X - EXi = 60'47803+63.9+49.5+77.5+ 66.3 = 66.317 = Mean temperature Sample mean (*) = 66.317 sample. variance = (xi-x) 52 = 644.248 - 128.850 nal | Standard deviation (S) - vs = 128.850 S = 31.351 : 100(1-0)% Confidence interval estimate for population mean is P/ X - tx S M L X + tx 1 = 1 V

0.1 Here CL = 0.90 . CL=1-a a = 1-CL => 1-0.90 critical value i ta, df = t o4,5 '=62.015 76 V6 .: P[66-317 -2015 X 351 <H <66317 + 20158 11-351] = 1-0 1. P[66:317 – 9:338 <p <66:3177 9:338) = 1-4 iP [ 56.9793 <HL < 75.6547] - 10 lie P 56.979344< 75.6547] OR (56.98 ML 75.65) · The 90% confidence interval estimate 56:98 to 75:85 (in degrees Fahrenheit).