Question

rom a population that is normally distributed. The sample mean is found to be x-69 and he sample standard deviation is found to be s-12. Construct a 90% confidence interval A simple random sample of size n-16 s drawn about the population mean. The 90% confidence interval is OD (Round to two decimal places as needed.)

Answer 1

Confidence Interval Formula for u is as follows R. ts core, * s/vn« μ eX + tscoreo" s/n where: sample mean, s : sample standard deviation, tscore-statistic with (n-1) Degrees of Freedom and a-1-confidence Percentage Degrees of Freedom n-1 Degrees of Freedom 16-1 Degrees of Freedom15 Calculate a a = 1-Confidence% a1-0.9 a=0.1 Find a spread range a4(0.1) a = 0.05 Find t-score for do.o5 using 15 degrees of freedom: tscoreoo_ 1.7531 <-Value can be found on Excel using =TINV(0.115) tscore 1.7531

andar An SEM12 SEM12 Calculate high end confidence interval total High End X tscoreg s/Nn High End 69 1.7531 x 12 16 High End 69 1.7531 x 3 High End 69 5.2593 High End 74.26 Low End X - tscoreg sNn Low End 69-1.7531 x 12 T6 Low End 69 1.7531 x 3 Low End 69 5.2593 Low End 63.74 Now we have everything, display our 90% confidence interval: 63.74 < μ < 74.26