6 please 400b ,543 Interpret (explain) the above calculated confidence interval

Question

6 please

400b ,543 Interpret (explain) the above calculated confidence interval Suppose a student measuring the boiling temperature ofcertain lapid ohnerves de m.inos (in degrees Celsius) 102.5, 101.7. 103.1. 100.9. 100.5, and 102.2 on 6 different semples the liquid. He calculates the sample moan to be 101,82. Ibe knews that the tendard deviation this procedure is1.2 degrees, construct the 95% confidence interval for the baling temperatie" of the ahove liquid. Interpnet the interval 6) 110 peoinds

Explanation / Answer

here sample size=n=6

let X be the random variable denoting the boiling temperature

let E[X]=mu given that SD[X]=1.2 and X follows a normal distribution

we have the sample values as 102.5,101.7,103.1,100.9,100.5,102.2

then sample mean Xbar=101.8167

then we know Xbar~N(mu,1.22/6)

let T=(xbar-mu)/(1.2/(sqrt(6))~N(0,1)

here we are to construct a 95% confidence interval we have

P[-taoalpha/2<(xbar-mu)/(1.2/(sqrt(6))<taoalpha/2]=0.95 where taoalpha/2 is the upper alpha/2 point of N(0,1) distribution

here alpha=0.05 and tao0.05/2=1.96

or, P[xbar-taoalpha/2*(1.2/(sqrt(6))<mu<xbar+taoalpha/2*(1.2/(sqrt(6))]=0.95

hence the confidenc interval is [xbar-taoalpha/2*(1.2/(sqrt(6)),xbar+taoalpha/2*(1.2/(sqrt(6))]

now xbar=101.8167 tao0.05/2=1.96

hence the confidence interval is [101.8167-1.96*1.2/sqrt(6),101.8167+1.96*1.2/sqrt(6)]=[100.8565,102.7769] [answer]

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