You are g ven the sample mean and the population standard deviation. Use this in

Question

You are g ven the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence confidence intervals als or the population mean. Which interval is wider? te convenient, use technology o construct the A random sample of 36 gas grills has a mean price of $647.70. Assume the population standard deviation is $55.40. The 90% confidence interval is OD. (Round to one decimal place as needed.) The 95% confidence interval is OD (Round to one decimal place as needed.) Which interval is wider? Choose the correct answer below The 95% confidence interval The 90% confidence interval

Explanation / Answer

let the population mean be mu.

we have sample size=n=36 and population SD=56.4

and sample mean=Xbar=647.7

now E[Xbar]=mu and SD[Xbar]=population SD/sqrt(n)=56.4/sqrt(36)=56.4/6=9.4

so Xbar~N(mu,9.42)

so T=(Xbar-mu)/9.4~N(0,1)

now P[|t|<taoalph/2]=1-alpha

where t is the observed value of T and taoalpha/2 is the upper 100*(1-alpha)% point of a standard normal variate

now |t|=|(647.7-mu)/9.4|

so P[|(647.7-mu)/9.4|<taoalpha/2]=1-alpha

or, P[647.7-9.4*taoalpha/2<mu<647.7+9.4*taoalpha/2]=1-alpha

so [647.7-9.4*taoalpha/2,647.7+9.4*taoalpha/2] is the 100*(1-alpha)% CI of mu

now for 90% confidence interval we have alpha=0.1

so tao0.05=1.64 [using MINITAB]

so the 90% confidence interval is [647.7-9.4*1.64,647.7+9.4*1.64]=[632.3,663.1] [answer]

for 95% confidence iterval we have alpha=0.05

so tao0.025=1.96

so 95% confidence interval is [647.7-9.4*1.96,647.7+9.4*1.96]=[629.3,666.1] [answer]

now the width of the first CI is 663.1-632.3=30.8

width of the second CI is 666.1-629.3=36.8

so the 95% confidence interval is wider [answer]

now for indicated confidence interval we have

xbar=12.2   sample standard deviation=s=4 and n=10  

population mean be mu

so T=(xbar-mu)/(s/sqrt(n))~tn-1

so P[|t|<tn-1;alpha/2]=1-alpha

or P[xbar-s/sqrt(n)*tn-1;alpha/2<mu<xbar+s/sqrt(n)*tn-1;alpha/2]=1-alpha

now alpha=0.10

xbar=12.2 s=4 n=10 and t9;0.05=1.83311 [by minitab]

so the CI is

[12.2-4/sqrt(10)*1.83311,12.2+4/sqrt(10)*1.83311]=[9.9,14.5] [answer]

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