A consumer magazine has contacted a simple random sample of 33owners of a certai

Question

A consumer magazine has contacted a simple random sample of 33owners of a certain model of automobile and asked each owner howmany defects had to be corrected within the first two months ofownership. The average number of defects was xbar = 3.7, with a standard deviation of 1.8defects.

a.      Use the t distribution toconstruct a 95% confidence interval for mean = the average numberof defects for this model.

b.     Use the z distribution to construct a95% confidence interval for mean = the average number of defectsfor this model.

c.      Given that the popular standarddeviation is not known, which of these two confidence intervalsshould be used as the interval estimate for mean?

Explanation / Answer

n=33, xbar=3.7, s=1.8 (a)=0.05, t(0.025, df=n-1=32)=2.04 (check student ttable) The 95% CI is xbar±t*s/n --> 3.7±2.04*1.8/sqrt(33) --> ( 3.06, 4.34) (b)=0.05, Z(0.025)=1.96 (check normal table) The 95% CI is xbar±Z*s/n --> 3.7±1.96*1.8/sqrt(33) --> ( 3.09, 4.31) (c) use t distribution

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