A manager of a cafeteria wants to estimate the average time customers wait befor

Question

A manager of a cafeteria wants to estimate the average time customers wait before being served. A random sample of 49 customers has an average waiting time of 8.4 minutes with a standard deviation of 3.5 minutes. a) With 90% confidence, what can the manager conclude about the possible size of his error in using 8.4 minutes to estimate the true average waiting time? b) Find a 90% confidence interval for the true average customer waiting time. c) With 98% confidence, what can the manager conclude about the possible size of his error in using 8.4 minutes to estimate the true average waiting time? d) Find a 98% confidence interval for the true average customer waiting time.

Explanation / Answer

a) here std error of mean =std deviation/(n)1/2 =3.5/(49)1/2 =0.5

for 90% CI and (49-1=48) degree of freedom ; critiical value of t =1.6772

hence possible size of his error in using 8.4 minutes to estimate the true average waiting time =margin of error

=t*std error =0.8386

b)90% confidence interval for the true average customer waiting time =sample mean -/+ margin of error

=7.5614 to 9.2386

c)

for 98% CI and (49-1=48) degree of freedom ; critiical value of t =2.4066

hence possible size of his error in using 8.4 minutes to estimate the true average waiting time =margin of error

=t*std error =1.2033

d)

98% confidence interval for the true average customer waiting time =sample mean -/+ margin of error

=7.1967 to 9.6033

please revert for any clarification required,

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