Suppose the mean per-person check for brunch at a restaurant chain is approximat

Question

Suppose the mean per-person check for brunch at a restaurant chain is approximately $19.25. Complete parts (a) through (d) below. a) Assuming a standard deviation of $5.00, what sample size is needed to estimate, with 99% confidence, the mean per-person check for brunch to within ±$0.25? b) Assuming a standard deviation of $5.50, what sample size is needed to estimate, with 99% confidence, the mean per-person check for brunch to within ±$0.25? c)Assuming a standard deviation of $6.00, what sample size is needed to estimate, with 99% confidence, the mean per-person check for brunch to within ±$0.25? d) Discuss the effect of variation on the sample size needed. -For the same level of confidence and sampling error, how are variation and sample size needed related? a) Populations with greater variability require smaller samples to estimate the mean. b) There is no relation between variation and sample size needed. c) Populations with greater variability require larger samples to estimate the mean.

Explanation / Answer

a) margin of error E =0.25

for 99% CI ; critical value of z =2.5758

hence required sample size n=(z*Std deviaiton/E)2 =~2654

b)required sample size n=(z*Std deviaiton/E)2 =3212

c)

required sample size n=(z*Std deviaiton/E)2 =3822

d)

c) Populations with greater variability require larger samples to estimate the mean.

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