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 $11.75. Complete parts (a) through (d) below. Click the icon to view a table of values for the standardized normal distribution a. Assuming a standard deviation of $5.00, what sample size is needed to estimate, with 90% confidence, the mean per-person check for brunch to within ± $0.50? The sample size needed is 269 Round up to the nearest integer.) b. Assuming a standard deviation of $5.50, what sample size is needed to estimate, with 90% confidence, the mean per-person check for brunch to within $0.50? The sample size needed is 325 (Round up to the nearest integer.) C. Assuming a standard deviation of $6.00, what sample size is needed to estimate, with 90% confidence, the mean per-person check for brunch to within ±$050? The sample size needed is (Round up to the nearest integer.)

Explanation / Answer

a)

(Note: for z with 2 decimal places n=269 is correct)

b)

(Note: for z with 2 decimal places n=326 is correct)

c)

(Note: for z with 2 decimal places n=388 is correct)

for 90 % CI value of z= 1.6449 standard deviation ?= 5 margin of error E = 0.5 required sample size n=(z?/E)2                                         = 271

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