Question

In an effort to estimate the mean dollars spent per visit by customers of a food store, the manager has selected a random sample of 100 cash register receipts. The mean of these was $45.67 with a population standard deviation equal to $12.30. The manager wants to develop a 99% confidence interval estimate.

a) State the best point estimate to use as an estimate of the true mean

b) Find the margin of error that will be reported

c) Find the lower limit of the confidence interval estimate

d) Find the upper limit of the confidence interval estimate

Answer 1

(a)

the best point estimate to use as an estimate of the true mean = Sample Mean = = 45.67

(b)

= 45.67

= 12.30

n = 100

= 0.01

From Table, critical values of Z = 2.576

the margin of error that will be reported = Z X SE =

So,

the margin of error that will be reported = 3.1685

(c)

the lower limit of the confidence interval estimate = 45.67 - 3.1685 = 42.5015

So,

the lower limit of the confidence interval estimate = 42.5015

(d)

the upper limit of the confidence interval estimate = 45.67 + 3.1685 = 43.8385

So,

the upper limit of the confidence interval estimate = 43.8385