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 sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, which of the following is the margin of error that will be reported?

Name: 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 4567 with 60271
Uploaded: 2022-03-16T01:36:27-08:00
Duration: 2 min 13 s
Channel: David Nguyen
Description: 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 4567 with 60271

          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 sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, which of the following is the margin of error that will be reported?

Added by Jonathan M.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert David Nguyen

03/16/2022

Step 1

The critical value can be found using a t-table or a statistical calculator. In this case, the critical value is approximately 1.645. Show more…

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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 sample standard deviation equal to $12.30. Assuming that he wants to develop a 90 percent confidence interval estimate, which of the following is the margin of error that will be reported?