Ask Question
19 March, 23:30

A random sample of 12 lunch orders at noodles and company showed a mean bill of $12.99 with a standard deviation of $4.6. find the 98 percent confidence interval for the mean bill of all lunch orders. (round your answers to 4 decimal places.)

+3
Answers (1)
  1. 19 March, 23:46
    0
    12.99 + 4.6 = 17.59 / 98
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A random sample of 12 lunch orders at noodles and company showed a mean bill of $12.99 with a standard deviation of $4.6. find the 98 ...” in 📘 Business if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers