53.4k views
4 votes
A random sampling of a company's monthly operating expenses for n=36 months produced a sample mean of $5474 and a stan-dard deviation of S 764. Find a 90% upper confidence bound for the company's mean monthly expenses.

User RaptoX
by
7.0k points

1 Answer

4 votes

Answer with explanation:

Mean of the sample(m) = $ 5474

Standard deviation of the sample (S)=764

Number of observation(n)=36


Z_{90 \text{Percent}}=Z_(0.09)=0.5359


z_(score)=(\Bar x-\mu)/((S)/(√(n)))\\\\0.5359=(5474- \mu)/((764)/(√(36)))\\\\0.5359=(5474- \mu)/((764)/(6))\\\\764 * 0.5359=6 * (5474- \mu)\\\\409.4276=32844-6 \mu\\\\6 \mu=32844 -409.4276\\\\ 6 \mu=32434.5724\\\\ \mu=(32434.5724)/(6)\\\\ \mu=5405.76

So, Mean Monthly Expenses of Population =$ 5405.76, which is 90% upper confidence bound for the company's mean monthly expenses.

User Mayank Chandak
by
7.7k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories