SOLUTION: Suppose we want a 90% confidence interval for the average amount of time (in minutes) spent per week on homework by the students in a large introductory statistics course at a larg

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Question 866660: Suppose we want a 90% confidence interval for the average amount of time (in minutes) spent per week on homework by the students in a large introductory statistics course at a large university. The interval is to have a margin of error of 2 minutes, and the amount spent has a Normal distribution with a standard deviation = 30 minutes.
The number of observations required is
Question 5 options:

25

30

608

609
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Use this table to find the critical value to be z = 1.645 (look in the row that starts with infinity, look above the 90%)


So we know the following info


ME = 2 (given margin of error)
s = 30 (given standard deviation)
z = 1.645 (found using the table given above)


Margin of Error (ME)


ME = z*s/sqrt(n)


2 = 1.645*30/sqrt(n)


2 = 49.35/sqrt(n)


2*sqrt(n) = 49.35


sqrt(n) = 49.35/2


sqrt(n) = 24.675


n = (24.675)^2


n = 608.855625


n = 609 round UP to the nearest whole number


Why do we round up? Well if we round down to 608, then we will have a margin of error (ME) larger than 2. We round up to clear this hurdle (since larger n leads to smaller ME). This happens whenever you get a decimal result for these type of problems.


So you need a sample size of at least 609 students to make sure the margin of error (ME) is 2 or less.