Question 1162493: 2. You want to determine if the average number of hours that a college student worked in a week . You take a simple random sample of 1100 students and find that the members of the sample work an average of 24 hours per week with a standard deviation of 11 hours.
a. Construct a 99% confidence interval, by hand (mean using the equations in the section). Show your work.
b. Write the conclusion to the confidence interval (follow the examples in the modules) in a complete sentence.
c. If you wanted to construct a confidence interval with a margin of error of 1 hour, how many students would you have to survey? You may use technology. Assume that the standard deviation for the population is 11 hours.
d. Suppose that last year at this time the average number of hours a student worked was 22 hours per week. According to the results of your confidence interval, are students working more hours this year than last? Explain your reasoning in complete sentences.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! half-interval is a t but with df=1099, t is about 2.581 * s/sqrt(n)
using 2.581*11/sqrt(1100)
=0.8560
the 99% CI is (23.14, 25.86)
I wouldn't fault anyone for using 2.576 the z-value, but it is strictly speaking a t.
This means we don't know the the true value of the number of hours a college student worked in a week, but we are very highly confident that the true number will be found somewhere in the interval above. It is confidence, because the true number is either in the interval or not, and such is not a useful probabilistic statement.
CI with 1 hour margin of error (half-interval)
1=2.576*11/sqrt(n)
sqrt(n)=2.576*11
n=802.92 or 803 people. Here, 2.576 may be used because assuming sd of the population is known.
They are working more, for 22 hours is not within the confidence interval, and it was just stated that we are very highly confident that the true mean, the parameter, is in the interval given.
|
|
|
