Statistics Question?

A nationally distributed college newspaper conducts a survey among students nationwide every year. This year, responses from a simple random sample of 204 college students to the question “About how many CDs do you own?” resulted in a sample mean = 72.8. Based on data from previous years, the editors of the newspaper will assume that = 7.2.


Reference: Ref 6-6





Use the information given to obtain a 95% confidence interval for the mean number of CDs owned by all college students.





a. (65.6, 80.0)


b. (71.8, 73.8)


c. (72.0, 73.6)


d. (72.3, 73.3)

Statistics Question?
the answer is b. here is the work....





Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval.





For large sample confidence intervals about the mean you have:





xBar ± z * sx / sqrt(n)





where xBar is the sample mean


z is the zscore for having α% of the data in the tails, i.e., P( |Z| %26gt; z) = α


sx is the sample standard deviation


n is the sample size





The sample mean xbar = 72.8


The sample standard deviation sx = 7.2


The sample size n = 204





The z score for a 0.95 confidence interval is the z score such that 0.025 is in each tail.


z = 1.959964


The confidence interval is:





( xbar - z * sx / sqrt( n ) , xbar + z * sx / sqrt( n ) )


( 71.81198 , 73.78802 )
Reply:The answer is a. 7.2 is your margin of error so all you have to do is subtract 7.2 from 72.8, and then add 7.2 to 72.8.

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