Question
Hi George, thanks for your last reply, your answer really makesense. for my next problem is the following: A college admistrator wishes to estimate the average amountstudents spend on living expenses, exluding rent, per month. Thesample mean and standard deviation of living expenses for a sampleof 81 students were found to be $210 and $20, respectively. a) Find a 95% confidence interval for the average amount spendmonthly on living expenses for this population of students. Well, what I did here is the following C.I = 210 ± Z/2SE SE= 20/81 = 2.22 Z/2 = 1.96 therefore, C.I= 210 ±1.96(2.22) Now for b) it asks to find the sample size necessary toestimate (the population mean) to within 5 (i.e. estimationerror e=0.05) with probability 0.90. [Use an optimisticapproach.] So what I did here is e=Z/2 SE Z/2 = 1.645 (since we have 90% probability) therefore 0.05=1.645 20/n in solving for n i get a really huge number which doesnt makeany sense. any suggestion?? Hi George, thanks for your last reply, your answer really makesense. for my next problem is the following: A college admistrator wishes to estimate the average amountstudents spend on living expenses, exluding rent, per month. Thesample mean and standard deviation of living expenses for a sampleof 81 students were found to be $210 and $20, respectively. a) Find a 95% confidence interval for the average amount spendmonthly on living expenses for this population of students. Well, what I did here is the following C.I = 210 ± Z/2SE SE= 20/81 = 2.22 Z/2 = 1.96 therefore, C.I= 210 ±1.96(2.22) Now for b) it asks to find the sample size necessary toestimate (the population mean) to within 5 (i.e. estimationerror e=0.05) with probability 0.90. [Use an optimisticapproach.] So what I did here is e=Z/2 SE Z/2 = 1.645 (since we have 90% probability) therefore 0.05=1.645 20/n in solving for n i get a really huge number which doesnt makeany sense. any suggestion??
Explanation / Answer
Hey Fisher, could explain to me why did you have your evalue as 5 and not 0.05? thanks