Question

When a very large STAT 200V class was surveyed and asked, “About how many hours do you

study during a typical week?” The mean response of 450 students were ?̅ = 13 hours. Supposed that we know that the study time follows a normal distribution with standard deviation ? = 7.4 hours in the population of all STAT 200V students at this University.

a. Use the survey result to give a 99% confidence interval for the mean study time of all the STAT 200V students.

b. What condition that is not yet mentioned must be met for your confidence interval to be valid?

Answer 1

sol:

z crit for 99%=2.576

n=450

since popualtion standard deviation is given use z critical

sigma=7.4

xbar=13

99% confidence interval for mean study time of all the STAT 200V students

xbar-z*sigma/sqrt(n),xbar+z*sigma.sqrt(n)

13-2.576*7.4/sqrt(450),13+2.576*7.4/sqrt(450)

12.10139,13.89861

we are 99% confident that the true mean  study time of all the STAT 200V students lies in between

12.10139 and 13.89861

b. What condition that is not yet mentioned must be met for your confidence interval to be valid

sample should follow normal distribution

sample should be simple random samplethen only confidence interval calculated above is vaid

here n=450,large sample according to central limit theorem (for n>30) large sample,sample follows normal distribution