# Stuck on a confidence interval question:

#### nathinho

##### New member
An interval estimate for the average distance car tyres lasted during their "lifetime" was reported to be 33112 km to 36775 km. This interval estimate was based on a sample of 47 tyres. The variance of the lifetime was determined from previous studies to be 44885212 km squared. What level of confidence can be attributed to this interval estimate? State your answer as a percentage, correct to the nearest whole number.

I've tried using the formula n = (z-crit*standard deviation/standard error)^2 but it didn't seem to work. Not sure what else I can try? Would working backwards from the formula: point estimate +/- z-crit * standard error work?

Thanks in advance!

#### tkhunny

##### Moderator
Staff member
An interval estimate for the average distance car tyres lasted during their "lifetime" was reported to be 33112 km to 36775 km. This interval estimate was based on a sample of 47 tyres. The variance of the lifetime was determined from previous studies to be 44885212 km squared. What level of confidence can be attributed to this interval estimate? State your answer as a percentage, correct to the nearest whole number.

I've tried using the formula n = (z-crit*standard deviation/standard error)^2 but it didn't seem to work. Not sure what else I can try? Would working backwards from the formula: point estimate +/- z-crit * standard error work?

Thanks in advance!
Yes, and that should lead you to $$\displaystyle n = \dfrac{s^{2}t^{2}}{d^{2}}$$

#### Harry_the_cat

##### Senior Member
Nathino, pease show your work and we can see where you have gone wrong. Your overall method is correct.
What is the point estimate? What is the standard error? What is the z-crit?