Question

Assume that 9 mechanics are randomly selected to measure the time (in seconds) they take in rotating a tire of a certain car model. It is known that distribution of all such times approximately normal. What is the probability that the average time of these 9 mechanics exceeded the population mean time by 5 seconds (the sample variance is 40 seconds)?

Answer 1

Solution:
No. of sample = 9
Sample variance = 40
Sample standard deviation = sqrt(40) = 6.32
Standard error of mean = Standard deviation/sqrt(n) = 6.32/sqrt(9) = 2.11
test stat = (Difference b/w Population and Sample mean)/Standard error of mean = 5/2.11 =2.37
Here we will use t test as sample size is less than 30 and population standard deviation is unknown
so df = 8,
P-value = 0.0226
So there is 2.26% probability that the average time of these 9 machines exceeded the population mean time by 5 seconds.