Question 983332: a transport company purchases tyres from a certain manufacturer.From each tyre a life of 24000 kilometres is required.A new manufacturer tenders to supply tyres and a sample of 50 of its tyres has an average life of 25000 kilometres with a standard deviation of 700 kilometres when tested.
a)estimate the proportion of tyres supplied by this company that will have an average life of less than 24000 kilometres even if the most optimistic estimate of the mean life at 95% confidence is assumed.
b)explain how you reached this estimate.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! t df=49 = (xbar-mu)/s/sqrt(n)
=(25000-24000)/*sqrt(50)/700
=10.10
SE=98.99
z or t this number essentially means that the confidence interval is about 2*(98.99) or 198 km.
Every estimate of the mean would say that the tire life is greater than 24,000. 95%CI (24802,25198)
Indeed, if this were a single tire, the probability would be 92% its life is greater than 24,000 km. With a sample of 50 tires, the probability is essentially 100%.
I have difficulty understanding the second part of the question. If I use the most optimistic mean of 25,020, essentially no proportion of the tires will be below 24,000. The optimistic estimate of the mean is an estimate of one point. The CI is the confidence we have in the mean. If we use the higher limit of the CI, we still have one point. To use that to determine proportion of tires that fall below the standard isn't clear to me. I can answer the question if this were a population of tires with mean 25,000 km and standard deviation of 700 km. Now, I can estimate the proportion that fall below the standard, and it is about 7.7%. Another way of stating the above is that the most pessimistic estimation of the mean, not the most optimistic, still has the mean greater than what is required. The mean of any sample of size 50 will be greater than 24,000 km. Perhaps others will interpret this differently from me.
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