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Womble548
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Report Thread starter 16 years ago
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I have started revising S3 and have got stuck on this question. We use the old T books so I cant giuve a page reference . It is Ex 2B q 12 in T2 if you use those too. Anyway, ill type the question.

An electrical company repairs very large numbers of televisions and wishes to estiomate the mean time taken to repair a particular fault.
It is known from previous research that the standard deviation of the time taken to repair this fault is 2.5 mins. THe manager wiushes to ensure that the probability that the estimate differs from the true mean by less than 30 seconds is 0.95. Find how large a sample is required.

The question is in the topic introducing standard error if that helps. Also the answer is 97. If you know how to do this could you please explain your working becuase I find stats confusing at the best of times. Probably becuase our teacher is useless

Thanks, there will probably be more questions coming soon!
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theone
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(Original post by Womble548)
I have started revising S3 and have got stuck on this question. We use the old T books so I cant giuve a page reference . It is Ex 2B q 12 in T2 if you use those too. Anyway, ill type the question.

An electrical company repairs very large numbers of televisions and wishes to estiomate the mean time taken to repair a particular fault.
It is known from previous research that the standard deviation of the time taken to repair this fault is 2.5 mins. THe manager wiushes to ensure that the probability that the estimate differs from the true mean by less than 30 seconds is 0.95. Find how large a sample is required.

The question is in the topic introducing standard error if that helps. Also the answer is 97. If you know how to do this could you please explain your working becuase I find stats confusing at the best of times. Probably becuase our teacher is useless

Thanks, there will probably be more questions coming soon!
Let the sample mean be x and the population mean y. Work in minutes. Now, by the central limit theorem, x has the normal distribution (y,(2.5/root(n))^2).

Now the 95 percent confidence intervals are given by y +/- 1.96 x 2.5/root(n). We want the difference to be less than 30 seconds, i.e 0.5 mins.

Hence 1.96x2.5/root(n) < 1/2 so 5x1.96 < root(n). So n > 96.04, so n > 97.
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Womble548
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Report Thread starter 16 years ago
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Thanks, I knew someone would be able to do it
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