How Large Should n Be for a Specific Confidence Interval Width?

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To achieve a confidence interval width of at most 0.2 for a normal distribution with an unknown mean and variance of 100, the sample size n must be calculated based on the desired precision. For a 95 percent confidence interval, the formula involves the critical value from the Z-distribution and the standard error. If the interval length is doubled to 0.4, the required sample size n can be reduced, as a larger width allows for less precision. The discussion emphasizes the importance of balancing sample size with the desired confidence interval width. Ultimately, the calculations ensure that n is minimized while meeting the specified criteria.
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You have a random sample of size n from a normal distribution with unknown mean μ and variance 100. You want to find n large enough so that the length of the confidence interval (from left endpoint to right endpoint) is at most .2. Find such a value of n so that n is as small as possible. If you instead were willing to have a confidence interval with twice this length, what would you need to do to n? (Note: All confidence intervals in this problem are 95 percent confidence intervals.)

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Looks right.
 
pasmith said:
Looks right.

Thanks :)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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