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14 August, 22:21

The owner of touchdown sports bar wants to develop a time standard for the task of mixing a specialty cocktail. in a preliminary study, he observed one of his bartenders perform this task seven times with an average of 90 seconds and a standard deviation of five seconds. how many observations should be made if he wants to be 95.44 percent confident that the maximum error in the observed time is one second?

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  1. 14 August, 22:47
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    100 observations needed for desired accuracy and confidence. The formula for the confidence of a sampling is: ME = z*d/sqrt (n) where ME = Margin of error z = z score for desired level of confidence d = standard deviation n = number of samples The z score desired is calculated as follows. If you want a 95% confidence, you calculate 1 - 0.95 = 0.05, then you divide the result by 2, getting 0.025, and finally you use a standard normal table to get the z score for the desired probability. So for this problem of 95.44% we get (1 - 0.9544) / 2 = 0.0456/2 = 0.0228 Looking up a standard normal table, the value of 0.0228 is found to have a z-score of 2.0, a. k. a. 2 standard deviations from the normal. So let's substitute the known values into the formula and solve for n. ME = z*d/sqrt (n) 1 = 2*5/sqrt (n) sqrt (n) = 2*5 sqrt (n) = 10 n = 100 So the owner needs at least 100 samples to be 95.44% certain that his measurement error is within 1 second of the correct time.
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