A scout is trying to determine the average velocity of pitchers in college. The scout took a random sample of 250 pitches on opening day. The mean was 88mph with a standard deviation of 3.7mph. The scout wants to decrease the margin of error on the next sample. Which choice would be his best option? (Margin of error is 2(sn√)2(𝑠𝑛), where n is the sample size and s is standard deviation.)
Question
A scout is trying to determine the average velocity of pitchers in college. The scout took a random sample of 250 pitches on opening day. The mean was 88mph with a standard deviation of 3.7mph. The scout wants to decrease the margin of error on the next sample. Which choice would be his best option? (Margin of error is 2(sn√)2(𝑠𝑛), where n is the sample size and s is standard deviation.)
Solution
The best option for the scout to decrease the margin of error would be to increase the sample size. The margin of error is directly related to the sample size (n) as seen in the formula. By increasing the number of pitches he samples, the scout can get a more accurate average velocity. This is because the larger the sample size, the closer the sample mean gets to the population mean, thus reducing the margin of error.
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