A sample of 83 golfers showed that their average score on a particular golf course ...
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A sample of 83 golfers showed that their average score on a particular golf course was 84.81 with a standard deviation of 5.99. Answer each of the following (show all work and state the final answer to at least two decimal places.): (A) Find the 90% confidence interval of the mean score for all 83 golfers. (B) Find the 90% confidence interval of the mean score for all golfers if this is a sample of 130 golfers instead of a sample of 83. (C) Which confidence interval is larger and why?
8 years ago
n = 83 x-bar = 84.81 s = 5.99 % = 90 ...
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