Steve's average golf score at his local course is 93.6 from a random sample of 34 rounds of golf. Assume that the population standard deviation for his golf score is 4.2. The margin of error for a 95% confidence interval around this sample mean is---------- .

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Quizplus · Accepted Answer

The margin of error (ME) for a 95% confidence interval can be calculated using the formula: ME = Z * (σ/√n), where Z is the Z-score corresponding to the confidence level (for 95%, Z ≈ 1.96), σ is the population standard deviation, and n is the sample size. Plugging in the values: ME = 1.96 * (4.2/√34) ≈ 1.41.

Steve's Average Golf Score at His Local Course Is 93