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 90% confidence interval around this sample mean is ------.

Question

Quizplus · Accepted Answer

The 90% confidence interval for a population mean, when the population standard deviation is known, is calculated using the formula: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 90% confidence level, the Z-score is approximately 1.645. Plugging in the given values: $93.6 \pm 1.645 \frac{4.2}{\sqrt{34}}$, we get the interval (92.4, 94.8).

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