Final answer:
To calculate the 95% confidence interval for the population mean height of undergraduates with n=74, we would need the population standard deviation value, which is not provided. Under the assumption that σ is known (e.g., 3 inches), we use the z-score for a 95% confidence interval (1.96) to calculate the margin of error and create the confidence interval, which would hypothetically be (66.116, 67.484) for a standard deviation of 3 inches.
Step-by-step explanation:
To calculate a 95% confidence interval for the population mean height of PLNU undergraduates with a sample size of n=74 and a sample mean (¯x) of 66.8 inches, we first need information about the population standard deviation (σ). However, as it is not provided, let's assume a hypothetical value or that we are using the standard error based on the sample standard deviation.
If the population standard deviation (σ) were known, we could use the formula for the confidence interval: CI = (¯x - z* (σ / √n), ¯x + z* (σ / √n)), where z is the z-score corresponding to the 95% confidence level. For a 95% confidence level, the z-score is typically 1.96. If we knew σ was 3 inches, for instance, the calculations would be as follows:
-
- Calculate the standard error (SE) = σ / √n = 3 / √74 ≈ 0.349
-
- Multiply the standard error by the z-score to find the margin of error (MOE) = 1.96 * SE ≈ 1.96 * 0.349 ≈ 0.684
-
- Determine the lower end of the confidence interval (low) = ¯x - MOE ≈ 66.8 - 0.684 ≈ 66.116
-
- Determine the upper end of the confidence interval (high) = ¯x + MOE ≈ 66.8 + 0.684 ≈ 67.484
With our hypothetical standard deviation, the 95% confidence interval would be (66.116, 67.484).
A 95% confidence interval implies that there is a 95% chance that this interval contains the true population mean, μ. In other words, if we were to take many samples and construct confidence intervals in the same way, we would expect that approximately 95% of these intervals will contain the population mean.