A random sample of 8 observations was drawn from a normal population. The sample mean and
sample standard deviation are X = 40 and s = 10.
a) Estimate the population mean with 95% confidence.
b) Repeat part (a) assuming that you know that the population standard deviation is = 10.
c) Explain why the interval estimate produced in part (b) is narrower than that in part (a).
To estimate the population mean with 95% confidence, we will use the t-distribution and the formula for the confidence interval. Here's how you can calculate it:
a) Estimate the population mean with 95% confidence:
Step 1: Determine the critical value for a 95% confidence level. This involves finding the t-value for a sample size of 8 and a confidence level of 95%. Since the population standard deviation is unknown, we use the t-distribution. We can find the t-value in a t-table or use statistical software. For example, with a sample size of 8, the critical t-value at a 95% confidence level is approximately 2.306.
Step 2: Calculate the margin of error. The margin of error is defined as the critical t-value multiplied by the standard error of the mean. The standard error of the mean (SE) is calculated by dividing the sample standard deviation (s) by the square root of the sample size (n). In this case, SE = s / √n = 10 / √8 ≈ 3.54. The margin of error is then 2.306 * 3.54 ≈ 8.15.
Step 3: Find the confidence interval. Take the sample mean (X) and add/subtract the margin of error to create the range for the confidence interval. In this case, the confidence interval is X ± margin of error = 40 ± 8.15. Therefore, the 95% confidence interval estimate for the population mean is approximately (31.85, 48.15).
b) Repeat part (a) assuming that you know that the population standard deviation is σ = 10:
In this case, since we know the population standard deviation, we can use the z-distribution instead of the t-distribution. The formula for the confidence interval remains the same. However, the critical value now corresponds to the standard normal distribution at the chosen confidence level (95% confidence level corresponds to a critical z-value of approximately 1.96). Following the same steps as in part (a), we calculate the standard error of the mean as SE = σ / √n = 10 / √8 ≈ 3.54. The margin of error is then 1.96 * 3.54 ≈ 6.94. The confidence interval is X ± margin of error = 40 ± 6.94. Therefore, the 95% confidence interval estimate for the population mean is approximately (33.06, 46.94).
c) The interval estimate produced in part (b) is narrower than that in part (a) because when the population standard deviation (σ) is known, we can use the z-distribution, which has a narrower distribution compared to the t-distribution. The z-distribution assumes a known standard deviation, which reduces the uncertainty associated with estimating the population mean. In contrast, the t-distribution is used when the population standard deviation is unknown, resulting in a wider confidence interval to account for the additional uncertainty.