suppose x has a mound-shaped distribution. A random sample size 16 has a sample mean of 10 and a sample standard deviation of 2. Find 99% cinfidence interval for mue (mean) using a student's t distribution.
To find the 99% confidence interval for the population mean (μ) using a student's t-distribution, you can follow these steps:
Step 1: Determine the sample size (n) and sample mean (x̄) from the given information.
- Sample size (n) = 16
- Sample mean (x̄) = 10
Step 2: Determine the sample standard deviation (s) from the given information.
- Sample standard deviation (s) = 2
Step 3: Calculate the degrees of freedom (df) for the t-distribution.
- Degrees of freedom (df) = n - 1 = 16 - 1 = 15
Step 4: Look up the critical value for a 99% confidence level and the degrees of freedom in the t-table.
- For a 99% confidence level and 15 degrees of freedom, the critical value is approximately 2.947.
Step 5: Calculate the margin of error (E), which is defined as the critical value multiplied by the standard deviation divided by the square root of the sample size.
- Margin of error (E) = 2.947 * (2 / √16) ≈ 1.843
Step 6: Determine the lower and upper limits of the confidence interval.
- Lower limit = sample mean - margin of error = 10 - 1.843 = 8.157
- Upper limit = sample mean + margin of error = 10 + 1.843 = 11.843
Step 7: Assemble the confidence interval.
- The 99% confidence interval for the population mean (μ) is (8.157, 11.843).
Therefore, the 99% confidence interval for the population mean (μ) using a student's t-distribution is (8.157, 11.843).