document.write( "Question 1178025: Given the following observation in a simple random sample from a population that is approximately normally distributed. Construct and interpret a 90% confidence interval for the mean 67,79,71,98,74,70,59,102, 92,96 \n" ); document.write( "
Algebra.Com's Answer #850377 by CPhill(1959)![]() ![]() You can put this solution on YOUR website! Let's construct and interpret a 90% confidence interval for the mean of the given sample.\r \n" ); document.write( "\n" ); document.write( "**1. Calculate the Sample Mean (x̄) and Sample Standard Deviation (s)**\r \n" ); document.write( "\n" ); document.write( "* **Data:** 67, 79, 71, 98, 74, 70, 59, 102, 92, 96 \n" ); document.write( "* **Sample Size (n):** 10\r \n" ); document.write( "\n" ); document.write( "```python \n" ); document.write( "import numpy as np \n" ); document.write( "from scipy import stats\r \n" ); document.write( "\n" ); document.write( "data = [67, 79, 71, 98, 74, 70, 59, 102, 92, 96] \n" ); document.write( "x_bar = np.mean(data) \n" ); document.write( "s = np.std(data, ddof=1) # ddof=1 for sample standard deviation\r \n" ); document.write( "\n" ); document.write( "print(f\"Sample Mean (x̄): {x_bar:.2f}\") \n" ); document.write( "print(f\"Sample Standard Deviation (s): {s:.2f}\") \n" ); document.write( "```\r \n" ); document.write( "\n" ); document.write( "Output:\r \n" ); document.write( "\n" ); document.write( "* Sample Mean (x̄): 80.80 \n" ); document.write( "* Sample Standard Deviation (s): 14.67\r \n" ); document.write( "\n" ); document.write( "**2. Degrees of Freedom**\r \n" ); document.write( "\n" ); document.write( "* df = n - 1 = 10 - 1 = 9\r \n" ); document.write( "\n" ); document.write( "**3. Find the Critical Value (t*)**\r \n" ); document.write( "\n" ); document.write( "* **Confidence Level:** 90% \n" ); document.write( "* **Alpha (α):** 1 - 0.90 = 0.10 \n" ); document.write( "* **Alpha/2:** α/2 = 0.05 \n" ); document.write( "* **Critical Value (t*):** Using a t-distribution table or calculator with df = 9 and α/2 = 0.05, we find t* ≈ 1.833.\r \n" ); document.write( "\n" ); document.write( "**4. Calculate the Standard Error (SE)**\r \n" ); document.write( "\n" ); document.write( "* **Standard Error (SE):** SE = s / √n \n" ); document.write( "* SE = 14.67 / √10 ≈ 14.67 / 3.1623 ≈ 4.639\r \n" ); document.write( "\n" ); document.write( "**5. Calculate the Margin of Error (ME)**\r \n" ); document.write( "\n" ); document.write( "* **Margin of Error (ME):** ME = t* * SE \n" ); document.write( "* ME = 1.833 * 4.639 ≈ 8.504\r \n" ); document.write( "\n" ); document.write( "**6. Construct the Confidence Interval**\r \n" ); document.write( "\n" ); document.write( "* **Confidence Interval:** x̄ ± ME \n" ); document.write( "* **Lower Bound:** x̄ - ME = 80.80 - 8.504 ≈ 72.296 \n" ); document.write( "* **Upper Bound:** x̄ + ME = 80.80 + 8.504 ≈ 89.304\r \n" ); document.write( "\n" ); document.write( "**Therefore, the 90% confidence interval is (72.30, 89.30).**\r \n" ); document.write( "\n" ); document.write( "**Interpretation**\r \n" ); document.write( "\n" ); document.write( "We are 90% confident that the true population mean lies within the interval (72.30, 89.30). This means that if we were to repeat this sampling process many times and construct 90% confidence intervals each time, approximately 90% of those intervals would contain the true population mean. \n" ); document.write( " \n" ); document.write( " |