You intend to estimate a population mean mu with the following sample.

87.1
69.8
94.3
94.7
76.3
88.4
95.3
87.8
109.5
85.8
102.5
85.6

You believe the population is normally distributed. Find the 80% confidence interval. Enter your answer as an open-interval (i.e., parentheses) accurate to two decimal places (because the sample data are reported accurate to one decimal place).

80% C.I. =

1 answer

To calculate the 80% confidence interval for the population mean \(\mu\), we will follow these steps:

  1. Calculate the Sample Mean (\(\bar{x}\)) and Sample Standard Deviation (s).
  2. Find the critical value (t) for 80% confidence level*.
  3. Calculate the Margin of Error.
  4. Determine the Confidence Interval.

Step 1: Calculate the Sample Mean and Sample Standard Deviation

Given the sample values:

  • 87.1
  • 69.8
  • 94.3
  • 94.7
  • 76.3
  • 88.4
  • 95.3
  • 87.8
  • 109.5
  • 85.8
  • 102.5
  • 85.6

Calculate the sample mean (\(\bar{x}\)):

\[ \bar{x} = \frac{\sum x_i}{n} = \frac{87.1 + 69.8 + 94.3 + 94.7 + 76.3 + 88.4 + 95.3 + 87.8 + 109.5 + 85.8 + 102.5 + 85.6}{12} = \frac{ 1,030.0 }{12} \approx 85.83 \]

Calculate the sample standard deviation (s):

\[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \]

Where:

  • \(x_i\) are the sample values,
  • \(\bar{x}\) is the sample mean,
  • \(n\) is the number of samples.

Calculating the variance:

  • First calculate \( (x_i - \bar{x})^2 \):

\[ \begin{align*} (87.1 - 85.83)^2 & \approx 1.6349 \ (69.8 - 85.83)^2 & \approx 258.0369 \ (94.3 - 85.83)^2 & \approx 70.0039 \ (94.7 - 85.83)^2 & \approx 78.6249 \ (76.3 - 85.83)^2 & \approx 90.7689 \ (88.4 - 85.83)^2 & \approx 6.5049 \ (95.3 - 85.83)^2 & \approx 87.4569 \ (87.8 - 85.83)^2 & \approx 3.8469 \ (109.5 - 85.83)^2 & \approx 566.5449 \ (85.8 - 85.83)^2 & \approx 0.0009 \ (102.5 - 85.83)^2 & \approx 284.7849 \ (85.6 - 85.83)^2 & \approx 0.0529 \ \end{align*} \]

The sum of these squared differences is: \[ 1.6349 + 258.0369 + 70.0039 + 78.6249 + 90.7689 + 6.5049 + 87.4569 + 3.8469 + 566.5449 + 0.0009 + 284.7849 + 0.0529 \approx 1,470.40 \]

Calculating the sample variance: \[ s^2 = \frac{1,470.40}{12 - 1} \approx 133.67 \quad \Rightarrow \quad s \approx 11.56 \]

Step 2: Find the Critical Value (t*) for 80% Confidence Level

For an 80% confidence level and \(n - 1 = 11\) degrees of freedom, we refer to the t-distribution table:

\[ t^* \approx 1.363 (approx) \]

Step 3: Calculate the Margin of Error

The margin of error \(E\) is calculated using the formula: \[ E = t^* \frac{s}{\sqrt{n}} = 1.363 \cdot \frac{11.56}{\sqrt{12}} \approx 1.363 \cdot 3.344 \approx 4.56 \]

Step 4: Determine the Confidence Interval

The 80% confidence interval is: \[ (\bar{x} - E, \bar{x} + E) = (85.83 - 4.56, 85.83 + 4.56) \approx (81.27, 90.39) \]

Therefore, the 80% confidence interval for the population mean \(\mu\) is: \[ \text{80% C.I.} = (81.27, 90.39) \]

So, the final answer is: \[ (81.27, 90.39) \]