To compute the $99.9\%$ confidence interval for the population mean, we follow these steps:

Step 1: Compute the sample mean ($\bar{x}$) and sample standard deviation ($s$)

Given the sample: $84, 86.2, 83.1, 73.6, 82.5, 89.8, 74.5$

1. Compute $\bar{x}$: $\bar{x} = \frac{\text{Sum of all data points}}{\text{Number of data points}}$ $\bar{x} = \frac{84 + 86.2 + 83.1 + 73.6 + 82.5 + 89.8 + 74.5}{7}$

2. Compute $s$: The formula for the sample standard deviation is: $s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$

Step 2: Determine the critical value ($t^*$)

Since the population standard deviation is unknown and the sample size is $n=7$, we use the $t$-distribution. For $99.9\%$ confidence level and $n-1 = 6$ degrees of freedom, we find $t^*$.

Step 3: Compute the margin of error (ME)

$ME = t^* \cdot \frac{s}{\sqrt{n}}$

Step 4: Calculate the confidence interval

The confidence interval is: $\text{CI} = \left(\bar{x} - ME, \bar{x} + ME\right)$

Let me perform the calculations. numpy as np from scipy.stats import t

Given data

data = np.array([84, 86.2, 83.1, 73.6, 82.5, 89.8, 74.5]) n = len(data)

Step 1: Calculate the sample mean and standard deviation

mean = np.mean(data) std_dev = np.std(data, ddof=1) # Sample standard deviation

Step 2: Find the critical value (t*) for 99.9% confidence and df = n-1

alpha = 1 - 0.999 df = n - 1 t_critical = t.ppf(1 - alpha / 2, df)

Step 3: Calculate the margin of error

margin_of_error = t_critical * (std_dev / np.sqrt(n))

Step 4: Calculate the confidence interval

lower_bound = mean - margin_of_error upper_bound = mean + margin_of_error

(mean, std_dev, t_critical, margin_of_error, lower_bound, upper_bound)### Results:

1. Sample Mean ($\bar{x}$): $81.96$
2. Sample Standard Deviation ($s$): $5.93$
3. Critical Value ($t^*$): $5.96$
4. Margin of Error (ME): $13.35$

The $99.9\%$ confidence interval is: $\text{CI} = (68.61, 95.30)$

Final Answer:

$99.9\% \text{ CI } = (68.61, 95.30)$

Would you like a detailed step-by-step breakdown?

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Related Questions:

1. What is the difference between using the $t$-distribution and $z$-distribution for confidence intervals?
2. How would the confidence interval change if the sample size increased to 20?
3. What does a wider confidence interval indicate about the data?
4. Why is the $99.9\%$ confidence interval larger than the $95\%$ confidence interval?
5. How would outliers in the data affect the confidence interval calculation?

Tip:

Always check if the population distribution is approximately normal when the sample size is small ($n < 30$), as the $t$-distribution assumes normality.