To construct a 99.9% confidence interval for the mean reduction in total cholesterol, we follow these steps:

Formula:

$\text{CI} = \bar{x} \pm Z \cdot \frac{\sigma}{\sqrt{n}}$ Where:

• $\bar{x}$ = Sample mean = 0.81 mmol/L
• $Z$ = Critical value for 99.9% confidence level
• $\sigma$ = Population standard deviation = 0.19 mmol/L
• $n$ = Sample size = 290

Step 1: Find the critical value $Z$

For a 99.9% confidence level, the critical value $Z$ corresponds to the area in the tail of the standard normal distribution. Using a Z-table or calculator: $Z \approx 3.291$

Step 2: Calculate the standard error

$\text{Standard error} = \frac{\sigma}{\sqrt{n}} = \frac{0.19}{\sqrt{290}} \approx 0.0112$

Step 3: Compute the margin of error

$\text{Margin of error} = Z \cdot \text{Standard error} = 3.291 \cdot 0.0112 \approx 0.0369$

Step 4: Construct the confidence interval

$\text{CI} = \bar{x} \pm \text{Margin of error} = 0.81 \pm 0.0369$ $\text{Lower bound} = 0.81 - 0.0369 = 0.77$ $\text{Upper bound} = 0.81 + 0.0369 = 0.85$

Final Answer:

The 99.9% confidence interval for the mean reduction in total cholesterol is: $0.77 < \mu < 0.85$

Would you like further details or have any additional questions?

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Related Questions:

1. How is the Z-value determined for different confidence levels?
2. Why is the population standard deviation used instead of the sample standard deviation in this case?
3. How would the confidence interval change with a smaller sample size?
4. How does increasing the confidence level affect the margin of error?
5. Can this result be generalized to all patients or only the studied group?

Tip: A wider confidence interval reflects greater uncertainty about the population mean.