Questions: A researcher collected sample data for 16 middle-aged women. The sample had a mean serum cholesterol level (measured in milligrams per one hundred milliliters) of 191.3, with a standard deviation of 6.8. Assuming that serum cholesterol levels for middle-aged women are normally distributed, find a 95% confidence interval for the mean serum cholesterol level of all women in this age group. Give the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. (If necessary, consult a list of formulas.)

Solution Steps

The researcher collected sample data for 16 middle-aged women with the following statistics:

• Sample mean (\(\bar{x}\)): \(191.3\)
• Sample standard deviation (\(s\)): \(6.8\)
• Sample size (\(n\)): \(16\)
• Confidence level: \(95\%\)

For a \(95\%\) confidence level and \(n - 1 = 15\) degrees of freedom, the critical value \(t\) can be found from the t-distribution table. The critical value is approximately \(2.1\).

The standard error (\(SE\)) is calculated using the formula: \[ SE = \frac{s}{\sqrt{n}} = \frac{6.8}{\sqrt{16}} = \frac{6.8}{4} = 1.7 \]

The margin of error (\(ME\)) is calculated as: \[ ME = t \cdot SE = 2.1 \cdot 1.7 = 3.57 \]

The confidence interval is given by: \[ \bar{x} \pm ME = 191.3 \pm 3.57 \] Calculating the lower and upper limits:

• Lower limit: \(191.3 - 3.57 = 187.73\)
• Upper limit: \(191.3 + 3.57 = 194.87\)

Rounding the limits to one decimal place gives:

• Lower limit: \(187.7\)
• Upper limit: \(194.9\)

Final Answer