Problem 94 - Chemistry: An Atoms-Focused Approach

Question

Glucose concentrations in the blood above \(110 \mathrm{mg} / \mathrm{dL}\) can be an early indication of several medical conditions, including diabetes. Suppose analyses of a series of blood samples from a patient at risk of diabetes produce the following results: \(106,99,109,108,\) and \(105 \mathrm{mg} / \mathrm{dL}\) a. What are the mean and the standard deviation of the data? b. Patients with blood glucose levels above \(120 \mathrm{mg} / \mathrm{dL}\) are considered diabetic. Is this value within the \(95 \%\)confidence interval of these data?

Step-by-Step Solution

Answer

Answer: No, 120 mg/dL is not within the 95% confidence interval of the given data, which is 100.72 mg/dL to 110.08 mg/dL.

Step 1: Calculate the mean

To calculate the mean of the blood glucose concentration measurements, we should add all the measurements together and then divide by the number of measurements. So the mean, denoted by \(\bar{x}\), will be: \(\bar{x} = \frac{106 + 99 + 109 + 108 + 105}{5} = \frac{527}{5} = 105.4 \mathrm{mg/dL}\)

Step 2: Calculate the variance

The variance measures how spread out the data points are in the dataset. To calculate the variance (\(s^2\)), we will find the difference between each measurement and the mean, square them, sum these squared differences, and divide by the total number of measurements minus 1. \(s^2 = \frac{(106 - 105.4)^2 + (99 - 105.4)^2 + (109 - 105.4)^2 + (108 - 105.4)^2 + (105 - 105.4)^2}{5-1} = \frac{0.36 + 40.96 + 12.96 + 6.76 + 0.16}{4} = \frac{61.2}{4} = 15.3 \)

Step 3: Calculate the standard deviation

The standard deviation (\(s\)) is the square root of the variance: \(s = \sqrt{s^2} = \sqrt{15.3} \approx 3.91 \mathrm{mg/dL}\)

Step 4: Calculate the 95% confidence interval

To calculate the 95% confidence interval of the data, we will use the following formula: \(Confidence \, Interval = \bar{x} \pm t \times \frac{s}{\sqrt{n}}\) Here, \(t\) is the t-value which corresponds to a 95% confidence level. For 4 degrees of freedom (since we have 5 data points), the t-value is approximately 2.776. So, \(Confidence \, Interval = 105.4 \pm 2.776 \times \frac{3.91}{\sqrt{5}} = 105.4 \pm 4.68\) This means the confidence interval is 100.72 mg/dL to 110.08 mg/dL.

Step 5: Determine if 120 mg/dL is within the confidence interval

Since the confidence interval is 100.72 mg/dL to 110.08 mg/dL and 120 mg/dL is not within this range, we can conclude that 120 mg/dL is not within the 95% confidence interval of the given data.

Key Concepts

Mean Calculation

The mean calculation is a fundamental concept in statistics, especially when evaluating data like blood glucose levels. To find the mean of a dataset, we add all the individual data points together and then divide by the number of data points. In the context of blood glucose concentration, this provides a central value around which individual measurements are distributed. For example, considering blood glucose readings of 106, 99, 109, 108, and 105 mg/dL, the mean is computed using the formula: \[ \bar{x} = \frac{106 + 99 + 109 + 108 + 105}{5} = 105.4 \, \text{mg/dL} \]This tells us that, on average, the glucose concentration in these samples is 105.4 mg/dL, offering insight into the patient's typical glucose level.

Standard Deviation

Standard deviation is a measure that tells us how spread out the data points are around the mean. A smaller standard deviation indicates that the data points tend to be closer to the mean, while a larger standard deviation suggests more variability. Calculating the standard deviation involves a few steps:

• Subtract the mean from each data point and square the result.
• Sum these squared differences.
• Divide by the number of data points minus one to find the variance.
• Take the square root of the variance to find the standard deviation.

For the sample data:\[s^2 = \frac{(106 - 105.4)^2 + (99 - 105.4)^2 + (109 - 105.4)^2 + (108 - 105.4)^2 + (105 - 105.4)^2}{4} = 15.3\]The standard deviation, therefore, is:\[s = \sqrt{15.3} \approx 3.91 \, \text{mg/dL}\]This value gives us an idea of how much variation there is from the average glucose level.

Confidence Interval

A confidence interval provides a range within which we expect a population parameter, like the mean blood glucose level, to lie, with a certain level of confidence. A 95% confidence interval suggests that if we repeated our measurements numerous times, approximately 95% of the intervals calculated would contain the true mean.To compute the confidence interval for the blood glucose data, we use the formula:\[Confidence \, Interval = \bar{x} \pm t \times \frac{s}{\sqrt{n}}\]Here, \(t\) is the t-value that corresponds to a 95% confidence level with 4 degrees of freedom, approximately 2.776. With the mean (\(\bar{x}\)) at 105.4 mg/dL and standard deviation (\(s\)) of 3.91 mg/dL, the confidence interval is:\[105.4 \pm 2.776 \times \frac{3.91}{\sqrt{5}} = 105.4 \pm 4.68\]Thus, the confidence interval spans from 100.72 mg/dL to 110.08 mg/dL, providing a range within which the true mean likely falls.

Diabetes Diagnosis

Diabetes diagnosis involves assessing blood glucose levels to determine if they regularly exceed certain thresholds. A reading of blood glucose concentration above 120 mg/dL is typically used as an indicator of diabetes. In the context of the provided data, the mean glucose level is 105.4 mg/dL, with a calculated 95% confidence interval of 100.72 mg/dL to 110.08 mg/dL. When evaluating whether a given glucose concentration falls within this interval, it helps determine the likelihood of consistent high glucose readings. Since 120 mg/dL lies outside the calculated interval, this indicates that a reading of 120 mg/dL is not typical for this set of samples, suggesting that this particular dataset does not conform to a diabetes diagnosis. However, consistent readings above 120 mg/dL could lead to a positive diabetes diagnosis, highlighting the importance of regular monitoring and comprehensive data evaluation.