Question

A pediatrician randomly selected 50 six-month-old boys from her practice's database and recorded an average weight of 8.0 kilograms with a standard deviation of 0.30 kilogram. She also recorded an average length of 67.3 centimeters with a standard deviation of 0.64 centimeter. a. Find a \(95 \%\) confidence interval for the average weight of all six-month- old boys. b. Find a \(99 \%\) confidence interval for the average length of all six-month- old boys. c. What do you have to assume about the pediatrician's database in order to make inferences about all sixmonth-old boys?

Short answer

Answer: The \(95\%\) confidence interval for the average weight of six-month-old boys is approximately (7.915, 8.085) kilograms, and the \(99\%\) confidence interval for the average length of six-month-old boys is approximately (67.057, 67.543) centimeters.

Step-by-step solution

a. \(95\%\) Confidence Interval for Average Weight

First, find the t-value for a \(95\%\) confidence interval with a sample size of 50 (49 degrees of freedom). Using a t-table or calculator, we get a t-value of approximately 2.01. Now, we can plug the given values into the confidence interval formula: \(CI = \bar{X} \pm (t * \frac{s}{\sqrt{n}}) = 8.0 \pm (2.01 * \frac{0.30}{\sqrt{50}})\) Calculate the margin of error: \(2.01 * \frac{0.30}{\sqrt{50}} \approx 0.085\) So the \(95\%\) confidence interval for the average weight is: \(8.0 \pm 0.085 = (7.915, 8.085)\) kilograms

b. \(99\%\) Confidence Interval for Average Length

First, find the t-value for a \(99\%\) confidence interval with a sample size of 50 (49 degrees of freedom). Using a t-table or calculator, we get a t-value of approximately 2.68. Now, we can plug the given values into the confidence interval formula: \(CI = \bar{X} \pm (t * \frac{s}{\sqrt{n}}) = 67.3 \pm (2.68 * \frac{0.64}{\sqrt{50}})\) Calculate the margin of error: \(2.68 * \frac{0.64}{\sqrt{50}} \approx 0.243\) So the \(99\%\) confidence interval for the average length is: \(67.3 \pm 0.243 = (67.057, 67.543)\) centimeters

c. Assumptions for Inferences

In order to make inferences about all six-month-old boys from the pediatrician's database, we have to assume the following: 1. The sample of 50 six-month-old boys is random and representative of the entire population of six-month-old boys. 2. The distribution of weights and lengths in the population is approximately normal or the sample size is large enough for the Central Limit Theorem to apply. In this case, a sample size of 50 is generally considered large enough for the Central Limit Theorem to apply. 3. The weights and lengths of the boys in the sample are independent of each other.

Key concepts

Standard Deviation

When analyzing data, such as the weights and lengths of six-month-old boys, standard deviation is a critical statistical measure that quantifies the amount of variation within a set of values. A low standard deviation indicates that the values are close to the mean (average), while a high standard deviation means the values are spread out over a wider range.

For the pediatrician's study, a standard deviation of 0.30 kilogram for weight and 0.64 centimeter for length tells us about the consistency of the measurements around the average. In practical terms, it helps in understanding the reliability and precision of the averages being represented.

Understanding standard deviation is essential for calculating confidence intervals, as it is used to determine the margin of error which essentially provides a range where the true mean is expected to lie within a certain level of confidence. It's important for students to grasp that a smaller standard deviation would lead to a narrower confidence interval, indicating a more precise estimate of the population mean.

Sample Size

The sample size, often denoted as 'n', is the number of individuals or observations included in a statistical sample. The size of the sample impacts the precision of the statistical estimates and the width of the confidence intervals. In the textbook problem, the sample size is 50 six-month-old boys, which influences the calculation of the margin of error for the confidence interval.

The importance of sample size cannot be overstated in statistics: a larger sample size typically leads to a smaller margin of error, rendering a more accurate estimate of the population parameter. However, it's also vital to acknowledge that increasing the sample size involves more resources and may not always be feasible or practical.

For students struggling to understand why sample size matters, it's beneficial to note that a larger sample is more likely to capture the diversity of the population, thereby providing more reliable statistics that better reflect the entire population's characteristics.

t-Value

In the context of the pediatrician's study, the t-value, also known as the t-score, plays a pivotal role in constructing confidence intervals when the population standard deviation is unknown and the sample size is relatively small (typically less than 30) or when employing a sample from a population that follows a normal distribution.

The t-value is obtained from student's t-distribution, a probability distribution that estimates population parameters when the sample size is small and the population standard deviation is unknown. For larger sample sizes, like in our example of 50 observations, the t-distribution closely approximates the standard normal distribution (z-distribution).

For the pediatrician's case, different t-values for 95% and 99% confidence intervals reflect how confidence levels affect the range's width. A higher confidence level (99%) results in a wider interval as it aims to include the true mean with greater certainty. Teaching this concept often involves explaining that a higher t-value, which corresponds to a higher level of confidence, naturally yields a wider interval, acknowledging a trade-off between precision and confidence.