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Cholesterol The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 188 milligrams per deciliter \((\mathrm{mg} / \mathrm{dl})\) and standard deviation \(41 \mathrm{mg} / \mathrm{dl}\). For 14 -year-old boys, blood cholesterol levels follow a Normal distribution with mean \(170 \mathrm{mg} / \mathrm{dl}\) and standard deviation \(30 \mathrm{mg} / \mathrm{dl}\). Suppose we select independent SRSs of \(25 \mathrm{men}\) aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels \(\bar{x}_{M}\) and \(\bar{x}_{B}\) (a) What is the shape of the sampling distribution of \(\bar{x}_{M}-\bar{x}_{B} ?\) Why? (b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work.

Short Answer

Expert verified
(a) Normal, due to Central Limit Theorem. (b) Mean is 18 mg/dl. (c) Standard deviation is approximately 9.61 mg/dl.

Step by step solution

01

Identify the Central Limit Theorem

According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large, even if the population distribution is not perfectly normal. Here, the sample sizes are 25 and 36, which are generally considered large enough for the theorem to apply. Thus, the shape of the sampling distribution of \(\bar{x}_M - \bar{x}_B\) is approximately normal.
02

Calculate the Mean of the Sampling Distribution

The mean of the difference in sample means \(\bar{x}_M - \bar{x}_B\) is the difference in the population means. Formally, it can be expressed as:\[ \mu_{\bar{x}_M - \bar{x}_B} = \mu_M - \mu_B = 188 - 170 = 18 \]Hence, the mean of the sampling distribution of \(\bar{x}_M - \bar{x}_B\) is 18 mg/dl.
03

Calculate the Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of \(\bar{x}_M - \bar{x}_B\) is calculated using the formula:\[ \sigma_{\bar{x}_M - \bar{x}_B} = \sqrt{\left(\frac{\sigma_M^2}{n_M}\right) + \left(\frac{\sigma_B^2}{n_B}\right)} \]Substituting the given values:\[ \sigma_M = 41, \sigma_B = 30, n_M = 25, n_B = 36 \]\[ \sigma_{\bar{x}_M - \bar{x}_B} = \sqrt{\left(\frac{41^2}{25}\right) + \left(\frac{30^2}{36}\right)} = \sqrt{\left(\frac{1681}{25}\right) + \left(\frac{900}{36}\right)} \]\[ \sigma_{\bar{x}_M - \bar{x}_B} = \sqrt{67.24 + 25} \approx \sqrt{92.24} \approx 9.61 \]Therefore, the standard deviation of the sampling distribution is approximately 9.61 mg/dl.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
A Normal Distribution is a continuous probability distribution that is symmetrical and bell-shaped. This means that most of the data points cluster around the mean, and the probabilities for values gradually decrease as you move away from the mean. The thickness or spread of the distribution is dictated by its standard deviation.
Normal distribution is essential as it applies to many natural phenomena, including human biological measurements like cholesterol levels. For men aged 20 to 34, and 14-year-old boys, blood cholesterol levels follow a normal distribution with given means and standard deviations in the given problem.
  • The mean is the center of the distribution.
  • The standard deviation indicates how spread out the values are from the mean.
  • About 68% of values lie within one standard deviation of the mean in a perfect normal distribution.
Understanding normal distribution helps in calculating probabilities and making predictions about certain biological conditions.
It provides a framework for developing and testing various statistical methods, such as those employed in this problem.
Sampling Distribution
The concept of a Sampling Distribution is central to statistics and deals with the probability distribution of a statistic based on a random sample.
When multiple samples are drawn from a population, each sample mean could vary slightly, but they form their own distribution - this is the sampling distribution of the sample mean. Its mean equals the population mean, and its spread is influenced by the sample size.
  • In this exercise, when calculating the difference between sample means (\( \bar{x}_M - \bar{x}_B \)), the sampling distribution assists in determining how this difference is distributed.
  • The Central Limit Theorem plays a crucial role by suggesting that if the sample size is large enough, the sampling distribution of the sample mean will be approximately normal, regardless of the population distribution's shape.
  • This is why even with different population standard deviations, we calculated a single overall standard deviation for the difference between means.
The use of sampling distributions allows for understanding and inference about population parameters based on sample data rather than requiring access to the entire population.
Sample Mean
The Sample Mean (denoted as \( \bar{x} \)) is a fundamental concept in statistics. It is the average of all the data points in a specific sample.
Calculating a sample mean provides a single value that summarizes a larger set of data. This concept is especially useful in situations where it's impractical to collect data from the entire population.
  • In this problem, two independent sample means, \( \bar{x}_M \) and \( \bar{x}_B \), are calculated for the cholesterol levels of men and boys, respectively.
  • The sample mean gives an estimate of the population mean, and it is pivotal when calculating the mean of the sampling distribution, \( \mu_{\bar{x}_M - \bar{x}_B} \), which turned out to be 18 mg/dl.
  • This aids in understanding how different the two groups are on average based on their cholesterol levels.

Sample means are utilized to make plausible predictions about the larger population, aiding significantly in research and decision-making processes.
Standard Deviation
Standard Deviation is a critical statistical measure that indicates how much the values in a dataset vary from the mean. It is calculated as the square root of the variance and expressed in the same units as the original data.
  • In the context of this problem, the standard deviations for men (41 mg/dl) and boys (30 mg/dl) were provided, reflecting how cholesterol levels spread around their respective means.
  • When we compute the standard deviation of the difference between sample means (\( \sigma_{\bar{x}_M - \bar{x}_B} \)), it accounts for the variability inherent in both groups, using the formula combining both groups’ variances:
\[ \sigma_{\bar{x}_M - \bar{x}_B} = \sqrt{\left(\frac{\sigma_M^2}{n_M}\right) + \left(\frac{\sigma_B^2}{n_B}\right)} \]
  • The calculation returns approximately 9.61 mg/dl, shedding light on the potential variability of the difference between the sampled means of the two groups.
  • Understanding standard deviation is critical for interpreting how spread out the data points are, thus contributing to analyzing sample data and inferring characteristics about the overall population.

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    Most popular questions from this chapter

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