/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Cholesterol The serum cholestero... [FREE SOLUTION] | 91Ó°ÊÓ

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Cholesterol The serum cholesterol Elevels of 50 subjects randomly selected from the L.A. Heart Data, data from an epidemiological heart disease study on Los Angeles County employees, follow. \(\begin{array}{llllllllll}148 & 304 & 300 & 240 & 368 & 139 & 203 & 249 & 265 & 229 \\ 303 & 315 & 174 & 209 & 253 & 169 & 170 & 254 & 212 & 255 \\ 262 & 284 & 275 & 229 & 261 & 239 & 254 & 222 & 273 & 299 \\ 278 & 227 & 220 & 260 & 221 & 247 & 178 & 204 & 250 & 256 \\ 305 & 225 & 306 & 184 & 242 & 282 & 311 & 271 & 276 & 248\end{array}\) a. Construct a histogram for the data. Are the data approximately mound- shaped? b. Use a \(t\) -distribution to construct a \(95 \%\) confidence interval for the average serum cholesterol levels for L.A. County employees.

Short Answer

Expert verified
Answer: To determine if the data is approximately mound-shaped, we need to construct a histogram for the data. Additionally, we can use the t-distribution to find the 95% confidence interval for the average serum cholesterol levels by calculating the sample mean and sample standard deviation, and using the formula for the confidence interval.

Step by step solution

01

a. Construct a histogram

First, we need to arrange the given data in ascending order and select appropriate intervals for creating the histogram. After that, we can count the frequency of the data points in each interval and plot the histogram using those frequencies.
02

a. Check for mound-shape

Once the histogram is plotted, we can look at its shape to determine if the data is approximately mound-shaped or not. A mound-shaped distribution is symmetrical with a single peak.
03

b. Calculate the sample mean

To calculate the sample mean, we will add all the cholesterol levels together and then divide the sum by the total number of subjects (50). The formula for the sample mean, denoted by \(\bar{x}\), is as follows: $$ \bar{x} = \dfrac{\sum_{i=1}^{n} x_i}{n} $$
04

b. Calculate the sample standard deviation

Next, we need to find the sample standard deviation, denoted by \(s\). The formula for the sample standard deviation is as follows: $$ s = \sqrt{\dfrac{\sum_{i=1}^{n}{(x_i - \bar{x})^2}}{n-1}} $$
05

b. Use t-distribution to find the 95% confidence interval

After calculating the sample mean and sample standard deviation, we can now use the t-distribution to find the 95% confidence interval for the average serum cholesterol levels. The formula for the confidence interval using the t-distribution is as follows: $$ \bar{x} \pm t_{\frac{\alpha}{2}, n-1} \times \dfrac{s}{\sqrt{n}} $$ Here, \(t_{\frac{\alpha}{2}, n-1}\) represents the critical value from the t-distribution table with \(\frac{\alpha}{2}\) degrees of freedom and \((n-1)\) degrees of freedom, where \(\alpha = 0.05\) for a 95% confidence interval. Using the t-distribution table or a calculator, we can find the critical value, then plug in our values to calculate the confidence interval.

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Key Concepts

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

Understanding Histograms
A histogram is a type of bar graph used to represent the distribution of a dataset. To construct a histogram for the serum cholesterol levels of the LA County employees, we first organize the data into a frequency distribution table by selecting class intervals (ranges of cholesterol levels) and then counting how many observations fall into each interval (frequency). Once we have the frequencies, we draw a series of bars corresponding to each interval. The height of each bar reflects the number of subjects with cholesterol levels within that interval.

When analyzing the shape of a histogram, we can check for patterns such as symmetry or skewness. A mound-shaped histogram, also referred to as a bell-shaped or normal distribution, is symmetrical with most data points clustering around the central peak and fewer points towards the tails. This type of distribution is key in many statistical analyses because it has well-defined properties that we can use to make inferences about the larger population from which the sample is drawn.
Insights from a t-Distribution
The t-distribution, or Student's t-distribution, is a probability distribution that is used when the sample size is small and the population standard deviation is unknown. It resembles the normal distribution but has heavier tails, meaning it's more prone to producing values that fall far from its mean. As the sample size increases, the t-distribution approaches the normal distribution.

The t-distribution plays a crucial role in hypothesis testing and constructing confidence intervals when we base our calculations on a sample mean and sample standard deviation. The concept of degrees of freedom, denoted by \(n-1\), is integral to the t-distribution. Degrees of freedom account for the number of values that can vary given that the sample mean is used in calculations. In the context of the cholesterol levels exercise, the t-distribution would provide the best way to estimate the average cholesterol level for all LA County employees because it accounts for the additional uncertainty due to estimating the population standard deviation from the sample.
Constructing a Confidence Interval
A confidence interval provides a range of values within which we believe a population parameter (like the average serum cholesterol level) lies. It's constructed around a sample statistic (the sample mean) and expands outwards by a margin of error that depends on the desired level of confidence and the variability in the data.

The formula for a 95% confidence interval around a sample mean is \[ \bar{x} \pm t_{\frac{\alpha}{2}, n-1} \times \dfrac{s}{\sqrt{n}} \], where \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(t_{\frac{\alpha}{2}, n-1}\) is the t-value from the t-distribution corresponding to a 95% confidence level and \(n-1\) degrees of freedom. This range tells us that if we were to take many samples and compute a confidence interval for each, we would expect 95% of those intervals to contain the actual average cholesterol level for the population.
Calculating the Sample Mean
The sample mean, denoted by \(\bar{x}\), is the average value of all the observations in a sample. It is calculated by summing up all the individual data points and dividing this total by the number of observations. Mathematically, the sample mean is expressed as \[ \bar{x} = \dfrac{\sum_{i=1}^{n} x_i}{n} \].

In the exercise on cholesterol levels, we would add up all the serum cholesterol levels provided for the 50 subjects and then divide that sum by 50 to find the average serum cholesterol level for the sample. This statistic is vital as it serves as a central value around which other calculations, such as standard deviation and confidence intervals, are based.
Determining Sample Standard Deviation
The sample standard deviation (denoted by \(s\)) is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range. The formula for the sample standard deviation is \[ s = \sqrt{\dfrac{\sum_{i=1}^{n}{(x_i - \bar{x})^2}}{n-1}} \].

This calculation involves taking the difference between each data point and the sample mean (\bar{x}), squaring that difference, and then summing all the squared differences. This sum is then divided by the number of observations minus one (\(n-1\)) - which represents the degrees of freedom - and finally, the square root of this quotient is taken to get back to the original units of the data. In the cholesterol example, calculating the standard deviation helps us understand the variability of serum cholesterol levels among the LA County employees.

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Most popular questions from this chapter

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Measurements of water intake, obtained from a sample of 17 rats that had been injected with a sodium chloride solution, produced a mean and standard deviation of 31.0 and 6.2 cubic centimeters \(\left(\mathrm{cm}^{3}\right),\) respectively. Given that the average water intake for non injected rats observed over a comparable period of time is \(22.0 \mathrm{~cm}^{3},\) do the data indicate that injected rats drink more water than non injected rats? Test at the \(5 \%\) level of significance. Find a \(90 \%\) confidence interval for the mean water intake for injected rats.

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