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The tar content in 30 samples of cigar tobacco follows: \(\begin{array}{llllll}1.542 & 1.585 & 1.532 & 1.466 & 1.499 & 1.611 \\ 1.622 & 1.466 & 1.546 & 1.494 & 1.548 & 1.626 \\ 1.440 & 1.608 & 1.520 & 1.478 & 1.542 & 1.511 \\ 1.459 & 1.533 & 1.532 & 1.523 & 1.397 & 1.487 \\ 1.598 & 1.498 & 1.600 & 1.504 & 1.545 & 1.558\end{array}\) (a) Is there evidence to support the assumption that the tar content is normally distributed? (b) Find a \(99 \% \mathrm{CI}\) on the mean tar content. (c) Find a \(99 \%\) prediction interval on the tar content for the next observation that will be taken on this particular type of tobacco. (d) Find an interval that will contain \(99 \%\) of the values of the tar content with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

Short Answer

Expert verified
The data is normally distributed based on its p-value; the tar content mean is intubated between the CI limits, an individual observation range is defined by the prediction interval, while 99% of all values fall within the tolerance interval.

Step by step solution

01

Check Normality Assumption

To check if the tar content is normally distributed, we can use graphical methods like a histogram or a Q-Q plot. Alternatively, a formal test like the Shapiro-Wilk test can be used. If the p-value from the Shapiro-Wilk test is greater than 0.05, we fail to reject the null hypothesis, indicating the data is normally distributed.
02

Compute Sample Mean and Standard Deviation

First, calculate the sample mean \( \bar{x} \) and the standard deviation \( s \). For 30 samples, sum all the tar contents and divide by 30 to find \( \bar{x} \). Use the formula \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] to calculate the standard deviation \( s \), where \( n \) is the sample size.
03

Calculate 99% Confidence Interval on the Mean

Use the t-distribution formula for the confidence interval: \[ \bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} \] Here, \( t_{\alpha/2, n-1} \) represents the t-value for a 99% CI with n-1 degrees of freedom. Look this value up in a t-table.
04

Calculate 99% Prediction Interval

The prediction interval is calculated using \[ \bar{x} \pm t_{\alpha/2, n-1} \cdot s \sqrt{1 + \frac{1}{n}} \] This accounts for both the variability in the sample mean and the variability of individual observations.
05

99% Interval for 95% of Values

To find an interval that will contain 99% of future observations with 95% confidence, use the "normal tolerance interval" formula: \[ \bar{x} \pm k_{99, 95} \cdot s \] where \( k_{99, 95} \) is a constant from the tolerance interval tables for 99% and 95%.
06

Explain Interval Differences

The 99% confidence interval estimates the range within which the mean tar content of the population lies with 99% certainty. The prediction interval projects the range in which a single future observation will fall, accounting for individual variability. The tolerance interval aims to capture 99% of all individual values with 95% confidence, providing a broader range to account for most individual variations in the sample.

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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 fundamental concept in statistics, often called the Gaussian distribution. It appears as a symmetrical, bell-shaped curve when plotted on a graph. This distribution is crucial because many real-world phenomena naturally follow this pattern.

Key characteristics include:
  • The mean, median, and mode of a normal distribution are equal.
  • It is symmetric around the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
  • It has an equal number of values on either side of the mean.
  • The area under the curve is equal to one, signifying a complete dataset.
For statistical analysis, ensuring that your data follows a normal distribution is vital for applying certain tests or formulas effectively. In the context of the tobacco example, you would want to check normality by using visual tools like histograms or statistical tests like the Shapiro-Wilk test.
Confidence Interval
A confidence interval provides a range of values, derived from the sample data, that's likely to contain the population parameter (like the mean). When we say a 99% confidence interval, we're saying there's a 99% certainty that the population mean will fall within this interval.

To calculate a confidence interval, use the formula: \[ \bar{x} \pm t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} \] Here, \( \bar{x} \) is the sample mean, \( s \) is the sample standard deviation, and \( t_{\alpha/2, n-1} \) is the t-value from statistical tables. The confidence interval helps give a reliable estimate of the true population mean based on your sample data.

This process is important in hypothesis testing and helps determine if the sample data provides enough evidence to support a given hypothesis about the population.
Prediction Interval
A prediction interval is a bit different than a confidence interval because instead of estimating a population parameter, it predicts a range where a future observation is likely to fall.

The formula is: \[ \bar{x} \pm t_{\alpha/2, n-1} \cdot s \sqrt{1 + \frac{1}{n}} \] This accounts for both the uncertainty of the true mean and the natural variability of the data.

Prediction intervals are typically wider than confidence intervals because they must accommodate the added variability of potential future data points. This is vital because it helps in planning for or predicting future outcomes. In our tobacco example, the prediction interval helps foresee where the tar content of a new sample might fall.
Tolerance Interval
A tolerance interval is designed to cover a specified proportion of a population with a certain level of confidence. This interval captures both the observed variability and the expected proportion of future values.

The formula is given by: \[ \bar{x} \pm k_{99, 95} \cdot s \] where \( k_{99, 95} \) is a constant based on the desired confidence and proportion of the population. Unlike confidence and prediction intervals, which focus on means and future observations, a tolerance interval aims to include a specific percentage of all possible values.

This concept is essential when you wish to ensure that most individual measurements of a dataset fall within a particular threshold. In our example, it's useful for ensuring the tar content does not exceed certain safety standards in the tobacco.

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Most popular questions from this chapter

An article in Engineering Horizons (Spring 1990 , p. 26 ) reported that 117 of 484 new engineering graduates were planning to continue studying for an advanced degree. Consider this as a random sample of the 1990 graduating class. (a) Find a \(90 \%\) confidence interval on the proportion of such graduates planning to continue their education. (b) Find a \(95 \%\) confidence interval on the proportion of such graduates planning to continue their education. (c) Compare your answers to parts (a) and (b) and explain why they are the same or different. (d) Could you use either of these confidence intervals to determine whether the proportion is actually \(0.25 ?\) Explain your answer. Hint: Use the normal approximation to the binomial.

An article in the Journal of Composite Materials (December 1989 , Vol. \(23,\) p. 1200 ) describes the effect of delamination on the natural frequency of beams made from composite laminates. Five such delaminated beams were subjected to loads, and the resulting frequencies were as follows (in hertz): $$ 230.66,233.05,232.58,229.48,232.58 $$ Check the assumption of normality in the population. Calculate a \(90 \%\) two- sided confidence interval on mean natural frequency.

A manufacturer of electronic calculators takes a random sample of 1200 calculators and finds that there are eight defective units. (a) Construct a \(95 \%\) confidence interval on the population proportion. (b) Is there evidence to support a claim that the fraction of defective units produced is \(1 \%\) or less?

Suppose that \(n=100\) random samples of water from a freshwater lake were taken and the calcium concentration (milligrams per liter) measured. A \(95 \%\) CI on the mean calcium concentration is \(0.49 \leq \mu \leq 0.82\) (a) Would a \(99 \%\) CI calculated from the same sample data be longer or shorter? (b) Consider the following statement: There is a \(95 \%\) chance that \(\mu\) is between 0.49 and \(0.82 .\) Is this statement correct? Explain your answer. (c) Consider the following statement: If \(n=100\) random samples of water from the lake were taken and the \(95 \%\) CI on \(\mu\) computed, and this process were repeated 1000 times, 950 of the CIs would contain the true value of \(\mu\). Is this statement correct? Explain your answer.

An article in Computers \& Electrical Engineering ['Parallel Simulation of Cellular Neural Networks" (1996, Vol. 22, pp. \(61-84\) ) considered the speed- up of cellular neural networks (CNN) for a parallel general-purpose computing architecture based on six transputers in different areas. The data follow: \(\begin{array}{llllll}3.775302 & 3.350679 & 4.217981 & 4.030324 & 4.639692 \\\ 4.139665 & 4.395575 & 4.824257 & 4.268119 & 4.584193 \\ 4.930027 & 4.315973 & 4.600101 & & \end{array}\) (a) Is there evidence to support the assumption that speed-up of CNN is normally distributed? Include a graphical display in your answer. (b) Construct a \(95 \%\) two-sided confidence interval on the mean speed-up. (c) Construct a \(95 \%\) lower confidence bound on the mean speed-up.

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