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A confidence interval estimate is desired for the gain in a circuit on a semiconductor device. Assume that gain is normally distributed with standard deviation \(\mathrm{s}=20 .\) (a) Find a \(95 \% \mathrm{CI}\) for \(\mathrm{m}\) when \(n=10\) and \(\bar{x}=1000\). (b) Find a \(95 \%\) CI for \(m\) when \(n=25\) and \(\bar{x}=1000\). (c) Find a \(99 \% \mathrm{CI}\) for \(\mathrm{m}\) when \(n=10\) and \(\bar{x}=1000\). (d) Find a \(99 \%\) CI for \(m\) when \(n=25\) and \(\bar{x}=1000\). (e) How does the length of the CIs computed change with the changes in sample size and confidence level?

Short Answer

Expert verified
Larger sample sizes lead to narrower intervals, while higher confidence levels produce wider intervals.

Step by step solution

01

Define the Data and Formula

The problem provides us with four different scenarios to determine the confidence interval (CI) for the mean, \(m\), of a normally distributed variable with a known standard deviation \(s = 20\). The CI can be calculated using the formula: \[ \text{CI} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \] where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(s\) is the standard deviation, and \(n\) is the sample size.
02

Calculate 95% CI for n=10

For a 95% confidence interval, the z-score is approximately 1.96. Using \(\bar{x} = 1000\) and \(n = 10\), the CI is calculated as follows: \[ CI = 1000 \pm 1.96 \left( \frac{20}{\sqrt{10}} \right) = 1000 \pm 1.96 \times 6.3246 \approx 1000 \pm 12.40 \] Thus, the 95% CI is approximately \([987.60, 1012.40]\).
03

Calculate 95% CI for n=25

For \(n = 25\) with the same confidence level, the CI is calculated as: \[ CI = 1000 \pm 1.96 \left( \frac{20}{\sqrt{25}} \right) = 1000 \pm 1.96 \times 4 = 1000 \pm 7.84 \] Therefore, the 95% CI is approximately \([992.16, 1007.84]\).
04

Calculate 99% CI for n=10

For a 99% confidence interval, the z-score is approximately 2.576. Using \(n = 10\), the CI is calculated as: \[ CI = 1000 \pm 2.576 \left( \frac{20}{\sqrt{10}} \right) = 1000 \pm 2.576 \times 6.3246 \approx 1000 \pm 16.30 \] Thus, the 99% CI is approximately \([983.70, 1016.30]\).
05

Calculate 99% CI for n=25

For \(n = 25\) with a 99% confidence interval, the CI is calculated as: \[ CI = 1000 \pm 2.576 \left( \frac{20}{\sqrt{25}} \right) = 1000 \pm 2.576 \times 4 = 1000 \pm 10.30 \] Therefore, the 99% CI is approximately \([989.70, 1010.30]\).
06

Analyze the Effect of Sample Size and Confidence Level

The length of the confidence interval is influenced by both the sample size and the confidence level. In general, an increase in the sample size results in a narrower interval since the margin of error decreases. Increasing the confidence level from 95% to 99% results in a wider interval due to the larger z-score, reflecting a higher certainty about the range containing the true mean.

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Key Concepts

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

Normal Distribution
In the world of statistics, the normal distribution is a fundamental concept. Imagine a large population of data points forming a bell-shaped curve when charted on a graph. This shape is what we call a normal distribution, and it appears in numerous natural and social phenomena.
For instance, heights of people, test scores, and even certain types of financial data often follow a normal distribution, characterized by a symmetric curve centered around the mean. The mean, median, and mode all coincide at the center of this distribution.
The standard deviation, which we will explore later, measures the spread of data points around this mean. Properties of normal distribution make it ideal for calculating probabilities and making inferences, such as determining confidence intervals.
The concept of confidence intervals is deeply tied to the assumption of normal distribution, ensuring that certain statistical techniques work. The distribution allows for the estimation of population parameters even when only a sample is available. This estimation process is crucial for informed decision-making in fields ranging from medicine to engineering.
Sample Size
Sample size, denoted sometimes as 'n,' refers to the number of observations or data points collected from a larger population. It is a crucial component when conducting statistical analysis since it impacts the precision of results.
In the given exercise, different sample sizes (10 and 25) are explored, demonstrating how increasing the number of observations typically gives us more confidence in our results.
This is because a larger sample size tends to offer a more accurate snapshot of the entire population, reducing the margin of error in our confidence intervals. Consider sample size as a magnifying glass—bigger lenses provide a clearer picture.
An adequately large sample size ensures that our statistical conclusions, like confidence intervals for the mean, are more reliable and meaningful. One must always be cautious; while a larger sample size improves accuracy, gathering too much data can be impractical or costly. Hence, finding a balance is essential.
Standard Deviation
The standard deviation is a measure of how spread out numbers are in a data set. It's a fundamental concept in statistics and provides insight into the variability of data points from the mean.
In a normal distribution, about 68% of data points lie within one standard deviation of the mean, 95% within two, and 99.7% within three standard deviations. This property is often called the empirical rule or 68-95-99.7 rule.
In the context of confidence intervals, the standard deviation helps determine the width of the interval by impacting the margin of error. The exercise specifies a standard deviation of 20 for the gain on a semiconductor device, showcasing its role in the overall calculation of confidence intervals.
Consider standard deviation as a measure of data `bounciness`; if the standard deviation is high, the data points are spread out. In contrast, a lower standard deviation indicates that data points are closely clustered around the mean. Understanding this can help anticipate data behavior and make more informed predictions.
Z-Score
The z-score is a statistical measure that tells us how many standard deviations a data point is from the mean. It is vital in the context of normal distribution and confidence intervals because it standardizes data, allowing for comparison across different scales.
To calculate a z-score, subtract the mean from the data point and then divide by the standard deviation. For example, a z-score of 1.96 represents the threshold for a 95% confidence level—meaning we are 95% confident that the true mean lies within that interval.
In problems involving confidence intervals, the z-score signifies the number of standard deviations we must extend the interval from the sample mean to reach our desired confidence level. A higher z-score means a wider confidence interval, reflecting more certainty.
Using a z-score guarantees that our confidence levels align with the properties of the normal distribution, facilitating accurate statistical inference and reducing the likelihood of erroneous conclusions.

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Most popular questions from this chapter

A particular brand of diet margarine was analyzed to determine the level of polyunsaturated fatty acid (in percentages). A sample of six packages resulted in the following data: 16.8,17.2,17.4,16.9,16.5,17.1 (a) Check the assumption that the level of polyunsaturated fatty acid is normally distributed. (b) Calculate a \(99 \%\) confidence interval on the mean \(\mu\). Provide a practical interpretation of this interval. (c) Calculate a \(99 \%\) lower confidence bound on the mean. Compare this bound with the lower bound of the two-sided confidence interval and discuss why they are different.

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 \(99 \%\) 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).

An article in The Engineer ("Redesign for Suspect Wiring," June 1990 ) reported the results of an investigation into wiring errors on commercial transport aircraft that may display faulty information to the flight crew. Such a wiring error may have been responsible for the crash of a British Midland Airways aircraft in January 1989 by causing the pilot to shut down the wrong engine. Of 1600 randomly selected aircraft, 8 were found to have wiring errors that could display incorrect information to the flight crew. (a) Find a \(99 \%\) confidence interval on the proportion of aircraft that have such wiring errors. (b) Suppose that you use the information in this example to provide a preliminary estimate of \(p .\) How large a sample would be required to produce an estimate of \(p\) that we are \(99 \%\) confident differs from the true value by at most \(0.008 ?\) (c) Suppose that you did not have a preliminary estimate of \(p\). How large a sample would be required if you wanted to be at least \(99 \%\) confident that the sample proportion differs from the true proportion by at most 0.008 regardless of the true value of \(p\) ? (d) Comment on the usefulness of preliminary information in computing the needed sample size.

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (38.02,61.98) and (39.95,60.05) (a) What is the value of the sample mean? (b) One of these intervals is a \(95 \% \mathrm{CI}\) and the other is a \(90 \%\) CI. Which one is the \(95 \%\) CI and why?

Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years. (a) Calculate a \(95 \%\) two-sided confidence interval on the death rate from lung cancer. (b) Using the point estimate of \(p\) obtained from the preliminary sample, what sample size is needed to be \(95 \%\) confident that the error in estimating the true value of \(p\) is less than \(0.03 ?\) (c) How large must the sample be if you wish to be at least \(95 \%\) confident that the error in estimating \(p\) is less than \(0.03,\) regardless of the true value of \(p ?\)

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