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Construct a 95\(\%\) Confidence Interval for the true mean age of winter Foothill College students by working out then answering the next seven exercises. Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why?

Short Answer

Expert verified
The error bound is smaller at 90% confidence level because the Z-score is lower than at 95%, making the interval narrower.

Step by step solution

01

Understanding the Confidence Interval

A 95\(\%\) confidence interval gives a range within which we are 95\(\%\) confident that the true population mean lies. This range is defined using the sample mean, the standard deviation, and the sample size.
02

Determine Z-Score for 95% Confidence Interval

The Z-score for a 95\(\%\) confidence interval is approximately 1.96. This comes from the standard normal distribution, where 95\(\%\) of data is within 1.96 standard deviations from the mean.
03

Calculate Error Bound for 95% Confidence Interval

The error bound (E) is given by \( E = Z \times \frac{\sigma}{\sqrt{n}} \), where \(Z\) is the Z-score, \(\sigma\) is the standard deviation, and \(n\) is the sample size. For a 95\(\%\) confidence interval, this error bound helps set the interval around the sample mean.
04

Assess Confidence Interval at 90% Confidence Level

At a 90\(\%\) confidence level, the Z-score is approximately 1.645. This value is less than 1.96, meaning that the 90\(\%\) confidence interval is narrower.
05

Calculate Error Bound for 90% Confidence Level

Use the formula \( E = Z \times \frac{\sigma}{\sqrt{n}} \) again, but with \(Z = 1.645\) for a 90\(\%\) confidence interval. This results in a smaller error bound than for a 95\(\%\) confidence interval.
06

Compare Error Bounds

Since the Z-score for a 90\(\%\) confidence level (1.645) is lower than for a 95\(\%\) confidence level (1.96), the error bound is smaller at 90\(\%\). This means the confidence interval is narrower at lower confidence levels.

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Key Concepts

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

Z-score
The Z-score is a critical statistic often used in constructing confidence intervals. It measures the number of standard deviations a data point lies from the mean. For a confidence interval, the Z-score helps determine the range where the true population mean is expected to lie.
  • For a 95% confidence interval, the Z-score is approximately 1.96. This means that 95% of the sample means will fall within 1.96 standard deviations of the true mean under the normal distribution.
  • When reducing the confidence level to 90%, the Z-score decreases to about 1.645. This indicates a smaller range, as only 90% of the data is captured.
The Z-score is crucial as it directly affects the width of the confidence interval. A higher Z-score indicates a wider interval, providing greater certainty that the true mean lies within this range, whereas a lower Z-score results in a narrower interval.
Error Bound
The error bound is a vital component in the calculation of a confidence interval. It represents the margin of error around the sample mean. The formula for the error bound (\(E\)) is:\[E = Z \times \frac{\sigma}{\sqrt{n}}\],
where:
  • \(Z\) is the Z-score,
  • \(\sigma\) is the standard deviation,
  • \(n\) is the sample size.
Changing the confidence level affects the Z-score, hence influencing the error bound. For example, at a 95% confidence level, the error bound is larger due to the Z-score of 1.96. However, decreasing the confidence level to 90% reduces the Z-score to 1.645, resulting in a smaller error bound.

This change illustrates a trade-off: higher confidence levels provide broader intervals with larger error bounds, offering more assurance that the true mean falls within the interval. Conversely, lower confidence levels produce a narrower interval with a smaller error bound, offering less certainty.
Sample Mean
The sample mean is a fundamental statistic used to estimate the population mean. It is the average value obtained from a sample, calculated by summing up all the sample data points and dividing by the number of data points.
For constructing a confidence interval, the sample mean serves as the midpoint.
  • The confidence interval is built around this value, extending equally in both directions based on the error bound.
  • As the sample size increases, the sample mean becomes a more reliable estimate of the population mean, due to the Central Limit Theorem suggesting that the distribution of the sample mean approaches a normal distribution.
The sample mean benefits from a larger sample size, which helps reduce the variability and make the estimate more precise. This is crucial for narrowing down the confidence interval, providing a more accurate reflection of the true population parameter.

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Most popular questions from this chapter

Use the following information to answer the next three exercises: According to a Field Poll, 79% of California adults (actual results are 400 out of 506 surveyed) feel that education and our schools is one of the top issues facing California. We wish to construct a 90% confidence interval for the true proportion of California adults who feel that education and the schools is one of the top issues facing California. A 90\(\%\) confidence interval for the population proportion is _____. a. \((0.761,0.820)\) b. \((0.125,0.188)\) c. \((0.755,0.826)\) d. \((0.130,0.183)\)

Use the following information to answer the next five exercises: of \(1,050\) randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks. We are interested in finding the 95\(\%\) confidence interval for the percent of executives who prefer trucks. Define random variables \(X\) and \(P^{\prime}\) in words.

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. In words, define the random variable \(\overline{X}\) .

Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours. Identify the following: a. \(x=\) b. \(s_{x}=\) C. \(n=\) d. \(n-1=\)

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds. In words, define the random variable \(X .\)

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