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Chapter 8: Problem 99

A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8. a. i. \(\overline{x}=\) _____ ii. \(\sigma=\) _____ iii. \(n=\) _____ b. Define the random variables \(X\) and \(\overline{X}\) in words. c. Which distribution should you use for this problem? Explain your choice. d. Construct a 90\(\%\) confidence interval for the population mean number of letters campers send home. i. State the confidence interval. ii. Sketch the graph. iil. Calculate the error bound. e. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?

Short Answer

Expert verified
The 90% confidence interval is (6.98, 8.82) with an error bound of 0.92. A larger sample size reduces error and narrows the interval.

Step by step solution

01

Identifying Sample Statistics

To solve the problem, we start by identifying the given variables from the sample. From the problem, we have:- The sample mean \( \overline{x} = 7.9 \).- The population standard deviation \( \sigma = 2.5 \).- The sample size \( n = 20 \).
02

Understanding Random Variables

The random variable \( X \) represents the number of letters a single child sends during the camp session. The random variable \( \overline{X} \) represents the average number of letters sent by a sample of campers.
03

Choosing the Distribution

Since the population standard deviation is known, and the sample size (n=20) is relatively small but considered large enough, we use the normal distribution. The sampling distribution of the sample mean is approximately normal because of the Central Limit Theorem, and the sample size is greater than 15.
04

Constructing the Confidence Interval

For a 90% confidence interval with the population standard deviation known, we use the formula: \[ \overline{x} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \]where \( Z \) is the Z-value for a 90% confidence level. For 90%, the Z-value is 1.645.
05

Confidence Interval Calculation

Substitute the values:\[ \overline{x} = 7.9 \]\[ \sigma = 2.5 \]\[ n = 20 \]Calculate the margin of error:\[ 1.645 \times \frac{2.5}{\sqrt{20}} \approx 0.9194 \]So the confidence interval is:\[ 7.9 \pm 0.9194 \rightarrow (6.9806, 8.8194) \]
06

Error Bound Calculation

The error bound, also known as the margin of error, is calculated as:\[ E = Z \left( \frac{\sigma}{\sqrt{n}} \right) = 0.9194 \]
07

Effect of Larger Sample Size

If the sample size increases to 500, the margin of error will decrease because the standard error \( \frac{\sigma}{\sqrt{n}} \) becomes smaller. The confidence interval will thus be narrower, providing a more precise estimate of the population mean.

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Key Concepts

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

Sample Statistics
Sample statistics refer to the analysis of a subset of data drawn from a larger population. In this exercise, the key sample statistics are:
  • Sample mean ( \( \overline{x} \) ): This is the average value of the data in the sample. We calculated this to be 7.9, representing the mean number of letters sent by a sample of 20 campers.
  • Sample size ( \( n \) ): This indicates how many individual observations there are in the sample. Here, it is 20 campers.
  • Sample standard deviation ( \( s \) ): Although not directly used in constructing the confidence interval, it indicates how spread out the values in the sample are, around the sample mean. In this case, it is 2.8 letters.
Understanding these sample statistics helps us make inferences about the whole population based on the sample data.
Normal Distribution
The normal distribution is a bell-shaped curve that is symmetrical about the mean. It is depicted with the mean of the distribution at the center and the standard deviations spreading outwards. In statistics, many variables naturally follow a normal distribution.
In this problem, the use of a normal distribution is justified because the population standard deviation is known. Furthermore, the sample size of 20 is large enough for the Central Limit Theorem to apply, allowing us to assume that the sampling distribution of the sample mean is approximately normal. This is crucial for constructing confidence intervals accurately at different confidence levels.
Error Bound
The error bound, often called the margin of error, provides the maximum expected difference between the true population parameter and the observed sample statistic, considering the confidence level.
  • It is computed as \( E = Z \left( \frac{\sigma}{\sqrt{n}} \right) \) where \( Z \) is the Z-score corresponding to the confidence level, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
  • For this exercise, with a 90% confidence level, we found \( E \approx 0.9194 \).
The smaller the error bound, the less uncertainty there is about the sample statistic. Larger samples or lower variance lead to a smaller error bound, making our estimates more precise.
Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental statistical principle that states, with a large enough sample size, the sampling distribution of the sample mean will be approximately normally distributed. This occurs regardless of the population's original distribution. The size needed for the sample to be 'large enough' is commonly considered to be greater than 30; however, in some cases, including this exercise, smaller sizes can suffice due to known parameters.
This theorem is crucial because it allows statisticians to make inferences about population means using normal distribution approaches, even if the actual population distribution is not normal. In our case, this enabled us to rely on the normal distribution to calculate a 90% confidence interval for the mean number of letters sent, even with a sample size of 20. This illustrates how CLT enables powerful statistical methods to operate in diverse real-world situations.

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Most popular questions from this chapter

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?

A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: \(2.7 ; 2.8 ; 3.0 ; 2.3 ; 2.3 ; 2.2 ; 2.8 ; 2.1 ;\) and 2.4 . a. i. \(\overline{x}=\) _____ ii. \(s_{x}=\) _____ iii. \(n=\) ____ iv. \(n-1=\) _____ b. Define the random variable \(X\) in words. c. Define the random variable \(X\) in words. d. Which distribution should you use for this problem? Explain your choice. e. Construct a \(95 \%\) confidence interval for the population mean length of time. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. What does it mean to be "\(95\%\) confident" in this problem?

Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student. Is \(\sigma_{x}\) known?

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.

In a recent Zogby International Poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as likely or very likely. Use the plus four鈥� method to create a 97% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. Explain what this confidence interval means in the context of the problem.
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