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91Ó°ÊÓ

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. Construct a 95% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.

Short Answer

Expert verified
The 95% confidence interval is (2.1123, 2.2877) pounds.

Step by step solution

01

Identify the Sample Statistics

First, we identify the sample statistics provided in the problem: the sample mean \( \bar{x} = 2.2 \) pounds, the sample size \( n = 20 \), and the known population standard deviation \( \sigma = 0.2 \) pounds.
02

Determine the Z-Score for Confidence Level

We need to find the Z-score that corresponds to the 95% confidence level. This Z-score represents the cutoff values for the middle 95% of the standard normal distribution. For a 95% confidence interval, the Z-score is approximately 1.96.
03

Calculate the Standard Error

The standard error (SE) of the mean is calculated using the formula: \( SE = \frac{\sigma}{\sqrt{n}} \). Using the values, \( SE = \frac{0.2}{\sqrt{20}} = 0.0447 \) pounds.
04

Calculate the Margin of Error

The margin of error (ME) is calculated by multiplying the Z-score by the standard error: \( ME = Z \times SE = 1.96 \times 0.0447 = 0.0877 \) pounds.
05

Construct the Confidence Interval

The 95% confidence interval is given by the formula: \( \text{Confidence interval} = (\bar{x} - ME, \bar{x} + ME) \). Thus, the interval is \( (2.2 - 0.0877, 2.2 + 0.0877) = (2.1123, 2.2877) \) pounds.

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Key Concepts

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

Sample Statistics
Sample statistics are numerical values that summarize or describe aspects of a sample. In this exercise, the sample statistics we are interested in are the sample mean and sample size.
- The **sample mean** (\( \bar{x} \)) is the average value of all the measurements in the sample. Here, it is 2.2 pounds, representing the average weight of the sample of lettuce heads.
- The **sample size** (\( n \)) is the number of observations in the sample, which in this case is 20.
Understanding these statistics is crucial because they serve as the foundation for estimating the population parameters. The sample mean is used to estimate the average weight of all heads of lettuce in the population, while the sample size helps determines the reliability of this estimation.
Standard Error
The standard error (SE) is an important concept when working with sample data. It measures how much the sample mean would be expected to vary if you took many samples from the same population.
- **Formula for SE:** The standard error of the mean is calculated as \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size.
In this situation, we calculated the SE as \( \frac{0.2}{\sqrt{20}} = 0.0447 \) pounds.
A smaller standard error indicates that the sample mean is closer to the real population mean. It gives you an idea of the precision of your sample statistic as an estimate of the true population parameter. A larger sample size typically leads to a smaller, more reliable standard error.
Margin of Error
The margin of error (ME) is crucial in determining the range within which we can expect the population parameter to lie. This concept is particularly important when constructing confidence intervals.
- **Formula for ME:** The margin of error is calculated with the formula \( ME = Z \times SE \), where \( Z \) is the Z-score corresponding to the desired confidence level.
In our example, the margin of error was determined to be \( 1.96 \times 0.0447 = 0.0877 \) pounds.
The Z-score value, in this example 1.96, comes from the confidence level (here 95%). The margin of error quantifies the uncertainty in our estimate. It tells us how much we can expect the sample mean to deviate from the true population mean.
Z-Score
The Z-score is a critical value that helps us determine the confidence interval for population parameters. It represents how many standard deviations an element is from the mean.
- **Understanding Z-Scores:** A Z-score corresponds to a point on the standard normal distribution.
For a 95% confidence interval, the Z-score typically used is 1.96 because it leaves 2.5% in each tail of the normal distribution.
- **Role in Confidence Intervals:** The Z-score determines how much spread or variability you can expect in your confidence interval from the sample mean.
The choice of Z-score directly affects the margin of error, which in turn impacts the width of the confidence interval. A higher confidence level would result in a higher Z-score, thus increasing the margin of error and expanding the confidence interval. This offers a higher expectation that the interval captures the true population parameter.

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Most popular questions from this chapter

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population. Using the same \(p^{\prime}\) and level of confidence, suppose that \(n\) were increased to \(100 .\) Would the error bound become larger or smaller? How do you know?

Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal. 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 99\(\%\) confidence interval for the population mean length of time using training wheels. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. Why would the error bound change if the confidence level were lowered to 90\(\%\)

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds. Construct a 95% confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph, and calculate the error bound.

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats. a. i. \(\overline{x}=\) _____ ii. \(s_{x}=\) _____ iii. \(n=\) _____ iv. \(n-1=\) _____ 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 92\(\%\) confidence interval for the population mean number of unoccupied seats per flight. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.

Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed. What is the error bound? a. 0.87 b. 1.98 c. 0.99 d. 1.74

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