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Chapter 23: Problem 3

You have ordered 10 bags of cement, which are supposed to weigh \(94 \mathrm{~kg}\) each. The average weight of the 10 bags is \(93.5 \mathrm{~kg}\). Assuming that the 10 weights can be viewed as a realization of a random sample from a normal distribution with unknown parameters, construct a \(95 \%\) confidence interval for the expected weight of a bag. The sample standard deviation of the 10 weights is \(0.75\).

Short Answer

Expert verified
The 95% confidence interval for the mean weight is (92.963 kg, 94.037 kg).

Step by step solution

01

Identify the Information Given

We know that there are 10 bags of cement, each bag supposedly weighs \(94\ kg\). However, the sample mean (average) weight of these bags is \(\bar{x} = 93.5\ kg\). The sample standard deviation is given as \(s = 0.75\ kg\). We are tasked to construct a 95% confidence interval for the expected weight of a bag, assuming weights follow a normal distribution.
02

Determine the Appropriate Formula

Since we are dealing with a small sample size (\(n = 10\)), we use the t-distribution to construct the confidence interval for the mean. The formula for the confidence interval is:\[ \bar{x} \pm t^* \left( \frac{s}{\sqrt{n}} \right) \]where \(t^*\) is the critical t-value that depends on the desired confidence level and the degrees of freedom \(df = n - 1\).
03

Find the Critical t-Value

To find the critical t-value for a 95% confidence interval with \(n - 1 = 9\) degrees of freedom, we refer to the t-distribution table or use a calculator. The critical t-value \(t^*\) for 9 degrees of freedom at 95% confidence is approximately \(2.262\).
04

Calculate the Margin of Error

The margin of error (ME) is calculated using:\[ ME = t^* \times \frac{s}{\sqrt{n}} \]Substitute the known values:\[ ME = 2.262 \times \frac{0.75}{\sqrt{10}} \approx 0.537 \]
05

Construct the Confidence Interval

Now, we construct the confidence interval by adding and subtracting the margin of error from the sample mean:\[ CI = \bar{x} \pm ME = 93.5 \pm 0.537 \]This yields:\[ CI = (92.963, 94.037) \]
06

Interpret the Confidence Interval

The 95% confidence interval for the expected weight of a bag is \((92.963, 94.037)\ kg\). This interval means that we are 95% confident that the true mean weight of the bags lies within this range.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics and is often referred to as a "bell curve" because of its distinct shape. It's a way to describe how data tends to be distributed around a mean (average) value.
For our cement bag example, we assume the weights follow a normal distribution, which lets us make inferences about the entire population from our sample. This is key when working with real-world data, as it is rarely possible to collect data from every possible member of a population.
  • The normal distribution is symmetrical around the mean.
  • It follows a particular mathematical formula involving the mean and standard deviation.
  • Most of the data will be within three standard deviations of the mean.
This assumption allows statisticians to apply tools and techniques, such as confidence intervals, to estimate population parameters.
t-Distribution
The t-distribution is closely related to the normal distribution. It's particularly useful when dealing with small sample sizes, like our 10 bags of cement. So why not use the normal distribution all the time? There are a couple of reasons:
  • The t-distribution is wider and flatter, which accounts for more variability in small samples.
  • It has heavier tails than the normal distribution, meaning there's a higher probability for extreme values.
When we don't know the population standard deviation and our sample size is small, the t-distribution gives us a more accurate assessment of our confidence intervals.
This is why we use it to find critical values that help determine the margin of error in our calculations.
Sample Standard Deviation
Sample standard deviation, denoted as \( s \), is a measure that indicates how much the individual observations in a sample deviate from the mean. For the cement bags, it is given as \( 0.75 \) kg. It plays a critical role in constructing confidence intervals because it represents the variability within our sample.Key points to understand about sample standard deviation:
  • It's calculated using the differences between each data point and the sample mean.
  • It's a "biased estimator" compared to the population standard deviation, which is why adjustments are made using the t-distribution for small samples.
  • The smaller the standard deviation, the closer the data points are to the mean, indicating less spread or variability.
Understanding the sample standard deviation helps assess the reliability of the sample mean as an estimate for the population mean.
Margin of Error
The margin of error (ME) is what statisticians add and subtract from the sample mean to form a confidence interval. It provides a range that we believe encompasses the true population mean. In our example, the margin of error is approximately \( 0.537 \) kg.Here's a breakdown of margin of error:
  • It is calculated using \( ME = t^* \times \frac{s}{\sqrt{n}} \), where \( t^* \) is the critical t-value, \( s \) is the sample standard deviation, and \( n \) is the sample size.
  • Larger margins of error are associated with wider confidence intervals, reflecting greater uncertainty.
  • It directly influences the width of the confidence interval and is affected by both the variability in the data (standard deviation) and the size of the sample (\( n \)).
The margin of error ensures that our confidence interval accounts for possible deviations from the sample mean, improving the accuracy of our population mean estimate.

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Most popular questions from this chapter

You are given a dataset that may be considered a realization of a normal random sample. The size of the dataset is 34 , the average is \(3.54\), and the sample standard deviation is \(0.13\). Construct a \(98 \%\) confidence interval for the unknown expectation \(\mu\).

Let \(Z_{1}, \ldots, Z_{n}\) be a random sample from an \(N(0,1)\) distribution. Define \(X_{i}=\mu+\sigma Z_{i}\) for \(i=1, \ldots, n\) and \(\sigma>0\). Let \(\bar{Z}, \bar{X}\) denote the sample averages and \(S_{Z}\) and \(S_{X}\) the sample standard deviations, of the \(Z_{i}\) and \(X_{i}\), respectively. a. Show that \(X_{1}, \ldots, X_{n}\) is a random sample from an \(N\left(\mu, \sigma^{2}\right)\) distribution. b. Express \(\bar{X}\) and \(S_{X}\) in terms of \(\bar{Z}, S_{Z}, \mu\), and \(\sigma\). c. Verify that $$ \frac{\bar{X}-\mu}{S_{X} / \sqrt{n}}=\frac{\bar{Z}}{S_{Z} / \sqrt{n}} $$ and explain why this shows that the distribution of the studentized mean does not depend on \(\mu\) and \(\sigma\).

A bottling machine is known to fill wine bottles with amounts that follow an \(N\left(\mu, \sigma^{2}\right)\) distribution, with \(\sigma=5(\mathrm{ml})\). In a sample of 16 bottles, \(\bar{x}=743(\mathrm{ml})\) was found. Construct a \(95 \%\) confidence interval for \(\mu\).

A data set \(x_{1}, x_{2}, \ldots, x_{n}\) is given, modeled as realization of a sample \(X_{1}, X_{2}, \ldots, X_{n}\) from an \(N(\mu, 1)\) distribution. Suppose there are sample statistics \(L_{n}=g\left(X_{1}, \ldots, X_{n}\right)\) and \(U_{n}=h\left(X_{1}, \ldots, X_{n}\right)\) such that $$ \mathrm{P}\left(L_{n}<\mu

A new type of car tire is launched by a tire manufacturer. The automobile association performs a durability test on a random sample of 18 of these tires. For each tire the durability is expressed as a percentage: a score of \(100(\%)\) means that the tire lasted exactly as long as the average standard tire, an accepted comparison standard. From the multitude of factors that influence the durability of individual tires the assumption is warranted that the durability of an arbitrary tire follows an \(N\left(\mu, \sigma^{2}\right)\) distribution. The parameters \(\mu\) and \(\sigma^{2}\) characterize the tire type, and \(\mu\) could be called the durability index for this type of tire. The automobile association found for the tested tires: \(\bar{x}_{18}=195.3\) and \(s_{18}=16.7\). Construct a \(95 \%\) confidence interval for \(\mu\).
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