Problem 3
Question
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 \(0.75\).
Step-by-Step Solution
Verified Answer
The 95% confidence interval for the mean weight is (92.963 kg, 94.037 kg).
1Step 1: 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.
2Step 2: 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 \(df = n - 1\).
3Step 3: 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 \(2.262\).
4Step 4: Calculate the Margin of Error
The margin of error (ME) is calculated using:\[ ME = t^* \times \frac{s}{\sqrt{n}} \]Substitute the known values:\[ ME = 2.262 \times \frac{0.75}{\sqrt{10}} \approx 0.537 \]
5Step 5: 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:\[ CI = (92.963, 94.037) \]
6Step 6: 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.
Key Concepts
Normal Distributiont-DistributionSample Standard DeviationMargin of Error
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.
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.
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:
This is why we use it to find critical values that help determine the margin of error in our calculations.
- 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.
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.
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 (\( n \)).
Other exercises in this chapter
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