Question

Here are the prices per ounce of \(n=13\) different brands of individually wrapped cheese slices: $$ \begin{array}{lllll} 29.0 & 24.1 & 23.7 & 19.6 & 27.5 \\ 28.7 & 28.0 & 23.8 & 18.9 & 23.9 \\ 21.6 & 25.9 & 27.4 & & \end{array} $$ Construct a \(95 \%\) confidence interval estimate of the underlying average price per ounce of individually wrapped cheese slices.

Short answer

Based on the provided step-by-step solution, please calculate the 95% confidence interval estimate of the average price per ounce for the given sample of 13 different brands of individually wrapped cheese slices.

Step-by-step solution

Calculate the sample mean

First, we want to find the sample mean of the given prices. To do this, we will sum the prices and divide by the number of samples (\(n=13\)): $$ \bar{x} = \frac{29.0 + 24.1 + 23.7 + \ldots + 25.9 + 27.4 }{13}$$ Calculate the sum of the given numbers and divide by the sample size (\(13\)) to find the sample mean.

Calculate the sample standard deviation

Next, we need to calculate the sample standard deviation, which is defined as follows: $$ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}$$ Substitute the values calculated in Step 1 to find the sample standard deviation.

Find the t-value

Since we want to construct a 95% confidence interval, and we have a sample size of 13, we need to find the t-value corresponding to the 95% confidence level, with a degree of freedom (\(df\)) of \(12\) (since \( df = n - 1\) ). You can look up this value in a t-table or use a calculator that has a t-distribution function. You will get the t-value approximately equal to \(2.179\).

Calculate the margin of error

To compute the margin of error, we will use the following formula: $$ Margin\_of\_error = t \times \frac{s}{\sqrt{n}}$$ Substitute the t-value found in step 3, the sample standard deviation calculated in step 2, and the sample size into the formula and calculate the margin of error.

Construct the 95% confidence interval

Finally, we can create our confidence interval by adding and subtracting the margin of error to the sample mean: $$ Confidence\_interval = (\bar{x} - Margin\_of\_error, \bar{x} + Margin\_of\_error)$$ Plug in the values from the previous steps and calculate the lower and upper bounds of the 95% confidence interval estimate of the underlying average price per ounce of individually wrapped cheese slices.

Key concepts

Sample Mean

The sample mean is a fundamental concept in statistics. It represents the average of a set of numbers and provides a central value for data distribution.

In our exercise, calculating the sample mean is the first step towards constructing a confidence interval for cheese prices. To find the sample mean, we add up all the prices of the cheese slices and divide by the number of data points, which is 13.

Here's how it's calculated: \( \bar{x} = \frac{29.0 + 24.1 + 23.7 + \ldots + 27.4}{13} \). This formula helps us summarize the data into a single, meaningful statistic.

Remember:

• A higher sample mean indicates that the average price per ounce is higher.
• A lower sample mean means a lower average price.

Understanding the sample mean is crucial as it forms the basis for further statistical calculations, like variability and confidence intervals.

Sample Standard Deviation

Sample standard deviation helps measure the amount of variation or dispersion in a set of data values. It's a key concept for understanding how data points differ from the mean.

Calculating it involves these steps:

• Subtract the sample mean from each data point.
• Square each result to eliminate negative numbers.
• Find the average of these squared differences, then take the square root.

For our set: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \). Here, \(n-1\) is used to correct the bias in the estimation of the population standard deviation.

Key takeaways:

• A small standard deviation means the data points are close to the mean.
• A large standard deviation indicates more spread out data.

This statistic is pivotal when determining the reliability and precision of the sample mean and affects the width of our confidence interval.

t-value

The t-value is a critical part of constructing confidence intervals, especially when the sample size is small. It reflects how many standard deviations the sample mean is from the hypothesized population mean under the t-distribution model.

For a 95% confidence interval with a sample size of 13, the degrees of freedom is \(12\) (calculated as \(df = n - 1\)). With these parameters, a t-table or calculator will give a t-value of approximately \(2.179\).

Understanding t-values is essential because:

• They help in estimating the margin of error and hence, the confidence interval.
• It accounts for variability in smaller samples, more than what a normal distribution would.

Choosing the right t-value ensures our confidence interval accurately represents the data.

Margin of Error

The margin of error quantifies the uncertainty in the sample mean's estimate of the population mean. It provides a range within which the true mean is expected to lie.

To find it, we use the formula: \( \text{Margin of Error} = t \times \frac{s}{\sqrt{n}} \). Here:

• \(t\) is the t-value.
• \(s\) is the sample standard deviation.
• \(n\) is the sample size.

This equation combines the sample's variability and theoretical t-distribution to give us a precise range.

Key points about the margin of error:

• A smaller margin indicates higher confidence in the sample mean's approximation of the population mean.
• A critical component in constructing the confidence interval as it determines the width around the sample mean.

Learning to calculate and interpret the margin of error is vital in ensuring the reliability of statistical analyses.