Problem 28 - Basic Statistics for Business and Economics with Student

Question

A processor of carrots cuts the green top off each carrot, washes the carrots, and inserts six to a package. Twenty packages are inserted in a box for shipment. To test the weight of the boxes, a few were checked. The mean weight was 20.4 pounds, the standard deviation 0.5 pounds. How many boxes must the processor sample to be 95 percent confident that the sample mean does not differ from the population mean by more than 0.2 pound?

Step-by-Step Solution

Answer

25 boxes should be sampled to achieve the desired confidence level.

Step 1: Identify the Known Values

We are given the standard deviation \( \sigma = 0.5 \) pounds, and need a confidence level of 95%, which corresponds to a \( z \)-score of approximately 1.96 for a two-tailed test. The margin of error \( E \) is 0.2 pounds.

Step 2: Use the Formula for Sample Size

The formula to determine the necessary sample size \( n \) for a given margin of error \( E \) is:\[ n = \left( \frac{z \times \sigma}{E} \right)^2 \]Substitute the known values: \( z = 1.96 \), \( \sigma = 0.5 \), and \( E = 0.2 \).

Step 3: Calculate the Sample Size

Insert the values into the formula:\[ n = \left( \frac{1.96 \times 0.5}{0.2} \right)^2 \ = \left( \frac{0.98}{0.2} \right)^2 \ = (4.9)^2 = 24.01\]Since the sample size cannot be a fraction, we round up to the nearest whole number, which is 25.

Key Concepts

Confidence Interval

When we talk about confidence intervals in statistics, we're referring to a range of values within which we expect the true population parameter to lie. The idea is to capture the true parameter with a certain level of confidence. This means that if we were to take many samples and calculate confidence intervals from them, a certain percentage of those intervals would contain the true population mean.
To compute a confidence interval for a population mean when the population standard deviation is known, we often use the formula:

• The sample mean (\( ar{x} \))
• The z-score corresponding to the desired level of confidence (for instance, 1.96 for 95% confidence)
• The standard deviation (\( \σ \)) divided by the square root of the sample size (\( \))

This formula helps in determining the upper and lower bounds of the interval using the expression:\[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{\sigma}{\sqrt{n}} \right)\]Understanding and calculating confidence intervals is crucial in making informed decisions from sample data, providing insight into the reliability of our estimates.

Standard Deviation

Standard deviation (\( \σ \)) is a key concept in statistics that measures the amount of variation or dispersion in a set of values. A small standard deviation indicates that the values tend to be close to the mean, while a large standard deviation shows that the values are spread out over a wider range. Standard deviation is calculated as the square root of variance, which is the average of the squared differences from the mean. Here's what you need to know:

• It helps to quantify the amount of variation or spread in a data set.
• A high standard deviation indicates that data points are spread out over a larger range of values.
• A low standard deviation indicates that data points tend to be clustered close to the mean.

Standard deviation plays a critical role in the confidence interval calculation as it helps in understanding the spread of the data. A small standard deviation will lead to a narrower confidence interval, implying more precision, while a larger standard deviation will result in a wider confidence interval.

Margin of Error

The margin of error is a crucial concept when we talk about sample estimation, as it defines the range within which we expect the true population parameter to lie, considering the sample data. It reflects how precise we want our estimate to be and is influenced by the confidence level, standard deviation, and the sample size.
The formula for calculating the margin of error (\( E \)) when estimating a population mean is:\[ E = z \left( \frac{\sigma}{\sqrt{n}} \right)\]

• The confidence level affects the z-score. Higher confidence levels require larger z-scores, increasing the margin of error.
• A smaller margin of error implies more precision but may require a larger sample size.
• The margin should balance precision needs with practicality, as overly precise margins may demand impractically large samples.

In the original exercise, the processor wants to be 95% confident the sample mean differs by no more than 0.2 pounds from the population mean, which defines a specific margin of error for the sample size calculation.