Problem 11 - Basic Statistics for Business and Economics with Student

Question

The owner of Britten's Egg Farm wants to estimate the mean number of eggs laid per chicken. A sample of 20 chickens shows they laid an average of 20 eggs per month with a standard deviation of 2 eggs per month. a. What is the value of the population mean? What is the best estimate of this value? b. Explain why we need to use the \(t\) distribution. What assumption do you need to make? c. For a 95 percent confidence interval, what is the value of \(t ?\) d. Develop the 95 percent confidence interval for the population mean. e. Would it be reasonable to conclude that the population mean is 21 eggs? What about 25 eggs?

Step-by-Step Solution

Answer

a. The best estimate for the population mean is 20 eggs. b. Use the t-distribution because the sample size is small and the population standard deviation is unknown. c. t-value is approximately 2.093 for 95% confidence. d. The interval is (19.065, 20.935). e. It is not reasonable to conclude the mean is 21 or 25 eggs.

Step 1: Identify Population Mean Estimate

The problem states that we want to estimate the population mean number of eggs laid per chicken. Since we only have a sample, the best estimate for the population mean, \( \mu \), is the sample mean. Therefore, the estimated population mean is 20 eggs.

Step 2: Why Use t Distribution?

We use the \( t \) distribution because the sample size is less than 30 and we do not know the population standard deviation, only the sample standard deviation. The assumption is that the distribution of eggs laid per chicken per month is approximately normal.

Step 3: Determine t-Value for Confidence Interval

To find the \( t \) value for a 95 percent confidence interval, we need to determine the degrees of freedom, which is \( n - 1 = 20 - 1 = 19 \). Using a \( t \) distribution table, we find that the \( t \) value at 95% confidence level and 19 degrees of freedom is approximately 2.093.

Step 4: Calculate Standard Error

The standard error \( SE \) is calculated using the formula \( SE = \frac{s}{\sqrt{n}} \), where \( s \) is the sample standard deviation and \( n \) is the sample size. So, \( SE = \frac{2}{\sqrt{20}} \approx 0.4472 \).

Step 5: Develop Confidence Interval

With the \( t \) value and standard error calculated, we can find the 95 percent confidence interval for the population mean using the formula: \( \text{CI} = \bar{x} \pm t \cdot SE \). Here, it is \( 20 \pm 2.093 \times 0.4472 \). This gives the interval approximately \( (19.065, 20.935) \).

Step 6: Assess Population Mean Reasonability

We check if 21 eggs is within the confidence interval \((19.065, 20.935)\). Since 21 is outside the interval, it is not considered a reasonable estimate for the population mean. Similarly, 25 eggs is far outside the interval, so it is also not reasonable.

Key Concepts

Population Mean

The population mean, represented by the symbol \( \mu \), is the average value of a given characteristic you want to estimate from a population. In the context of Britten's Egg Farm, it refers to the mean number of eggs laid per chicken across the entire farm. Since we cannot measure every chicken, we use a sample to estimate this mean. The best estimate of the population mean when using a sample is the sample mean. In the exercise, the sample mean was 20 eggs per chicken per month. This is the starting point for understanding the broader behavior of the entire chicken population on the farm.

t Distribution

The \( t \) distribution is essential for making inferences about a population mean, especially when dealing with small sample sizes and unknown population standard deviation. In situations where the sample size is less than 30, as in this problem, the normal distribution is not appropriate unless the population standard deviation is known. Instead, the \( t \) distribution is used because it accounts for the extra uncertainty introduced by estimating the population standard deviation from a small sample. This distribution is similar to the normal distribution but has heavier tails, providing more probability for extreme values. An additional assumption for using the \( t \) distribution is that the data comes from a population with an approximately normal distribution.

Standard Error

The standard error (SE) measures the variability or dispersion of the sample mean from the true population mean. It’s like a yardstick showing how much error might be in the sample estimate of the mean. In our egg farm scenario, the standard error is calculated using the formula:

• \( SE = \frac{s}{\sqrt{n}} \)

where \( s \) represents the sample standard deviation and \( n \) the sample size. For this exercise, the sample standard deviation is 2 eggs, and the sample size is 20. Therefore, the standard error is approximately 0.4472. This value helps in constructing a confidence interval, giving insights into how well the sample mean reflects the true population mean.

Sample Size

Sample size, denoted as \( n \), is the number of observations in a sample. It has a critical impact on statistical analysis, including the precision of estimates like the mean and the width of confidence intervals. In this scenario, there are 20 chickens in the sample, which is relatively small.

• Smaller sample sizes result in more variability and a wider confidence interval, reflecting more uncertainty about the true population mean.
• Larger samples generally provide more accurate estimates of the population mean because they tend to capture more of the population's characteristics.

Choosing the correct sample size is important for the reliability of statistical inferences you make about a population. In practice, the choice often involves a balance between resources available for sampling and the desired precision of the results.