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What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of \(\bar{x}=\$ 6.88\) per 100 pounds of watermelon. Assume that \(\sigma\) is known to be \(\$ 1.92\) per 100 pounds (Reference: Agricultural Statistics, U.S. Department of Agriculture). (a) Find a \(90 \%\) confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (b) Sample Size Find the sample size necessary for a \(90 \%\) confidence level with maximal margin of error \(E=0.3\) for the mean price per 100 pounds of watermelon. (c) A farm brings 15 tons of watermelon to market. Find a \(90 \%\) confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds.

Step by step solution

01

Calculate the confidence interval for population mean

For part (a), we need to find a 90% confidence interval for the population mean price of watermelon. The formula for a confidence interval is given by: \[ \bar{x} \pm z_{\frac{\alpha}{2}} \left( \frac{\sigma}{\sqrt{n}} \right) \]Where:- \( \bar{x} = 6.88 \) is the sample mean,- \( \sigma = 1.92 \) is the population standard deviation,- \( n = 40 \) is the sample size, and- \( z_{\frac{\alpha}{2}} \) is the z-score corresponding to 90% confidence level. For a 90% confidence level, \( z_{\frac{\alpha}{2}} \approx 1.645 \).Plugging in these values, the confidence interval is:\[ 6.88 \pm 1.645 \left( \frac{1.92}{\sqrt{40}} \right) \approx 6.88 \pm 0.498 \\approx (6.382, 7.378) \]Thus, the 90% confidence interval for the population mean is \((6.382, 7.378)\). The margin of error (E) is \(0.498\).

02

Calculate required sample size

In part (b), we need to find the sample size required for a 90% confidence interval with a margin of error of 0.3. The formula for sample size (n) is:\[ n = \left( \frac{z_{\frac{\alpha}{2}} \cdot \sigma}{E} \right)^2 \]Using \( z_{\frac{\alpha}{2}} = 1.645 \), \( \sigma = 1.92 \), and \( E = 0.3 \):\[ n = \left( \frac{1.645 \times 1.92}{0.3} \right)^2 \approx \left( \frac{3.1584}{0.3} \right)^2 \approx (10.528)^2 = 110.847 \]Thus, rounding up, the sample size required is 111.

03

Calculate confidence interval for total crop value

For part (c), we need to find the confidence interval for the cash value of 15 tons of watermelon. First, convert the weight into pounds:\[ 15 \text{ tons} = 15 \times 2000 = 30,000 \text{ pounds} \]Then, find the value per 100 pounds in the interval, multiplying by 300 (since 30,000 pounds is 300 times 100 pounds):- Lower value: \( 6.382 \times 300 = 1914.6 \)- Upper value: \( 7.378 \times 300 = 2213.4 \)Thus, the 90% confidence interval for the crop's cash value is \((1914.6, 2213.4)\). The margin of error for this cash value confidence interval is: \[ (2213.4 - 1914.6) / 2 = 149.4 \].

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Key Concepts

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

Sample Mean

The sample mean, denoted by \( \bar{x} \), is a fundamental concept in statistics. It represents the average value of a set of observations from a sample. For example, in the watermelon pricing exercise, the sample mean is \( \$6.88 \) per 100 pounds of watermelon.
This figure is calculated by summing up all observed values, and then dividing by the total number of observations in the sample. Mathematically, it is given by the formula: \( \bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i \), where \( n \) is the sample size and \( x_i \) represents each observed value.

• It helps provide an estimate of the average outcome for the larger population.
• The sample mean in statistics acts as a point estimate for the population mean when the population itself is too large to be measured in its entirety.

Understanding the sample mean enables us to analyze data and draw conclusions about a broader group from which our sample is taken.

Population Standard Deviation

The population standard deviation, represented by \( \sigma \), reflects the amount of variability or dispersion in a set of population data. It shows how much individual data points differ from the mean of the population.
In our example, the population standard deviation is given as \( \\(1.92 \) per 100 pounds of watermelon. This value describes how price data points are spread around the average price of \( \\)6.88 \).

• A small \( \sigma \) suggests that the data points tend to be close to the mean.
• A larger \( \sigma \) shows a wider spread, indicating more variability in data.

The importance of the population standard deviation lies in its role in estimating confidence intervals, as it helps determine how well the sample mean represents the population mean. It is crucial for statistical methods that require understanding the spread of data.

Margin of Error

The margin of error (E) quantifies the range within which we can expect the true population parameter (like the mean) to fall with a specific level of confidence. It provides a measure of the precision of our estimate.
In the watermelon pricing scenario, with a 90% confidence level, the margin of error was calculated to be \( 0.498 \), indicating a reasonably precise estimate for the actual population mean price.

• The margin of error is influenced by three key factors:
• Standard deviation of the population (\( \sigma \))
• The confidence level (such as 90%)
• Sample size (\( n \))
• It decreases with larger sample sizes and lower standard deviations, or by accepting a lower confidence level.

To calculate the margin of error, the formula used is \( E = z_{\frac{\alpha}{2}} \left( \frac{\sigma}{\sqrt{n}} \right) \), where \( z_{\frac{\alpha}{2}} \) is the z-score. Understanding the margin of error helps in assessing the reliability of the confidence interval in statistical analyses.