91Ó°ÊÓ

Aphid infestation of fruit trees can be controlled either by spraying with pesticide or by inundation with ladybugs. In a particular area, four different groves of fruit trees are selected for experimentation. The first three groves are sprayed with pesticides 1,2 , and 3 , respectively, and the fourth is treated with ladybugs, with the following results on yield: \begin{tabular}{llll} \hline Treatment & \multicolumn{1}{l}{\(n_{i}\) (number of } & \({\bar{x}_{i}}\) (bushels/ \\ & trees) & tree) & \\ \hline 1 & 100 & \(10.5\) & \(1.5\) \\ 2 & 90 & \(10.0\) & \(1.3\) \\ 3 & 100 & \(10.1\) & \(1.8\) \\ 4 & 120 & \(10.7\) & \(1.6\) \\ \hline \end{tabular} Let \(\mu_{i}=\) the true average yield (bushels/tree) after receiving the \(i\) th treatment. Then $$ \theta=\frac{1}{3}\left(\mu_{1}+\mu_{2}+\mu_{3}\right)-\mu_{4} $$ measures the difference in true average yields between treatment with pesticides and treatment with ladybugs. When \(n_{1}, n_{2}, n_{3}\), and \(n_{4}\) are all large, the estimator \(\hat{\theta}\) obtained by replacing each \(\mu_{i}\) by \(\bar{X}_{i}\) is approximately normal. Use this to derive a large-sample \(100(1-\alpha) \%\) CI for \(\theta\), and compute the \(95 \%\) interval for the given data.

Step by step solution

01

Understand the given estimator for \( \theta \)

We are provided with \( \theta = \frac{1}{3}(\mu_1 + \mu_2 + \mu_3) - \mu_4 \), which compares the average yields of pesticide treatments to ladybug treatment. To estimate \( \theta \), replace \( \mu_i \) with \( \bar{x}_i \) to get \( \hat{\theta} = \frac{1}{3}(\bar{x}_1 + \bar{x}_2 + \bar{x}_3) - \bar{x}_4 \).

02

Compute \( \hat{\theta} \) using the given data

Substitute the provided values into the equation: \( \hat{\theta} = \frac{1}{3}(10.5 + 10.0 + 10.1) - 10.7 \). Calculate each component: - Average of treatments: \( \frac{1}{3}(10.5 + 10.0 + 10.1) = 10.2 \)- Subtract the ladybug treatment: \( 10.2 - 10.7 = -0.5 \)Hence, \( \hat{\theta} = -0.5 \).

03

Calculate the variance of \( \hat{\theta} \)

The variance of the estimator \( \hat{\theta} \) is computed as: \( \text{Var}(\hat{\theta}) = \frac{1}{9}(\sigma_1^2/n_1 + \sigma_2^2/n_2 + \sigma_3^2/n_3) + \sigma_4^2/n_4 \).Use given variances: \( \text{Var}(\hat{\theta}) = \frac{1}{9}(\frac{1.5^2}{100} + \frac{1.3^2}{90} + \frac{1.8^2}{100}) + \frac{1.6^2}{120} \).

04

Calculate numerical values for variance

Calculate each part:- \( \frac{1.5^2}{100} = 0.0225 \), \( \frac{1.3^2}{90} = 0.0188 \), \( \frac{1.8^2}{100} = 0.0324 \)- Sum for pesticides: \( \frac{1}{9}(0.0225 + 0.0188 + 0.0324) = 0.0081 \)- Add ladybug component: \( \frac{1.6^2}{120} = 0.0213 \)- Total variance: \( \text{Var}(\hat{\theta}) = 0.0081 + 0.0213 = 0.0294 \)

05

Find standard deviation and CI critical value

Standard deviation \( \sigma_{\hat{\theta}} = \sqrt{0.0294} \approx 0.1716 \).For a 95% confidence interval, \( z_{0.025} \approx 1.96 \).

06

Determine the confidence interval for \( \theta \)

The 95% CI for \( \theta \) is given by:\( \hat{\theta} \pm z_{0.025} \cdot \sigma_{\hat{\theta}} = -0.5 \pm 1.96 \times 0.1716 \).Calculate the margin of error: \( 1.96 \times 0.1716 \approx 0.3363 \).Hence, the CI is: \((-0.8363, -0.1637)\).

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Key Concepts

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

Estimator

An estimator is a rule or formula used to make estimates about unknown parameters based on observed data. In statistics, we use different kinds of estimators to infer properties about a population given sample data. The goal is to derive estimates that give us the best possible guesses of the true population parameters.

Consider an estimator as a mathematical formula. This formula takes the sample data and outputs an estimated value for a particular parameter. In the context of our exercise, \(\theta\) is the parameter we aim to estimate, defined as the average yield difference between pesticide and ladybug treatments. The estimator \(\hat{\theta}\) is calculated by substituting sample means \(\bar{x}_i\) for the true means \(\mu_i\) in the given formula. This substitution forms the basis of our estimate for \(\theta\).

A good estimator is unbiased, meaning it accurately targets the true parameter. It also has a small variance, indicating precision and consistency, providing reliable results across different samples.

Normal Distribution

The normal distribution is a fundamental concept in statistics, often referred to as the bell curve due to its characteristic shape. It describes how data points are distributed and is defined by its mean and standard deviation.

In many cases, especially with large samples, the distribution of sample means tends to approximate a normal distribution, regardless of the population's original distribution. This is due to the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normal if the sample size is large enough.

Why is this important? By assuming our estimator is approximately normal, we can apply statistical techniques to calculate confidence intervals and make inferences about the population parameter. This assumption allows us to use the \(z\)-distribution to find the critical values needed for the confidence interval calculation in our problem.

Sample Variance

Sample variance measures the variability or spread of the data in a sample. It tells us how much individual data points deviate from the sample mean on average and is denoted by \(s^2\). In practice, we compute this by taking the average of the squared differences between each data point and the sample mean.

Variance is crucial when assessing how precise our estimator is. A smaller variance signifies that data points are closely clustered around the mean, leading to a more accurate estimation of the population parameter. In our exercise, the variance of the estimator \(\hat{\theta}\) is needed to calculate the confidence interval.

The variance formula presented in the exercise combines the variances from different treatments, broken down into contributions from the methods used (pesticides vs. ladybugs). Understanding these components helps calculate the overall variability of our estimator, informing us about the reliability of our confidence interval.

Large-Sample Theory

Large-sample theory, or asymptotic theory, plays an essential role in statistics, providing useful approximations when sample sizes are big. This theory helps statisticians understand the behavior of many statistical estimators and tests as the number of observations approaches infinity.

As sample size increases, certain properties of the sample distribution become more predictable. For instance, estimators such as the sample mean become more precise and normally distributed due to the Central Limit Theorem. This predictability allows us to perform approximate inference even when the population distribution is unknown.

Applying large-sample theory in our exercise means we use the normal approximation for the estimator \(\hat{\theta}\) to derive a confidence interval. The assumption of a large sample lets us comfortably use the normal distribution to approximate the behavior of \(\hat{\theta}\), knowing we can expect results in alignment with the continuous data pattern as described by these theorems.